Elementary Number Theory and Its Applications 9780201065619, 0201065614

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Elementary Number Theory andlts Applications KennethH. Rosen AT&T Informotion SystemsLaboratories (formerly part of Bell Laborotories)

A YY

ADDISON-WESLEY PUBLISHING COMPANY Read ing, Massachusetts Menlo Park, California London Amsterdam Don Mills, Ontario Sydney

Cover: The iteration of the transformation

T(n) :

if n is even \ n/2 if n is odd l Qn + l)/2

is depicted. The Collatz conjecture assertsthat with any starting point, the iteration of ?"eventuallyreachesthe integer o n e . ( S e eP r o b l e m 3 3 o f S e c t i o n l . 2 o f t h e t e x t . )

Library of Congress Cataloging in Publication Data Rosen, Kenneth H. Elementary number theory and its applications. Bibliography: p. Includes index. l. Numbers, Theory of.

QA24l.R67 1984 rsBN 0-201-06561-4

I. Title.

512',.72

8 3 - l1 8 0 4

Reprinted with corrections, June | 986 Copyright O 1984 by Bell Telephone Laboratories and Kenneth H. Rosen. All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted, in any form or by any means, electronic, mechanical,photocopying, recording, or otherwise, without prior written permission of the publisher. printed in the United States of America. Published simultaneously in Canada. DEFGHIJ_MA_8987

Preface

Number theory has long been a favorite subject for students and teachersof mathematics. It is a classical subject and has a reputation for being the "purest" part of mathematics, yet recent developments in cryptology and computer science are based on elementary number theory. This book is the first text to integrate these important applications of elementary number theory with the traditional topics covered in an introductory number theory course. This book is suitable as a text in an undergraduatenumber theory course at any level. There are no formal prerequisitesneeded for most of the material covered, so that even a bright high-school student could use this book. Also, this book is designed to be a useful supplementarybook for computer science courses,and as a number theory primer for computer scientistsinterested in learning about the new developmentsin cryptography. Some of the important topics that will interest both mathematics and computer sciencestudents are recursion,algorithms and their computationai complexity, computer arithmetic with large integers, binary and hexadecimal representations of integers, primality testing, pseudoprimality,pseudo-randomnumbers, hashing functions, and cryptology, including the recently-invented area of public-key cryptography. Throughout the book various algorithms and their computational complexitiesare discussed.A wide variety of primality tests are developedin the text. Use of the Book The core material for a course in number theory is presentedin Chapters 1, 2, and 5, and in Sections 3.1-3.3 and 6.1. Section 3.4 contains some linear algebra; this section is necessary background for Section 7.2; these two sections can be omitted if desired. Sections 4.1, 4.2, and 4.3 present traditional applications of number theory and Section 4.4 presents an application to computer science; the instructor can decide which of these sectionsto cover. Sections 6.2 and 6.3 discussarithmetic functions. Mersenne primes, and perfect numbers; some of this material is used in Chapter 8. Chapter 7 covers the applications of number theory to cryptology. Sections 7.1, 7.3, and 7.4, which contain discussionsof classical and public-key

vt

Preface

cryptography,should be included in all courses.Chapter 8 deals with primitive roots; Sections 8.1-8.4 should be covered if possible. Most instructors will want to include Section 8.7 which deals with pseudo-randomnumbers. Sections 9.1 and 9.2 are about quadratic residues and reciprocity, a fundamental topic which should be covered if possible;Sections 9.3 and 9.4 deal with Jacobi symbols and Euler pseudoprimesand should interest most readers. Section 10.1, which covers rational numbers and decimal fractions. and Sections I 1.1 and I 1.2 which discussPythagoreantriples and Fermat's last theorem are coveredin most number theory courses. Sections 10.2-10.4 and I 1.3 involve continued fractions; these sectionsare optional. The Contents The reader can determine which chapters to study based on the following descriptionof their contents. Chapter I introduces two importants tools in establishing results about the integers, the well-ordering property and the principle of mathematical induction. Recursive definitions and the binomial theorem are also developed. The concept of divisibility of integers is introduced. Representations of integers to different bases are described, as are algorithms for arithmetic operations with integers and their computational complexity (using big-O notation). Finally, prime numbers, their distribution, and conjectures about primes are discussed. Chapter 2 introduces the greatest common divisor of a set of integers. The Euclidean algorithm, used to find greatest common divisors, and its computational complexity, are discussed, as are algorithms to express the greatest common divisor as a linear combination of the integers involved. The Fibonacci numbers are introduced. Prime-factorizations, the fundamental theorem of arithmetic, and factorization techniques are covered. Finally, linear diophantine equationsare discussed. Chapter 3 introduces congruences and develops their fundamental properties. Linear congruencesin one unknown are discussed,as are systems of linear congruences in one or more unknown. The Chinese remainder theorem is developed,and its application to computer arithmetic with large integers is described. Chapter 4 developsapplicationsof.congruences. In particular, divisibility tests, the perpetual calendar which provides the day of the week of any date, round-robin tournaments,and computer hashing functions for data storage are discussed.

Preface

vtl

Chapter 5 developsFermat's little theorem and Euler's theorem which give some important congruencesinvolving powers of integers. Also, Wilson's theorem which gives a congruencefor factorials is discussed. Primality and probabilistic primality tests based on these results are developed. Pseudoprimes, strong pseudoprimes, and Carmichael numbers which masquaradeas primes are introduced. Chapter 6 is concernedwith multiplicative functions and their properties. Special emphasisis devotedto the Euler phi-function, the sum of the divisors function, and the number of divisors function and explicit formulae are developed for these functions. Mersenne primes and perfect numbers are discussed. Chapter 7 gives a thorough discussionof applicationsof number theory to cryptology, starting with classical cryptology. Character ciphers based on modular arithmetic are described,as is cryptanalysisof these ciphers. Block ciphers based on modular arithmetic are also discussed. Exponentiation ciphers and their applications are described, including an application to electronic poker. The concept of a public-key cipher system is introduced and the RSA cipher is describedin detail. Knapsackciphers are discussed,as are applicationsof cryptographyto computer science. Chapter 8 includes discussionsof the order of an integer and of primitive roots. Indices, which are similar to logarithms, are introduced. Primality testing basedon primitive roots is described. The minimal universalexponent is studied. Pseudo-random numbers and means for generating them are discussed.An applicationto the splicingof telephonecablesis also given. Chapter 9 covers quadratic residues and the famous law of quadratic reciprocity. The Legendreand Jacobi symbolsare introduced and algorithms for evaluating them are developed. Euler pseudoprimesand a probabilistic primality test are covered. An algorithm for electronically flipping coins is developed. Chapter l0 coversrational and irrational numbers,decimal representations of real numbers,and finite simple continuedfractionsof rational and irrational numbers. Special attention is paid to the continued fractions of the square roots of po"itive integers. Chapter 1l treats some nonlinear diophantine equations. Pythagorean triples are described. Fermat's last theorem is discussed. Finallv. Pell's equation is covered.

vill

P reface

Problem Sets After each sectionof the text there is a problem set containing exercisesof various levelsof difficulty. Each set containsproblemsof a numerical nature; these should be done to develop computational skills. The more theoretical and challenging problems should be done by studentsafter they have mastered the computationalskills. There are many more problemsin the text than can be realistically done in a course. Answers are provided at the end of the book for selectedexercises,mostly those having numerical answers. Computer Projects After each section of the text there is a selectionof computer projects that involve concepts or algorithms discussedin that section. Students can write their programs in any computer language they choose, using a home or personal computer, or a minicomputer or mainframe. I encouragestudents to use a structured programming languagesuch as C, PASCAL, or PL/ 1, to do these projects. The projects can serve as good ways to motivate a student to learn a new computer language, and can give those students with strong computer science backgrounds interesting projects to tie together computer scienceand mathematics. Unsolved Problems In the text and in the problem setsunsolvedquestionsin number theory are mentioned. Most of these problems have eluded solution for centuries. The reader is welcome to work on these questions,but should be forewarned that attempts to settle such problems are often time-consuming and futile. Often people think they have solved such problems,only to discover some subtle flaw in their reasoning. Bibliography At the end of the text there is an extensivebibliography,split into a section for books and one for articles. Further, each section of the bibliography is subdivided by subject area. In the book section there are lists of number theory texts and references, books which attempt to tie together computer scienceand number theory, books on some of the aspectsof computer science dealt with in the text, such as computer arithmetic and computer algorithms, books on cryptography, and general references.In the articles section of the bibliography, there are lists of pertinent expository and research papers in number theory and in cryptography. These articles should be of interest to the reader who would like to read the original sources of the material and who wants more details about some of the topics coveredin the book.

Preface

tx

Appendix A set of five tables is included in the appendix to help studentswith their computations and experimentation. Students may want to compile tables different than those found in the text and in the appendix; compiling such tables would provide additional computer projects. List of Symbols A list of the svmbols used in the text and where they are defined is included. Acknowledgments I would like to thank Bell Laboratoriesand AT&T Information Systems Laboratories for their support for this project, and for the opportunity to use the UNIX system for text preparation. I would like to thank George Piranian for helping me develop a lasting interest in mathematics and number theory. Also I would like to thank Harold Stark for his encouragementand help, starting with his role as my thesisadvisor. The studentsin my number theory courses at the University of Maine have helped with this project, especially Jason Goodfriend, John Blanchard, and John Chester. I am grateful to the various mathematicians who have read and reviewed the book, including Ron Evans, Bob Gold, Jeff Lagarias and Tom Shemanske. I thank Andrew Odlyzko for his suggestions,Adrian Kester for his assistancein using the UNIX system for computations, Jim Ackermann for his valuable comments, and Marlene Rosen for her editing help. I am particularly grateful to the staff of the Bell Laboratories/American Bell/AT&T Information Services Word ProcessingCenter for their excellent work and patience with this project. Special thanks go to Marge Paradis for her help in coordinating the project, and to Diane Stevens, Margaret Reynolds, Dot Swartz, and Bridgette Smith. Also, I wish to express my thanks to Caroline Kennedy and Robin Parson who typed preliminary versions of this book at the University of Maine. Finally, I would like to thank the staff of Addison-Wesley for their help. I offer special thanks to my editor, Wayne Yuhasz, for his encouragement,aid, and enthusiasm.

Lincroft, New Jersey December.1983

Kenneth H. Rosen

Contents

Chapterl. l.l 1.2 1.3 t.4 1.5

The Integers The well-ordering Divisibility Representations of int;;;;;....-'.....-'-.'......... Computer operationswith integers............ Prime numbers...

Chapter2.

Greatest Common Divisors and Prime Factorization

2.1 2.2 2.3 2, 4 2.5

Greatest common divisors The Euclideanalgorithm ........... The fundamentaltheorem of arithmetic ............ Factorization of integers and the Fermat numbers Linear diophantineequations...............

Chapter3. 3.1 3.2 3.3 3.4 Chapter4. 4.1 4.2 4.3 4. 4

4 l8 24 33 45

53 58 69 79 87

Congruences Introduction to congruences Linearcongruences.............. The Chinese remainder theorem Systemsof linear congruences..............

9l 102 107 I 16

Applications of Congruences D i v i s i b i l i t yt e s t s . . . . . . . . . T h e p e r p e t u a cl a l e n d a r . . . . . . . . . . . . . R o u n d - r o b i nt o u r n a m e n t s . . . . . . . . . . Computer file storageand hashingfunctions...............

.. ..

129 134 139 l4l

xl

Contents

Chapter 5. 5.1 5.2 5.3 Chapter6. 6.1 6.2 6.3 Chapter 7. 7 .l 7 .2 7.3 7.4 7.5 7.6 Chapter 8. 8.1 8.2 8.3 8.4 8.5 8.6 8.7 8.8 Chapter 9. 9.I 9.2 9.3 9.4

Some Special Congruences Wilson's theorem and Fermat's little theorem Pseudoprimes.............. Euler's theorem

..

147 152 16l

MultiplicativeFunctions E u l e r ' sp h i - f u n c t i o n. . . . . . . . . . . . . . . T h e s u m a n d n u m b e ro f d i v i s o r s . . . . . . . . . . . . . . Perfect numbersand Mersenneprimes

166 174 180

Cryptology Character ciphers Block ciphers Exponentiation ciphers............... Public-keycryptography............. Knapsack ciphers Some applicationsto computer science

..

188 198 205 212 219 227

Primitive Roots The order of an integer and primitive roots Primitive roots for primes Existenceof primitive roots Index arithmetic Primality testing using primitive roots......... Universal exponents. Pseudo-random numbers............ The splicingof telephonecables

.. ..

232 238 243 252 263 268 275 280

Quadratic Residuesand Reciprocity Quadratic residues Quadratic reciprocity The Jacobi symbol Euler pseudoprimes.............

..

288 304 314 325

xtl

Contents

Chapter 10. 10.1 10.2 10.3 10.4 Chapter I l.

l.l t.2 1.3

Decimal Fractions and Continued Fractions Decimal fractions... Finite continuedfractions Infinite continued fractions Periodic continued fractions

336 350 361 315

Some Nonlinear Diophantine Equations Pythagoreantriples.... F e r m a t ' sl a s t t h e o r e m. . . . . . . . . . . . . Pell'sequations

391 397 401

Appendix.. Answers to selected problems Bibliography............. List of symbols.... Index

410 426 438 445 447

lntroduction

Number theory, in a general sense, is the study of numbers and their p r o p e r t i e s .I n t h i s b o o k ,w e p r i m a r i l y d e a l w i t h t h e i n t e g e r s , 0 ,+ 1 , + 2 , . . . . We will not axiomatically define the integers, or rigorously develop integer arithmetic.l Instead, we discussthe interestingpropertiesof and relationships between integers. In addition, we study the applicationsof number theory, particularly thosedirected towardscomputer science. As far back as 5000 years ago, ancient civilizations had developedways of expressingand doing arithmetic with integers. Throughout history, different methods have been used to denote integers. For instance, the ancient Babyloniansused 60 as the base for their number system and the Mayans used 20. Our method of expressing integers, the decimal system,was first developed in India approximately six centuries ago. With the advent of modern computers, the binary system came into widespreaduse. Number theory has been used in many ways to devise algorithms for efficient computer arithmetic and for computer operationswith large integers. The ancient Greeks in the school of Pythagoras, 2500 years ago, made the distinction betweenprimes and composites. A prime is a positive integer with no positive factors other than one and the integer itself. In his writings, Euclid, an ancient Greek mathematician, included a proof that there are infinitely many primes. Mathematicians have long sought formulae that generate primes. For instance, Pierre de Fermat, the great French number theorist of the seventeenthcentury, thought that all integers of the form 22' + 1 are prime; that this is false was shown, a century after Fermat made this claim, by the renowned Swiss mathematician Leonard Euler, who demonstratedthat 641 is a factor of 22' + | . The problem of distinguishing primes from compositeshas been extensively studied. The ancient Greek scholarEratosthenesdeviseda method, now called l.

S u c h a n a x i o m a t i c d e v e l o p m e n to f t h e i n t e g e r sa n d t h e i r a r i t h m e t i c c a n b e f o u n d i n L a n d a u

t6ll.

Introduction

the sieve of Eratosthenes, that finds all primes less than a specified limit. It is inefficient to use this sieve to determine whether a particular integer is prime. The problem of efficiently determining whether an integer is prirne has long challengedmathematicians. Ancient Chinese mathematiciansthought that the primes were precisely those positive integers n such that n divides 2' - 2. Fermat showed that if n is prime, then n does divide 2n - 2. However, by the early nineteenth century, it was known that there are compositeintegersn such that n divides 2n - 2, such as n : 341 . These compositeintegers are called pseudoprimes Becausemost compositeintegers are not pseudoprimes,it is possibleto develop primality tests based on the original Chinese idea, together with extra observations. It is now possibleto efficiently find primes; in fact, primes with as many as 200 decimal digits can be found in minutes of computer time. The fundamental theorem of arithmetic, known to the ancient Greeks, says that every positive integer can be written uniquely as the product of primes. This factorization can be found by trial division of the integer by primes less than its square-root; unfortunately, this method is very timeconsuming. Fermat, Euler, and many other mathematicians have produced imaginative factorization techniques. However, using the most efficient technique yet devised, billions of years of computer time may be required to factor an integer with 200 decimal digits. The German mathematician Carl Friedrich Gauss, consideredto be one of the greatest mathematicians of all time, developed the language of congruences in the early nineteenth century. When doing certain computations,integers may be replaced by their remainders when divided by a specific integer, using the language of congruences. Many questions can be phrased using the notion of a congruencethat can only be awkwardly stated without this terminology. Congruenceshave diverse applications to computer science,including applications to computer file storage, arithmetic with large integers,and the generationof pseudo-randomnumbers. One of the most important applications of number theory to computer science is in the area of cryptography. Congruencescan be used to develop various types of ciphers. Recently, a new type of cipher system, called a public-key cipher system, has been devised. when a public-key cipher is used, each individual has a public enciphering key and a private deciphering key. Messagesare encipheredusing the public key of the receiver. Moreover, only the receiver can decipher the message,since an overwhelming amount of computer time is required to decipher when just the enciphering key is known. The most widely used public-key cipher system relies on the disparity in computer time required to find large primes and to factor large integers. In

lntrocluction

particular, to produce an enciphering key requires that two large primes be found and then multiplied; this can be done in minutes on a computer. When these large primes are known, the decipheringkey can be quickly found. To find the deciphering key from the enciphering key requires that a large integer, namely the product of the large primes, be factored. This may take billions of years. In the following chapters,we discussthese and other topics of elementary number theory and its applications.

1 The Integers

1.1 The Well-OrderingProperty In this section,we discussseveral important tools that are useful for proving theorems. We begin by stating an important axiom, the well-ordering property. The Well-Ordering Property. Every nonempty set of positive integers has a least element. The principle of mathematical induction is a valuable tool for proving results about the integers. We now state this principle, and show how to prove it using the well-ordering property. Afterwards, we give an example to demonstrate the use of the principle of mathematical induction. In our study of number theory, we will use both the well-ordering property and the principle of mathematical induction many times. The Principle of Mathematical Induction. A set of positive integers that contains the integer I and the integer n I I whenever it contains n must be the set of all positive integers. Proof. Let S be a set of positive integers containing the integer I and the integer n * | whenever it contains n. Assume that S is not the set of all positive integers. Therefore, there are some positive integers not contained in .S. By the well-ordering property, since the set of positive integers not contained in S is nonempty, there is a least positive integer n which is not in . S . N o t e t h a t n 1 1 , s i n c el i s i n S . N o w s i n c en ) l , t h e i n t e g e r n - 1 i s

l.l

The Well-Ordering ProPertY

a positive integer smaller than n, and hence must be in S. But since S contains n - l, it must also contain (n-t) + | : n, which is a contradiction, since n is supposedlythe smallest positive integer not in S. This shows that S must be the set of all positive integers. tr To prove theorems using the principle of mathematical induction, we must show two things. We must show that the statement we are trying to prove is true for l, the smallest positive integer. In addition, we must show that it is true for the positive integer n * I if it is true for the positive integer n. By the principle of mathematical induction, one concludes that the set S of all positive integers for which the statement is true must be the set of all positive integers. To illustrate this procedure, we will use the principle of mathematical induction to establish a formula for the sum of the terms of a geometric progression. Definition. Given real numbers 4 and r. the real numbers a , a r , e r 2 ,o t 3 r . . . are said to form a geometric progression. Also, a is called the initial term and r is called the common ratio. Exa m ple. T he num b e rs 5 , -1 5 ,4 5 , -1 3 5 ,... fo rm a geometri c progressi on with initial term 5 and common ratio -3. In our discussion of sums, we will find summation notation useful. The following notation representsthe sum of the real numberse1, o2,...,on. lan

2oo:er*az*

k-l

We note that the letter k, the index of summation, is a "dummy variable" and can be replaced by any letter, so that

5,

ak:

k-l

Example. We see that

nn

2 oi

j-t

i-l

The Integers

) 2j:I+2+3+4+5:15,

j-r

) 2t2:2+2+2+2+2:10,

j-r

and ) 2 2i : 2 * 22+ 23+ 24+ 2s : 62 .

j-1

We also note that in summation notation, the index of summation may range betweenany two integers,as long as the lower limit does not exceedthe upper limit. If m and h are integers such that z ( n, then b

oo:am*a^a1*

*an.

k-m

For instance.we have 5

> k 2 : 3 3+ 4 2+ 5 2 : 5 0 ,

k;t

> 3k:30 + 3t + 32: 13,

fr:0 and I k--2

We now turn our attention to sums of terms of geometricprogressions.The su m of t he t er m s e ) e r, o r2 ,...,a rn i s n

2ori:e*ar*ar2+

*arn,

j-0

where the summation beginswith 7 : g. We have the following theorem. Theorem l.l.

If a and r ^re real numbersand r *

l. then

1.1 The Well-OrderingProperty

n),,narn*l-Q

(1.1)

r* a arn r ' ' : T: T

: a * ar i*a rar2 -t + E ori j:o

.

Proof. To prove that the formula for the sum of terms of a geometric progressionis valid, we must first show that it holds for n : l. Then, we must show that if the formula is valid for the positive integer n, it must also be true for the positive integer n * l. To s t ar t t hings o ff, l e t n : l . T h e n , th e l e ft si de of (t.t) i s a * ar, w hi l e o n t he r ight s ideof (1 .1 ) w e h a v e arL-a _ a?z-t) r-l r-l

_ ab*l)(r-1) T:

a(r*l) : a * ar

So the formula is valid when n : l. N ow we as s um eth a t (1 .1 ) h o l d s for the positive integer n. assumethat

0.2)

That is, we

'tar'-arn*l-Q

alar+arz+

I

We must show that the formula also holds for the positive integer n * l. What we must show is that

(t.:)

a*ar+ar2+

* arn * arn*l :

or@+t)+t_o

ar'+2-e

r-l

r-l

To show that (1.3) is valid, we add orn*r to both sidesof (1.2), to obtain (t.+)

(a*ar*ar2+...+arn)

*

arn+t:o + arr+t, r-l

a r ' + r-

The left side of (t.+) is identical to that of (1.3). To show that the right sides are equal, we note that arn*l-a r-

1 T A ^r - n r r _

I

arn+l-e

r-l

, or'*l (r- I )

T-

orn*l-a*ar'+Z

r-1 arn*l

: r-l

Since we have shownthat 0.2) i m p l i e s (t.:), w e can concl udethat (t.t)

The Integers

holds for all positive integers n. tr Example. Let n be a positive integer. To find the sum

*2',

bro:r*2+22+

k:0

we use Theorem l.l with e : I and r : 2, to obtain

l+2+22+

.

J- 1n

1n*l _ I :

rn*l_r

2-l

Hence, the sum of consecutivenonnegative powers of 2 is one less than the next largest power of 2. A slight variant of the principle of mathematical induction is also sometimes useful in proofs. The Second Principle of Mathematical Induction. A set of positive integers which contains the integer 1, and which has the property that if it contains all th e pos it iv eint eg e rs1 ,2 ,..., k , th e n i t a l s o c ontai nsthe i nteger k + l , must be the set of all positive integers. Proof. Let T be a set of integers containing I and containing k + I if it co nt ains 1, 2, . . . , k . L e t S b e th e s e t o f a l l p osi ti vei ntegersn such that al l the positive integers less than or equal to n are in Z. Then I is in S, and by the hypotheses,we see that if k is in S, then k + | is in S. Hence, by the principle of mathematical induction, S must be the set of all positive integers, so clearly T is also the set of all positive integers. tr The principle of mathematical induction provides a method for defining the values of functions at positive integers. Definition. We say the function f is defined recursively if the value of f at I from f h) . is specifiedand if a rule is providedfor determiningf h*l) If a function is defined recursively, one can use the principle of mathematical induction to show it is defined uniquely at each positive integer. (See problem 12 at the end of this section.) We now give an example of a function defined recursively. We define the factorial function f fu) : nt . First, we specify that

1.1 The Well-Ordering ProPertY

f(r): I , and then we givethe rule for finding f h*1) from f fu), namely

f h+r) : (n+r)'ffu). These two statementsuniquely define r!. To find the value of f G) : 6! from the recursive definition of f h) : nl, use the secondproperty successively,as follows

(2) :6's'4'3'2f0). f 6) :6.f (5) : 6.5.f(4) : 6.s.4'f(3) : 6's'4'3'f We now use the first statement of the definition to replacef 0) by its stated value l. to concludethat 6 l : 6 ' 5 ' 4 ' 3 ' 2 ' :l 7 2 0 . In general, by successivelyusing the recursive definition, we see that n! is the product of the first n positive integers,i.e. n! : l'2'3

n

For convenience,and future use, we specify that 0! : l. We take this opportunity to define a notation for products, analogous to summation notation. The product of the real numbers a1, a2,...,a, is denoted by

ft o, : ere2

j -r

an

The letter 7 above is a "dummy variable", and can be replaced arbitrarily. Example. To illustrate the notation for products we have ) fI j:l'2'3'4'5:120. j-r 5

I I 2 : 2 . 2 . 2 . 2 . 22: 5: 3 2 .

j-r 5 fI Zi : j-r

2.22.23.24.2s:

2r5

l0

The Integers

We note that with this notation, n ! :

fI . j -r ,r

Factorials are used to define binomial cofficients. Definition. Let m and k be nonnegativeintegers with k 4 m. The

r) binomial cofficien,lT I isoenneo uy (^ / r)

mt kt(m_k)t

l*| lk J t r t : -

l^)

In computing we see that there is a good deal of cancellation,because lO ,J,

l^) : - lk )

m; kt@_k)l

t . 2 . 3. . . @ - k ) @ - k + t ) . . . t u - t ) m k! t.2.3 fu-k) (m-k+r) ( m - r )m kt

fzl

Example.To evaluatethe binomialcoefficien, we notethat L, ,J, r\ 1 7| : 7 t : 1 . 2 . 3 . 4 . s . 6 . 7s . 6 . 7

f3J

3t4t r23.r234:E:i)'

We now prove some simple propertiesof binomial coefficients. Proposition 1.2. Let n and k be nonnegativeintegerswith k ( n . Then (i) [;]:[;]:,

r) r ) ( i i ) l l l : -l ' . 1 fkj l,-t,)' Proof. To see that (i) is true, note that

11

1.1 The Well-OrderingProperty

[;]:#

:n'':l nt

and

t;]

_n,._ n !0!

\:t

To verify (ii), we seethat

frl

n;

| . kJ

kth-k)t

l,l:

:-:l

nt

lr

t u - k ) r ( n -h - k ) ) t

,l

l-

ln-* )'

tr

An important property of binomial coefficientsis the following identity. Theorem 1.2. Let n and k be positive integers with n > k. Then

|',]*, I n I _ |,,*'l r loj [o-,J:I

)

Proof. We perform the addition

[;]. lr:, by using the c om m o nd e n o mi n a to rftl (n -k + t)!.

t. + Uc lr\,

Thi s gi ves

n th - k t l ) n tk ktfn-k+l\ ktJtt-t(+il nl((n-k +r) +k) k th - k + t ) t ntfu*l) klfu-k+r)t (n+l)! kth-k +r)t

[l l nn + rI f k )

u

t2

The Integers

Using Theorem 1.2, we can easily construct Pascal's triangle, which displavsthe binomial coefficients. In this triangle, the binomial coefficient |,,]

rs t he |.r,l

( k + t)ttr n u m b e r i n th e (n + l )th

row . The fi rst ni ne row s of

Pascal'st r ianglea re d i s p l a y e di n F i g u re l .l . I ll

r2l l33l r4641 15101051 1615201561 172135352171 18285670562881 'Plr"urt Figure1.1.

triangle.

We see that the exteriornumbersin the triangleare all l. To find an interiornumber,we simplyadd the two numbersin the positionsabove,and to either side,of the positionbeing filled. From Theorem1.2, this yieldsthe correctinteger. of powersof sums. Exactly occur in the expansions Binomial coefficients how they occuris describedby the binomial theorem. The BinomialTheorem. Let x and y be variablesand n a positiveinteger. Then

: y'+ (x*y)n l:).. [;]".. [T]".-',. + l,:r)*r.-,+ [,:,]'y n - +l:),' -2

or using summation notation,

2

l3

1.1 The Well-Ordering ProPertY

G + y ) n: 2

^

j-0

(n]

l\ J; ll * " - t y t

We prove the binomial theorem by mathematical induction. In the proof we make use of summation notation. Proof. We use mathematical induction. When n : l, according to the binomial theorem. the formula becomes

(x*y)r-frlfrl + loj"'.yoI,,J"or' lrlfrl

But because lnl: t"J

lil:t,this

s t a t e st h a t ( x + y ) r : x

*y,

w h i c hi s

\^/

obviously true. We now assume the theorem is valid for the positive integer n, that is, we assumethat

^ fn) G+ y ) n: 2 l , l r ' - i r i . \r ) j-0 We must now verify that the correspondingformula holds with n replaced by n * l, assumingthe result holds for n. Hence, we have (x+y)n+r - (xty)"(x+y)

'l I : l, |,,.l l a l\ri l)" - t ' l l ( x + r ) |.i:o

, lnl j-0

\r )

J

, fr) j:0

\J ./

We see that by removing terms from the sums and consequently shifting indices.that

t4

The Integers

21,).'-'."' 2l;).'-'.',' :'Al,).'-'''.' 3l:).'-'''*' : In+l +

and

* yn+t

:21'!'1"-'*' yj + yn*t

Hence, we find that 't

( x *Y )' + r -

I

n

xn+r

+>

lxn-i+tri I yn+t I

j-r

By Theorem 1.2, we have

t;l+ [,1']:

[';']

,

so we conclude that

- ,,*, + bl':'fx,-i*,ri k+y),,'+r I r )

* yn+r

i-t

-

n * t [ n + rI S

I

l*n+t-iri

t 1 ^l . j )

This establishesthe theorem. u We now illustrate one use of the binomial theorem. If we let x : y : l. we see from the binomial theorem that

^ lrl

2 n: ( t + t ) , : )

j-0

, rl

lnl : l\ r r )l t , - r l i j -)o LJ,l

This formula showsthat if we add all elementsof the fu+l)th row of Pascal's triangle, we get 2n. For instance,for the fifth row, we find that

15

1.1 The Well-OrderingProPertY

:, +4+6+4+,:,6:24 . . . . [;] [l] [l] [l] [l]

Problems

l.l l.

Find the values of the following sums l0

l0

a) >2 j-r

c) 2j' j-r

l0

t0

u) 2i

o) 22i. j-r

j-l

2.

Find the values of the following products 55

c) r. j'

i l j -rl r 2 ) b) trj

0) il2i

j-t

j-l

3 . Find n ! for n equal to each of the first ten positive integers. 4.

fro)frolfrolfrol frol Find lo,|'|.,.l'I r.l'I tJ'^na lroJ'

5 . Find the binomial coefficients

fnl , fnl

f,ol

|'qI fgI

froI

and o andverirvthat l',l' loJ' I ,J'

lrj*loj: loJ 6 . Show that a nonempty set of negative integers has a largest element. 7 . Use mathematical induction to prove the following formulae.

a) >,i:t+2+3+

+ ,:n(nlD. L

j-l

U) 2i': j-l

12+22+32+

+

.t , a

n (n+l) (2n+l) 6

t6

The Integers

|

c ) i . r ' : t ' + 2 3+ 3 3+ i-tt2l

* n3: | 't'ftl

12

I

8.

Finda formula rcrjft Zi. -l

9.

Use the principle of mathematical induction to show that the value at each positive integer of a function defined recursivelyis uniquely determined.

r0.

what function f (n) is defined recursively by f 0) : 2 and for n)l?

ll.

I f g i s d e f i n e d r e c u r s i v e l yb y g ( l ) : 2 what is S(02

t2.

The second principle of mathematical induction can be used to define functions recursively. We specify the value of the function at I and give a rule for finding from the values of f at the first n positive integers. Show that the f h+l) values of a function so defined are uniquely determined.

t3.

We define a function recursively for all positive integers n bV (l) : l, "f and for n 2 2, f h+t):f Show that f (n) : h) + 2f (n-t). f (2):5, 2^ + el)n, using the secondprinciple of mathematical induction.

14. a)

g(n) :2sb-D

and

Let n be a positive integer. By expanding (l+(-l))'with theorem. show that

f (n+D : 2f (n) for

n 7 2,

the binomial

,

fr) : o. ) (-r)o lrJ b) usepart(a),andthefactthat > f;l :2' , to find \'' J t-o

f,l* f,l* l,l * loj IrJ loj and

[,lf,l|,,l ['J*l,J* I'J* c)

Findthesuml -2+22-23

+

15. Show by mathematical induction that (2n)t < 22'(nl)z.

+2too. if

n

is a

positive integer, then

t7

1.1 The Well-Ordering ProPertY

16. The binomial coefficients

x is a variable, and n is a positive integer, [;],*nr." : x and

can be defined recursivelyby the equations

[l ]

| .I ,_n [,1 In+tJ:R l;l |.".l

x!

,

a)Showthatifxisapositiveinteger,then[oJ:ffi,wherekisan integerwithl(k(x.

.

b)

["]

S h o w t h a tl - l + l,?J

[*l 1.,,

| : lt?+rj

f'+rl l--*, ln,'t

l,whenevernisapositiveinteger.

of inclusion - exclusion. Suppose P2,.,., P, be t different properties Pr, that S is a set with n elements and let that an element of S may have. Show that the number of elements of S possessingnone of the / properties is

t 7 . In this problem, we develop the principle

+ n@)l n -ln(rr) + n(p) + + n(P,-r,P,)l + l n ( P t , P z+ ) n ( P t , P r+) - { n ( P r , P z , P t )* n ( P r P z , P q ) + + + (-l)'n (P1,P2,...,P,),

* n(P,-2,P,4,P,)|

where n(Pi,,Pi,,..., P,,) is the number of elements of S possessingall of the properties Pi,,P;,,...,P;,.The first expressionin brackets contains a term for each property, the secondexpressionin brackets contains terms for all combinations of two properties, the third expressioncontains terms for all combinations of three properties,and so forth. (Hint: For each element of S determine the number of times it is counted in the above expression. If an element has k of the

properties, showit is counted t-

lrl + lpllrJ

Itl

+ (-l)ft

ltl ,i-.t. lrJ

This

equals zeroby problem la(a).) 1 8 . The tower of Hanoi was a popular puzzle of the late nineteenth century. The puzzle includes three pegs and eight rings of different sizes placed in order of size, with the largest on the bottom, on one of the pegs. The goal of the puzzle is to move all the rings, one at a time without ever placing a larger ring on top of a smaller ring, from the first pbg to the second,using the third peg as an auxiliary peg.

l8

The Integers

a)

Use mathematicalinduction to show that the minimum number of movesto transfer n rings, with the rules we have described,from one peg to another is 2n - 1.

b)

An ancient legend tells of the monks in a tower with 64 gold rings and 3 diamond pegs. They started moving the rings, one move per second, when the world was created. When they finish transferring the rings to the second peg, the world ends. How long will the world last?

19. Without multiplying all the terms, show that il 6! 7!: l0! b) l0!:7! 5! 3!

c) 16!: l4t 5t 2l d ) 9 t - 7 13 ! 3 ! 2 ! .

20. Let an : (af a2l. ar-1!) - l, and on+t: af. a2t o1,a2,...,etr-1 or€ positiveintegers. Show that an*1!: al. a2t 2 1 . F i n d a l l p o s i t i v ei n t e g e r sx , y , a n d z s u c h t h a t x t * y l : l.l

an_tl, onl.

where

z!.

Computer Projects Write programs to do the following:

l.

Find the sum of the terms of a geometric series.

2.

Evaluate n !

3.

Evaluate binomial coefficients.

4.

Print out Pascal'striangle.

5.

List the movesirr the Tower of Hanoi puzzle (see problem l8).

6.

Expand (x*y)",

where n is a positive integer, using the binomial theorem.

1.2 Divisibility When an integer is divided by a secondnonzerointeger, the quotient may or m ay not be an i n te g e r. F o r i n s ta n c e ,2 4 /8 : 3 i s an i nteger,w hi l e l 7/5:3.4 is not. This observationleads to the following definition. Definition. If a and b are integers, we say that a divides b if there is an integer c such that b : ac. lf a divides b, we also say that a is a divisor or factor of b.

t9

1.2 Divisibility

I f a d i v i d e sb w e w r i t e a l b , w h i l e i f a d o e s n o t d i v i d e b , w e w r i t e a t r U . Example. The following examples illustrate the concept of divisibility of

i n t e g e r s1:3| 1 8 2-,5 | 9 0 ,t 7 l 2 8 g , e t r q q , l t r s o-,l | : 1 , a n d1 71 0 .

Example. The divisorsof 6 are +1, *2, +3, and +6. The divisorsof 17 are +5, +10, The divisors of 100 are +1, *2,+4, tl and tI7. +20, +25, +50, and + 100. In subsequentsections,we will need some simple properties of divisibility. We now state and prove these properties. 1.3. If a,b,and c areintegerswitha Proposition l b a n db l r , t h e n a l c . Proof. Since a I b and b I c, there are integers e and f with ae : b and bf : ,. Hence, bf : be)f : aGf) : c, and we concludethat a I c. a Exa mple. S inc e 1l | 6 6 a n d 6 6 | tl a , P ro p o s i ti on1.3 tel l s us that 11 | 198. P r o p o s i t i o n1 . 4 . l f a , b , m , c | (ma+nb).

a n d n a r e i n t e g e r sa, n d i f c l a a n d c l D , t h e n

Proof. Since c I a and c | 6, there are integers e and / such that a : ce and b : c f . Henc e, m a * n b : m c e * n c f : c (me + nf). C onsequentl y, w e see th a t c | f ua+ nb) . E Exa mple. S inc e 3l2 l

a n d : I l l , Pro p o s i ti o n1 .4 tel l s us that

3 | 6 - z l - 3 . 3 3:) l o 5 - 9 9 : 6 . The following theorem states an important fact about division. The Divisionl$f$* If a and b are integers such that b > 0, then there are unique integers q and r such that a : bq * r with 0 ( r < b. In the equation given in the division algorithm, we call q the quotient and r the remainder. We note that a is divisible by b if and only if the remainder in the division algorithm is zero. Before we prove the division algorithm, consider the following examples.

20

The Integers

Example. If a-.133 and b:21, then Q:6 and r:7, since 133:21'6+7. L i k e w i s ei,f a : - 5 0 a n d b : 8 , t h e n q - - 7 and r:6, s i n c e- 5 0 : 8 ( - 7 ) + 6. For the proof of the division algorithm and for subsequent numerical computations,we need to define a new function. Definition. Let x be a real number. The greatest integer in x, denoted by [x ], is the largest integer lessthan or equal to x. Example.

We

have the following values for

: 2,131: 3, andI-t.sl : -2. x'. 12.21

the greatest integer in

The proposition below follows directly from the definition of the greatest integer function. Proposition 1.5. If x is a real number, then x-l

< [x] ( x.

We can now prove the division algorithm. Note that in the proof we give explicit formulae for the quotient and remainder in terms of the greatest integer function. Proof. Let q:la/bl a n d r : a - b l a / b l . C l e a r l ya : b q * r . T o s h o w r that the remainder satisfies the appropriate inequality, note that from Proposition1.5, it follows that G/b)-l

< ta/bl 4a/b.

We multiply this inequality by b, to obtain a - b < btalbl 4 a. Multiplying by -1, and reversingthe inequality,we find that

-a(-b[a/bl l. integer n can be written uniquely in the form n : a k b k * a p - 1 b k - rt

Then every positive

* a1b I oo,

w h e r e a; is an int eg e rw i th 0 ( o ; < b -l coefficientak I O.

fo r,/ :0,

1,..., k and the i ni ti al

Proof . We obtain an expressionof the desired type by successivelyapplying the division algorithm in the following way. We first divide n by b to obtain n:beo*oo,

0(ao "'>

0,

and any decreasing sequence of nonnegative integers must eventually terminate with a term equaling 0.

26

The Integers

From the first equation above we find that n:

beo* ao.

We next replace {6 using the secondequation, to obtain n : b(bqfta1) + as : bzqrI a1b I as, Successively substituting for qr, Q2,..., Qk_r,we have n:

b 3 q z + a 2 b 2* a 1 b * o r ,

: =i: ri::,-'**"::,t{,-'..**olr'u**ol' : a t b k + a 1 r - 1 b k -*r

t aft * ao.

w her e 0 ( a; < b -l fo r 7 : 0 ,1 ,...,ka n d a * I 0, si nceek : 4r-r i s the l ast nonzero quotient. Consequently,we have found an expansion of the desired type. To see that the expansion is unique, assume that we have two such expansionsequal to n, i.e. n : e k b k + a 1 r - y b k - *t : c * b k * c 1 r-1 b k -r*

t a1b * ao * cft * ro,

where 0 ( ar (b and 0 ( c1(b (and if necessarywe add initial terms with zero coefficients to have the number of terms agree). Subtracting one expansionfrom the other, we have (ar,-c)bk +(o,,-r-c1,-)bk-t *

*(a;cr)b

+ (as-ca):0.

If the two expansionsare different, there is a smallest integer j, O ( < k, "l such that ai # ci. Hence, .f

br

+ l(a*-c*)b(-r

* (ai+rci+r)b * G1-c1)] : o,

so that

Gr,-c)bk-i +

+ (a1+rci+)b

r (ai-c1) : O.

27

1.3 Representationsof Integers

Solving for ai-c; we obtain aj- c j:

(c rr-a r)b k -j +

* (c 7+ r-ai + )b

: bl(c1,-a1)bk-j-t +

* (c7+r-or*,) ].

Hence, we see that

bl

G 1 -c 1 ).

But since 0 ( a; < b and 0 ( c; < b, we know that -b < ai-c1 I b. implies that ej : cj. This contradicts the Consequently, b I h1-c) assumptionthat the two expansionsare different. We concludethat our base 6 expansionof n is unique. ! For b - 2 . we see from Theorem 1.3 that the following corollary holds. Corollary 1.1. Every positive integer may be represented as the sum of distinct powersof two. Proof. Let n be a positive integer. From Theorem 1.3 with b : 2, we know t h a t n : a t r T k * a 1 r - 1 2 k - t* + a Q * a s w h e r e e a c h a ii s e i t h e r 0 o r 1 . Hence, every positive integer is the sum of distinct powersof 2. tr In the expansionsdescribedin Theorem 1.3, b is called the base or radix of the expansion. We call base l0 notation, our conventionalway of writing integers, decimal notation. Base 2 expansionsare called binary expansions, base 8 expansionsare called octal expansions,and base 16 expansionsare called hexadecimal, or hex for short, expansions. The coefficients ai are called the digits of the expansion. Binary digits are called bits (binary digils) in computer terminology. To distinguish representationsof integers with different bases, we use a special notation. We write (apapa...aps) 6 to represent the expansion a*bklapabk-rl taft*ao. Example. To illustrate base b notation, note that Q3Ot : 2.72+ a n d ( 1 0 0 1 0 0 1 1 :) 2 1 . 2 7+ 1 . 2 4+ 1 . 2 r+ 1 .

3.7 + 6

Note that the proof of Theorem 1.3 gives us a method of finding the base b expansion of a given positive integer. We simply perform the division algorithm successively,replacing the dividend each time with the quotient, and

28

The Integers

stop when we come to a quotient which is zero. We then read up the list of remaindersto find the base b expansion. Example. To find the base 2 expansionof 1864, we use the division algorithm successively:

1 8 6 4: 2 . 9 3 2 + 0 , 932:2'466 +0, 466:2'233 +0 233-2'116+1, 1 1 6: 2 ' 5 8 + 0 , 58:2'29 +0, 29:2'14 +1, 14:2'7 +0, 7 : 2'3 + 1, 3 : 2'l + l, | : 2'O + 1. To obtain the base 2 expansionof 1984, we simply take the remaindersof t h e s ed i v i s i o n s .T h i s s h o w st h a t ( 1 8 6 4 ) r o : ( 1 1 1 0 1 0 0 1 0 0 0 ) 2 . Computers represent numbers internally by using a series of "switches" which may be either "on" or "off". (This may be done mechanically using magnetic tape, electrical switches, or by other means.) Hence, we have two possiblestates for each switch. We can use "on" to represent the digit I and "off" to representthe digit 0. This is why computers use binary expansionsto representintegers internally. Computers use base 8 or base 16 for display purposes. In base 16, or hexadecimal, notation there are l6 digits, usually denoted by 0 ,1, 2, 3, 4, 5, 6,, 8, 7 9 ,A,8 ,,C ,D ,,Ea n d F . T h e l e tters A ,B ,C ,D ,E , and F are and l5 (written used to representthe digits that correspondto 10,11,12,13,14 in decimal notation). We give the following example to show how to convert from hexadecimalnotation to decimal notation. Example. To convert (A35B0F) 16we write ( e l s n o r ) r e : 1 0 . 1 6 s + 3 ' 1 6 4+ 5 ' 1 6 3+ l l ' r c z + 0 ' 1 6 + 1 5 : ( t o7o5 679)rc.

29

1.3 Representationsof Integers

A simple conversionis possible between binary and hexadecimal notation. We can write each hex digit as a block of four binary digits according to the correspondencegiven in T a b l e l . l . Hex Digit

Binary Digits

Hex Digit

Binary Digits

0 I 2 3 4 5 6 7

0000 0001 0010 0 0 1l 0100 0101 0110 0l l1

8 9 A B C D E F

r000 1001 1010 1011 l 100 I l0l 1110 llll

Table1.1. Conversion from hex digits to blocksof binarydigits. Example. An example of conversionfrom hex to binary is (zFBrrc: (tOt t 1110110011)2 .E a c h h e x d i g i t i s c o n v e rt edto a bl ock of four bi nary digits (the initial zeros in the initial block (OOIO)2correspondingto the digit (2) rc are omitted). To convert from binary to hex, consider(t t t tOl I I101001)2. We break this into blocks of four starting from the right. The blocks are, from right to left, 1 0 0 1, 1110, 1101,an d 0 0 1 1 (w e a d d th e i n i ti a l z eros). Transl ati ngeach bl ock to hex, we obtain GOng)ru. We note that a conversionbetween two different basesis as easy as binary hex conversion,wheneverone of the basesis a power of the other.

1.3 Problems l.

Convert (1999)1sfrom decimal to base 7 notation. Convert (6tOS)t from base 7 to decimal notation.

2.

Convert (tOtOOtOOO),from binary to decimal notation and (tgg+),0 from decimal to binary notation.

30

The Integers

3 . c o n v e r t ( 1 0 0 0 1 II l 0 l 0 l ) 2 a n d ( l I 1 0 1 0 0 1 1 1 0 ) 2f r o m b i n a r y t o h e x a d e c i m a l . 4 . convert (ABCDEF)rc, @nrecnD)to,

and (9A08)rc from hexadecimal to

binary.

5 . Explain why we really are using base 1000 notation when we break large decimal integers into blocks of three digits, separatedby commas.

6 . a)

Show that if D is a negative integer less than -1, then every integer n can be uniquer';:.])::'::;' .

*

a 1 b*

oo,

where a1, I 0 and O lj-0 )

For i : 0, this is clearly correct, since R0 : a : qb + R. Now assumethat

38

The Integers

Rft:

Then Rt+r :

:

:|

Rft - bqn-*-rrn-k-l 'l (n-k-t .

I U l. .r-o

qirilb+R-bqn-*-rvn-k-l )

fn-(k+r)-r

>

.l

qi"lb+R'

Ij-0)

e s t a b l i s h i n( 1g . 5 ) . F r o m ( t . S ) , w e s e e t h a t 0 ( R i < r n - i b , f o r i : 1 , 2 , . . . ,f l , s i n c e

n-i -l i-0

O ( Ri < rn-tb, we see that the digit qn-i is given by lRi-r/brn-il and can be obtained by successivelysubtracting brn-t from Ri-1 until a negative result is obtained,and then qn-; is one lessthan the number of subtractions. This is how we find the digits of q. E x a m p l e .T o d i v i d e( t t t O l ) 2 b y ( t t t ) 2 , w e l e t q : ( q r q r q i r . W e s u b t r a c t Z2( t t l) z : ( t t t O O), o n c e fro m (t t tO t)z to obtai n (l )2, and once more to o b t a i na n e g a t i v er e s u l t s, o t h a t Q 2 : l . N o w R l : ( t t t O l ) t - ( t t t 0 0 ) t : (1)2. We find that ql:0, s i n c eR 1 - 2 ( 1 l l ) 2 i s l e s st h a n z e r o ,a n d l i k e w i s e q u o ti e n t Henc e t h e o f th e d i v i s i o ni s (1 00)2and the remai nderi s (l )2 Qz : 0. We will be interested in discussinghow long it takes a computer to perform calculations. We will measure the amount of time needed in terms of bit operations. By a bit operation we mean the addition, subtraction, or multiplication of two binary digits, the division of a two-bit integer by one-bit, or the shifting of a binary integer one place. When we describethe number of bit operations needed to perform an algorithm, we are describing the computational complexity of this algorithm. In describing the number of bit operations needed to perforrn calculations we will use big-O notation.

39

1.4 ComputerOperationswith Integers

Definition. If f and g are functions taking positive values, defined for all x in a set S, then we say f is OQ) if there is a positive constant K such that f G) < K g( x ) f or a l l x i n th e s e t S . Proposition 1.6. If / is OQ) and c is a positiveconstant,then cf is Ok). Proof . If / is Ok), then there is a constantK such that f G) < Kg(x) for Therefore, y' is all x under consideration. Hence cf G) < GK)gG).

oQ). n P r o p o s i t i o1n. 7 .l f f t i s O ( g r ) a n d f 2 i s O k z ) , t h e n" f t + - f z i s O Q f t g 2 ) andfJzisoQe). Proof . If / is OQr) and f2 is Okz), then there are constantsK1 and K2 su ch t hat - f , ( *) < ,< 1 g 1 (x ) a n d " f z (x ) 1 K2g2(x) for al l x under consideration. Hence f 1G) +f2G)

( Krsr(x) + x2g2k)

( Kkr(x) + sz?))

where K is the maximum of K1 and K2. Hencef r + -f zis Ok,

+ gz).

Also -f tk)f

so th at " f f z is 0( 96 ).

z(.x) ( Krsr G) K2s2G) : (KrK2)kt?)g2(x)),

tr

C o rollar y 1. 2. I f / 1 a n d f 2 a re O G), th e n -f r + -f zi s Ok). Proof . Proposition 1.7 tells us that But if "f t + f z is O QS). ( ( (z x )g , s o th at -f r + .f zi s Ok). a + K Q s ) , t h e nf t + f t "fz " fz Using the big-O notation we can see that to add or subtract two r-bit integers takes Ofu) bit operations,while to multiply two n-bit integers in the conventionalway takes OGz) bit operations(see problems 16 and 17 at the end of this section). Surprisingly, there are faster algorithms for multiplying large integers. To develop one such algorithm, we first consider the multiplication of two 2n-bit integers, say a : (a2n4a2n_2...eflo)z and 2 .e w r i t e a : 2 n A t f 4 6 a n d b : 2 n B r t B s , w h e r e b : ( b 2 , 6 b , 2 n - 2 . . . b f t iW -l

40

The Integers

A t : ( a 2 r - 1 a 2 n * 2 . . . a 1 7 1 1 eA1o7: ) 2 (, a n - 1 a n - 2 . . . a p g ) 2B, t : ( b 2 n - f t 2 r - z . . . b n + t br)2, and B0 : (br-t bn-z...brbiz. We will use the identity (t.e)

a b : ( 2 2 , + 2 , ) A r B r r 2 n( A r A i ( a o - n r )

+ (2,+l)AoB0.

To find the product of a and 6 using (t.0), requires that we perform three mu lt iplic at ions o f n -b i t i n te g e rs (n a me l y A r B r (A , - A d(B oB r), and AsBs), as well as a number of additions and shifts. If we let M(n) denote the number of bit operations needed to multiply two n -bit integers, we find from (t.0) t t r at

(r.z)

M (2n) < ru h) + Cn.

where C is a constant, since each of the three multiplications of n -bit integers takes M (n) bit operations,while the number of additions and shifts neededto compute a'b via (t.0) does not depend on n, and each of these operations takes O (n) bit operations. From (t.Z), using mathematical induction, we can show that

a(zk) ( c(3k -2k),

(1.8)

where c is the maximum of the quantities M Q) and C (the constant in (t.Z)). To carry out the induction argument, we first note that with k: l, we have MQ) ( c(3t -2t) : c, sincec is the maximum of M(2) and C. As the induction hypothesis,we assumethat MQk)

( c (3 ft - 2 k).

Then, us ing ( 1. 7), w e h a v e M (z k + t) ( ( ( (

3 u (z k ) 3c (lt c a k + t_ c ( 3 f t + l-

+ czk 2k) + c2k c . 3 . 2 k* c 2 k zk+t).

This establishesthat (1.8) is valid for all positive integers ft. Using inequality (t.8), we can prove the following theorem. Theorem 1.4. Multiplication of two n-bit integers can be performed using O(nto9'3) bit operations. (Note: log23 is approximately 1.585, which is

1.4 ComputerOperationswith Integers

4l

considerably less than the exponent 2 that occurs in the estimate of the number of bit operations needed for the conventional multiplication algorithm.) Proof . From (t.8) we have M h) : M (ztos'n)( lzlttloerl+t; < , (3ttot'nl+t_rltoe'nl+t; ( 3 c .rl l o g Irn( 3 c .3 l o sr,:3rnto93 (since 3lo8'n: ,'ot"). Hence, Mh)

:

glnroe'3l. tr

We now state, without proof, two pertinent theorems. Proofs may be found in Knuth [50] or Kronsjii tSgl. Theorem 1.5. Given a positive number e ) 0, there is an algorithm for multiplication of two n-bit integersusing O(nr+') bit operations. Note that Theorem 1.4 is a specialcaseof Theorem 1.5 with e : log23- l, which is approximately0.585. Theorem 1.6. There is an algorithm to multiply two n-bit integers using O(n log2n log2log2n)bit operations. Since log2n and log2log2nare much smaller than n' for large numbers n, Theorem 1.6 is an improvement over Theorem 1.5. Although we know that M h) : O (n log2n log2log2n), for simplicity we will use the obvious fact that M fu) : O (n2) in our subsequentdiscussions. The conventionalalgorithm described above performs a division of a 2n-bit integer by an n-bit integer with O(n2) bit operations. However, the number of bit operations needed for integer division can be related to the number of bit operations needed for integer multiplication. We state the following theorem, which is basedon an algorithm which is discussedin Knuth 1561. Theorem 1.7. There is an algorithm to find the quotient q:Ia/bl, when the 2n-bit integer a is divided by the integer b having no more than n bits, using O(M Q)) bit operations, where M fu) is the number of bit operationsneededto multiply two n-bit integers.

42

The Integers

1.4 Problems l.

Add (l0llll0ll)2 and(ttootll0ll)2.

2 . S u b t r a c t( t o t t l 0 l 0 l ) 2 f r o m ( 1 1 0 1 1 0 1 1 0 0 ) 2 . 3.

Multiply (t t rOr), and (l10001)2.

4.

F i n d t h e q u o t i e n ta n d r e m a i n d e rw h e n ( t t o t o o n l ) 2 i s d i v i d e db y ( 1 1 0 1 ) 2 .

5.

A d d ( A B A B ) 1 6a n d ( B A B A ) r c .

6.

Subtract (CAFE)16 from (rnno)ru.

7.

Multiply

8.

Find the quotient and remainder when Gneono),u

9.

Explain how to add, subtract, and multiply the integers 18235187and 22135674 on a computer with word size 1000.

(FACE) 16and (BAD)rc. is divided by (enn.n)ru.

10. Write algorithms for the basic operations with integers in base (-2) (see problem 6 of Section 1.3).

notation

11. Give an algorithm for adding and an algorithm for subtracting Cantor expansions (see problem l4 of Section 1.3). 12. Show that if f 1 and f 2 are O(St) and O(g2), respectively,and c1 and c2 are constants,then c;f1 * ,zf z is O(g1 * g). 13. Show that if f is O(g), then fr

it OQk) for all positiveintegersk.

14. Show that a function f is O(log2n) if and only if f is O(log,n) wheneverr ) (Hint: Recall that logon/log6n: logo6.)

l.

15. Show that the base b expansionof a positive integer n has llog6nl+t digits. 16. Analyzing the algorithms for subtraction and addition, show that with n-bit integers these operationsrequire O h) bit operations. 17. Show that to multiply an n-bit and an m-bit integer in the conventional manner requires OQm) bit operations. 18. Estimate the number of bit operationsneededto find l+2+ il

by performing all the additions.

b)

by using the identity l+2* shifting.

I n:

nh+l)/2,

* n

and multiplying and

43

1.4 Computer Operations with Integers

19. Give an estimate for the number of bit operationsneededto find

b)

a) n'.

["1 |.o,|

20. Give an estimate of the number of bit operations needed to find the binary expansionof an integer from its decimal expansion'

21.

22.

il

Show there is an identity analogousto (1.6) for decimal expansions.

b)

Using part (a), multiply 73 and 87 performing only three multiplications of one-digit integers,plus shifts and additions.

c)

Using part (a), reduce the multiplication of 4216 and 2733 to three multiplications of two-digit integers, plus shifts and additions, and then using part (a) again, reduce each of the multiplications of two-digit integers into three multiplications of one-digit integers, plus shifts and additions. Complete the multiplication using only nine multiplications of one-digit integers, and shifts and additions.

il

lf A and B are nxn

matrices, with entries aii and bii for I ( i ( n,

I ( f ( n, then AB

is the nxn matrix with entries cii :

Show that n3 multiplications of integers are used to find AB its definition. b)

2

ai*b*j.

dir:;;ly from

Show it is possible to multiply two 2x2 matrices using only seven multiplications of integers by using the identity

o,rf lb,, D'tl

lo,, l a z r o,,)

l"r r b r r*

II lx

lr,, t,,)

anbzt

( a r r l a 1 2 - a 2 1 - a 2 2 )b 2 2

* (as-a2)(bzz-bn) -

a 2 2 ( br - b z r - b e * b 2 2 )

w h e r ex : c)

* a22)(bn-b,+ , )l x I (a21 |

x * ( a n - a z t ) ( b r r - b r+r ) I ( a 2 1* a 2 ) ( b r z - b ' , - )

a r r b r ,- ( a t t - c t 2 r - a 2 ) ( b n -

bp*

|

b2).

Using an inductive argument, and splitting 2nx2n matrices into four nxn matrices, show that it is possibleto multiply two 2k x2k matrices using only 7ft multiplications, and less than 7ft+r additions.

44

The Integers

d)

23.

Conclude from part (c) that two nxn matrices can be multiplied using O(nt"c7) bit operations when all entries of the matrices have less than c bits, where c is a constant.

A dozen equals 12 and a gross equals 122. Using base 12, or duodecimal. arithmetic answer the following questions. il

If 3 gross, 7 dozen, and 4 eggs are removed from a total of l l gross and 3 dozen eggs, how many eggs are left?

b)

If 5 truckloads of 2 gross, 3 dozen, and 7 eggs each are delivered to the supermarket, how many eggs were delivered?

c)

If I I gross, I 0 dozen and 6 eggs are divided in 3 groups of equal size, how many eggs are in each group?

24.

A well-known rule used to find the square of an integer with decimal expansion (an-1...apJro with final digit ao:5 is to find the decimal expansionof the product (anan-1...a)rcl(anan-r...ar)ro* ll and append this with the digits (25)ro. For instance, we see that the decimal expansion of (tOS)2 begins with 16'17 :272, so that (165)2 :27225. Show that the rule just describedis valid.

25.

In this problem, we generalizethe rule given in problem 24 to find the squaresof integers with final base 28 digit 8, where I is a positive integer. Show that the base 28 expansion of the integer (ana,-1...afl0)z,astarts with the digits of the base 28 expansionof the integer (anana...aflo)zn l(anan-1...ap0)zn* ll and ends with the digits Bl2 and 0 when B is even, and the digits G-l)12 and.B when I is odd.

1.4 Computer Projects Write programs to do the following: l.

Perform addition with arbitrarily large integers.

2.

Perform subtraction with arbitrarily large integers.

3.

Multiply two arbitrarily large integers using the conventionalalgorithm.

4.

Multiply two arbitrarily laige integers using the identity (1.6).

5.

Divide arbitrarily large integers, finding the quotient and remainder.

6.

Multiply two n xn matrices using the algorithm discussedin problem 22.

45

1.5 Prime Numbers

1.5 Prime Numbers The positive integer I has just one positive divisor. Every other positive integer has at least two positive divisors, becauseit is divisible by I and by itself. Integers with exactly two positive divisors are of great importance in number theory; they are called primes. Definition. A prime is a positive integer greater than I that is divisible by no positive integers other than I and itself. Example. The integers2,3,5,13,101and 163 are primes. Definition. A positive integer which is not prime, and which is not equal to l, is called composite. Example. The integers 4:2'2,8:4'2, l 0 0 l : 7' ll' 13 ar e co m p o s i te .

3 3 : 3 ' 1 1 ,1 l l : 3 ' 3 7 , a n d

The primes are the building blocks of the integers. Later, we will show that every positive integer can be written uniquely as the product of primes. Here, we briefly discuss the distribution of primes and mention some conjecturesabout primes. We start by showing that there are infinitely many primes. The following lemma is needed. Lemma 1.1. Every positive integer greater than one has a prime divisor. Proof . We prove the lemma by contradiction; we assume that there is a positive integer having no prime divisors. Then, since the set of positive integers with no prime divisors is non-empty, the well-ordering property tells us that there is a least positive integer n with no prime divisors. Since n has no prime divisors and n divides n, we see that n is not prime. Hence, we can write n:ab with I 1 a 1 n and | < b 1 n. Becausea 1 n. a must have a prime divisor. By Proposition 1.3, any divisor of a is also a divisor of n, so that n must have a prime divisor, contradicting the fact that n has no prime divisors. We can conclude that every positive integer has at least one prime divisor. tr We now show that the number of primes is infinite. Theorem 1.8. There are infinitely many primes.

46

The Integers

Proof . Consider the integer Qn: nt t l,

n 2 l.

Lemma 1.1. tells us that Q, has at least one prime divisor, which we denote by gr. Thus, q, must be larger than n; for if 4, ( n, it would follow that Qn I n!, and then, by Propositionl.!, Q, | (er-rr) : l, which is impossible. Since we have found u priJ.''lur*r, tt* there must be infinitely many primes. tr

r, for every positive integer n,

Later on we will be interested in finding, and using, extremely large primes. We will be concerned throughout this book with the problem of determining whether a given integer is prime. We first deal with this question by showing that by trial divisions of n by primes not exceeding the square root of n, we can find out whether n is prime. Thedrem 1.9. If n is a composite integer, then n has a prime factor not exceeding..1n. Proof . Since n is composite, we can write n : ab, where a and b are ( D < n. we must have a 4 r/i, since otherwise integers with | 1a : n. Now, by Lemma I.l, a must have a b 7 a > ,/; and ab > '/i.,/i prime divisor, which by Proposition 1.3 is also a divisor of a and which is clearly less than or equal to ,/i . D We can use Theorem 1.9 to find all the primes less than or equal to a given positive integer n. This procedure is called the steve of Eratosthenes. We illustrate its use in Figure 1.2 by finding all primes less than 100. We first note that every composite integer less than 100 must have a prime factor less than J00-: 10. Since the only primes lessthan l0 are 2,3,4, and 7, we only need to check each integer less than 100 for divisibility by these primes. We first cross out, below by a horizontal slash -, all multiples of 2. Next we cross out with a slash / those integers remaining that are multiples of 3. Then all multiples of 5 that remain are crossedout, below by a backslash\. Finally, all multiples of 7 that are left are crossedout, below with a vertical slash l. ntt remaining integers (other than l) must be prime.

41

1.5 Prime Numbers

5

t23+ ++ ll

13

l+-

X

2{-*23+g-. 3l+2Ii+ 4r+43

1+

>{+*s3*r4*tr# 61 7t+73.+ y{ 83 t.> I

tlt

+>

yr

2
h -l )l o g l s a

> (C I-l ) /5.

Consequently, n-l(S'logleb.

63

2 .2 T he E uc lidean Al g o ri th m

Let b have k decimal {igits, so that b < 10ftand loglsb < k. Hence, we see that n - I < 5k and since /c is an integer, we can conclude that n < 5k. This establishesLam6's theorem. tr The following result is a consequence of Lam6's theorem. Corollary 2.1. The number of bit operations needed to find the greatest

integers of twopositive a and, yy divisor common

ir;;i.:f$;:ri?',

Proof. We know from Lam6's theorem that O Qogra) divisions, each taking O(log2a)2) bit operations,are neededto find fu, b). Hence, by Proposition 1.7, (a, b) may be found using a total of O((log2a)3) bit operations. D The Euclideanalgorithm can be used to expressthe greatestcommon divisor of two integers as a linear combination of these integers. We illustrate this by expressing(252, 198) : l8 as a linear combinationof 252and 198. Referring to the stepsof the Euclideanalgorithm used to find (252, 198), from the next to the last step, we seethat 18:54-l'36. From the secondto the last step, it follows that

36:198-3'54, which implies that

1 8: 5 4 - t . ( 1 9 8 - 3 . 5 4: ) 4 . 5 4 - 1 . 1 9 8 . Likewise, from the first stepwe have 54:252 - l'198. so that

l 8 - 4 ( 2 5 2 - 1 . 1 9 8- ) 1 . 1 9 8: 4 . 2 5 2 - 5 . 1 9 8 . This last equationexhibits l8 : (252, 198) as a linear combinationof 252 and l 98. In general,to see how d : (a, b) may be expressedas a linear combination of a and 6, refer to the series of equations that is generated by use of the Euclideanalgorithm. From the penultimateequation,we have rn: (a, b) : Th i s e x pr es s es b, b) ' a s

r n - 2 - r n - r Q n - .r

a l i n e a r c o mb i n a ti o no f rr-2e,fi drr-1. The secondto

GreatestCommonDivisorsand PrimeFactorization

64

the last equation can be used to expressr2-1 &S rn-3 -rn-zen-z . Using this last equation to eliminate rn-1 in the previousexpressionfor (4,6), we find that ln:

ln-3-

fn-24n-2,

so that b, b) : rn-2- (rn4-rn-zQn-z)en-r -- (l + q rn Q n -z )rn - zQn-rrn-3, which expressesb, b) as a linear combinationof rn-2 zfid r,4. We continue working backwards through the steps of the Euclidean algorithm to express G, b) as a linear combinationof each precedingpair of remaindersuntil we havefound (a, b) as a linear combinationof to: a and 11- b. Specifically, if we have found at a particular stagethat G,b):sriltrit, then, since ti:

ti_2- ri_tQi_r,

we have b,b) : s (ri-z*ri-g1-r) * tr1-r : Q-sqt-)ri-r * sri-2. This showshow to move up through the equationsthat are generatedby the Euclidean algorithm so that, at each step, the greatestcommon divisor of a and b may be expressedas a linear combination of a and b. This method for expressingG, b) as a linear combinationof a and b is somewhatinconvenientfor calculation, becauseit is necessaryto work out the steps of the Euclidean algorithm, save all these steps, and then proceed backwardsthrough the steps to write G,b) as a linear combinationof each successivepair of remainders. There is another method for finding b,b) which requires working through the steps of the Euclidean algorithm only once. The following theoremgivesthis method. Theorem 2.4. Let a and b be positive integers. Then fu,b):sna+tnb, defined for n:0,1,2,..., where,sn andtn are the nth terms of the sequences recursivelyby

65

2.2 The Euclidean Algorithm

SO: l, /0:0, sl :0, /l : l,

and si : Si*z- ?i-tsi-t, tj : tj-z - Q1-zt1-t for 7 :2,3, ..., fl, where the q;'s are the quotientsin the divisionsof the Euclideanalgorithm when it is usedto find G,b). Proof. We will prove that ri : sia + tjb

Q.D

for 7 : 0, I ,...,fl. Since G,b) : r, once we have established(2.2), we will know that G,b):sna+tnb. We prove (2.2) using the secondprinciple of mathematicalinduction. For :0 , we hav e a : r0 : l ' a * 0 ' b : s s a* ts b . H ence, Q.D i s val i d for l j : 0 . L i k e w i s eb, : r r : 0 ' a + l ' b : s l c + t f t , s o t h a t Q . D i s v a l i d f o r j : l. Now, assumethat ri:Sia+tjb for 7 : 1,2,..., k-1. Then, from the kth step of the Euclideanalgorithm,we have tk : rk-2 - r*_lQt-l . Using the inductionhypothesis,we find that r1 : (s1-2a*tp-2b) - (s1raa*t1r-1b) Q*-r : (s 1 -2 -s * -tq * -)a * Q p 2 -t* -rq* -)b :Ska+tkb. This finishesthe proof. tr The following example illustrates the use of this algorithm for expressing (a,b) as a linear combinationof a and b. Example. Let a :252 and D : 198. Then

GreatestCommonDivisorsand prime Factorization

66

so: l, sl :0, J2:S0-sql:l0'l:1, J 3 : S t - S Z Q z : 0- l ' 3 : - 3 , s 4 : s 2- s t Q t : I - ( - l ) ' t : 4 ,

lo:0, Ir : 1, tZ:tO-ttQt:01 . 1: - 1 , t 3 : t t - 1 Z Q Z :1 - ( - l ) 3 : 4 , t q : t z - t t Q z : - l - 4 . 1: - 5 .

S i n c e1 4 : 1 8 : ( 2 5 2 , 1 9 8 )a n d 1 4 : s 4 o+ t 4 b , w e h a v e 1 8 - ( 2 5 2 ,1 9 8 ): 4 . 2 5 2- 5 . 1 9 8. It should be noted that the greatestcommon divisor of two integersmay be expressedin an infinite number of different ways as a linear combination of theseintegers. To seethis, let d : (a,b) and let d : so I tb be one way to write d as a linear combination of a and b, guaranteed to exist by the previousdiscussion.Then d : (s - k(b/d))a + Q - kb/d))b for all integersk. Example. With a :252 and b : 198, lB: (-S - l4k)198 whcneverk is an integer.

(252, 198) :

(+ - t Ik)252 +

2.2 Problems l.

Use the Euclidean algorithm to find the following greatest common divisors

il (45,75)

c) (ooo, r+r+)

b) 002,22D

d) (2078S,44350).

2.

For each pair of integers in problem l, expressthe greatest common divisor of the integers as a linear combination of these integers.

3.

For each of the following sets of integers, expresstheir greatest common divisor as a linear combination of these integers il

6, 10,l5

b)

7 0 , 9 8 ,1 0 5

c)

2 8 0 ,3 3 0 , 4 0 5 , 4 9 0 .

4. The greatest common divisor of two integers can be found using only subtractions, parity checks, and shifts of binary expansions,without using any divisions. The algorithm proceedsrecursively using the following reduction

67

2.2 The Euclidean Algorithm

I,

G.b):

if a:b

)2 k l L ,b/2 ) if a and 6 are even

l{o/z,t) -D,b)

if a is even and b is odd if a and b are odd.

[(a a)

Find (2106,8318) usingthis algorithm.

b)

Show that this algorithm always produces the greatest common divisor of a pair of positiveintegers.

5. In problem 14 of Section 1.2, a modified division algorithm is given which says that if a and 6 > 0 are integers,then there exist unique integersq,r, and e such that a : bq * er, where e - tl,r ) 0, and -blz < er { bl2. We can set up an algorithm, analogous to the Euclidean algorithm, based on this modified division algorithm, called the least-remainder algorithm. It works as follows. Let rs: a and rr: b, where a ) b 7 0. Using the modified division algorithm repeatedly,obtain the greatest common divisor of a and b as the last nonzeroremainder rn in the sequenceof divisions ro :

rn-Z : fn-l :

rtQr * e2r2,

-rtlz

1 e2r2 4 ,tlz

ln-tQn-t I enrn, 7n4n'

-rn-tl2

I

enrn 4, rn-tl2

a)

Use the least-remainderalgorithm to find (384, 226).

b)

Show that the least-remainder algorithm always produces the greatest common divisorof two integers.

c)

Show that the least-remainderalgorithm is always faster, or as fast, as the Euclidean algorithm.

d)

Find a sequenceof integers v6, V1,v2,... such that the least-remainder algorithm takes exactly n divisionsto find (vn*,, vn+z).

e)

Show that the number of divisions needed to find the greatest common divisor of two positive integers using the least-remainderalgorithm is less than 8/3 times the number of digits in the smaller of the two numbers,plus

413. 6 . Let m and n be positive integers and let a be an integer greater than one. Show that (a^-1, an-l) - a(^' n)- l.

7 . In this problem, we discuss the game of Euclid. Two players begin with a pair of positive integers and take turns making movesof the following type. A player can move from the pair of positiveintegers{x,y} with x 2 y, to any of the pairs where / is a positive integer and x-ty 2 0. A winning move [x-ty,yl,

68

GreatestCommonDivisorsand PrimeFactorization

consistsof moving to a pair with one element equal to 0. a)

Show that every sequence of moves starting with the pair {a, bl must eventuallyend with the pair {0, (a, b)}.

b)

show that in a game beginning with the pair {a, b},1he first player may play a winning strategy if a - 6 or if a 7 b0+ Jil/z; otherwisethe second player mgr play a winning strategy. (Hint: First show that if y < x ( y(t+VS)/Z then thge is a unique move from l*,Ol that goes to a pair lt, r| with y > ze+Jil/z.)

In problems8 to 16, un refers to the nth Fibonaccinumber. 8. Show that if n is a positiveinteger,then rz1l u2 I

I ttr:

9. Show that if n is a positiveinteger, then unapn-r - u] :

GD'.

10. Show that if n is a pqsitive integer, then un: o : (t+.,6) /2 andp : Q-'./-il/2. ll.

un+z- l.

(c'n-0\/'..fs, where

Show that if m and n arepositiveintegerssuch that m I n, then u^ | un.

12. Show that if m and n are positiveintegers,then (u^, un) : u(m,il. 13. Show that un is even if and only if 3 | n.

(t 'l

t4. Letu: li i,. a)

Show that Un :

Irn*, Itn I lu,

b)

u^_r)

.

Prove the result of problem 9 by consideringthe determinant of Un.

15. We define the generalized Fibonacci numbers recursively by the equations gr- a, E2: b, and gn - gn-t* gr-zfor n 2 3. Showthat gn: oun-2* bun-1 for n )- 3. 16. The Fibonacci numbers originated in the solution of the following problem. Supposethat on January I a pair of baby rabbits was left on an island. These rabbits take two months to mature, and on March I they produce another pair of rabbits. They continually produce a new pair of rabbits the first of every succeeding month. Each newborn pair takes two months to mature, and producesa new pair on the first day of the third month of its life, and on the first day of every succeedingmonth. Show that the number of pairs of rabbits alive after n months is precisely the Fibonacci number un, assuming that no rabbits ever die. 17. Show that every positive integer can be written as the sum of distinct Fibonacci numbers.

2.3 The Fundamental Theorem of Arithmetic

69

2.2 Computer Projects Write programs to do the following: l.

Find the greatestcommondivisor of two integersusing the Euclideanalgorithm.

2.

Find the greatest common divisor of two integers using the modified Euclidean algorithm given in problem 5.

3.

Find the greatest common divisor of two integers using no divisions (see problem

0. 4.

Find the greatest common divisor of a set of more than two integers.

5.

Express the greatest common divisor of two integers as a linear combination of theseintegers.

6.

Express the greatest common divisor of a set of more than two integers as a linear combination of these integers.

7.

List the beginning terms of the Fibonacci sequence.

8.

Play the game of Euclid describedin problem 7.

2.3 The FundamentalTheoremof Arithmetic The fundamental theorem of arithmetic is an important result that shows that the primes are the building blocks of the integers. Here is what the theoremsays. The Fundamental Theorem of Arithmetic. Every positive integer can be written uniquely as a product of primes,with the prime factors in the product written in order of nondecreasing size. Example. The factorizationsof somepositive integersare given by 2 4 0: 2 . 2 . 2 . 2 . 3:. 5 2 4 . 3 . 5 , 2 8: 9 1 7 . 1 7: 1 i 2 . 1 0 0 1: 7 . 1 1 . 1 3 . Note that it is convenient to combine all the factors of a particular prime into a power of this prime, such as in the previous example. There, for the factorization of 240, all the fdctors of 2 were combined to form 24. Factorizationsof integers in which the factors of primes are combined to form powersare called prime-power factorizations. To prove the fundamental theorem of arithmetic, we need the following lemma concerningdivisibility. Lemma 2.3. lf a, b, and c are positive integers such that (a, b) : I and

70

GreatestCommonDivisorsand PrimeFactorization

a I bc , t hen a I c , Proof. Since G,b): 1, there are integersx and y such that ax * by : y. Multiplying both sides of this equation by c, we have acx * bcy: c. By Proposition1.4, a divides acx * 6cy, since this is a linear combinationof a and bc, both of which are divisibleby a. Hencea I c. a The following corollary of this lemma is useful. Corollary 2.2. If p dividasap2 an wherep is a prime and c r, a2,...,on are positive integers, then there is an integer i with I < t ( n such that p dividesa;. Proof. We prove this result by induction. The case where n : I is trivial. Assume that the result is true for n. Consider a product of n * t, integers, ar az aral that is divisibleby the prime p. Sincep I ar az on*t: (a1a2 an)ana1,we know from Lemma 2.3 that p I ar az en or p I ar+r. Now, it p I ar az a' from the induction hypothesisthere is an integer i with 1 < t ( n such Ihat p I ai. Consequentlyp I a; for some i w i t h l < t < n * 1 . T h i s e s t a b l i s h e s t h e r e s ut rl t . We begin the proof of the fundamental theorem of arithmetic. First, we show that every positive integer can be written as the product of primes in at least one way. We use proof by contradiction. Let us assume that some positive integer cannot be written as the product of primes. Let n be the smallest such integer (such an integer must exist from the well-ordering property). lf n is prime, it is obviously the product of a set of primes, namely t h e o n e p r i m e n .S o n m u s t b e c o m p o s i t Le e. t n : a b , w i t h | 1 a ( n a n d | 1 b I n. But since a and b are smaller than n they must be the product of primes. Then, since n : ab, we conclude that n is also a product of primes. This contradictionshowsthat every positiveinteger can be written as the product of primes. We now finish the proof of the fundmental theorem of arithmetic by showing that the factorization is unique. Supposethat there is a positive interger that has more than one prime factorization. Then, from the well-ordering property, we know there is a least integer n that has at least two different factorizationsinto primes: fl:PtPz

Ps:QtQz

Qt,

w h e r ep t , p 2 , . . . , p s , Q t , . . . , 4atr e a l l p r i m e s ,w i t h p r ( p z ( (q'. {r(42(

( p, and

71

2.3 The Fundamental Theorem of Arithmetic

We will s how t ha t p t: Qr,p 2 : Q 2 ,...,a n d c o nti nueto show that each of p's and q's are equal, and that the number of prime factors in the successive the two factorizations must agree, that is s : /. To show that pr: Qr, assumethat pr * qy Then, either pr ) 4r or pr 1 Qr By interchanging we can assumethat pr ( qr. Hence,pr 1q; for the variables,if necessary, i : 1, 2, . . . , ts inc e41 i s th e s m a l l e sot f th e q ' s . H e nce,pr tr qi for al l i . B ut, from Corollary 2.2, we see that pr I qflz et : tt. This is a pr : Qr contradiction. Hence, we can conclude that and p s : QzQ t n /p r: pz pt n l p l i s i nteger smal l er than S i n c e an Qt. n, and since n is the smallest positive integer with more than one prime factorization,nfpl con be written as a product of primes in exactly one way. Hence, each pi is equal to the correspondingq;, and s : /. This proves the uniquenessof the prime factorization of positive integers. tr The prime factorization of an integer is often useful. As an example, let us find all the divisorsof an integer from its prime factorization. Example. The positivedivisorsof 120 : 233'5 are thosepositiveintegerswith prime power factorizationscontaining only the primes 2,3, and 5, to powers lessthan or equal to 3, 1, and l, respectively.Thesedivisorsare

I 2 22: 4 23:8

3 2 ' 3: 6 22.3: 12 z3-3: 24

5 2 ' 5: 1 0 22.5: 20 23.5: 40

3'5:15 2 ' 3 ' 5: 3 0 223.5: 6o : l2o . 23.3.s

Another way in which we can use prime factorizations is to find greatest common divisors. For instance,supposewe wish to find the greatest common divisor of 720 : 2432'5and 2100 : 223'52'7. To be a commondivisor of both 720 and 2100, a positiveinteger can contain only the primes 2, 3, and 5 in its prime-power factorization, and the power to which one of these primes appears cannot be larger than either of the powersof that prime in the factorizations of 720 and 2100. Consequently,to be a common divisor of 720 and 2100, a positive integer can contain only the primes 2,3, and 5 to powers no larger than2, l, and l, respectively.Therefore,the greatestcommon divisor of 720 a n d 2100is 22. 3. 5: 6 0 . To describe, in general, how prime factorizations can be used to find greatestcommondivsors,let min(a, D) denotethe smaller or minimum, of the two numbers d and 6. Now let the prime factorizationsof a and b be

o : pi,pi2 .. . p:., b : p'r,plz.. . p:,, where each exponent is a nonnegativeinteger and where all primes occurring

72

GreatestCommonDivisorsand PrimeFactorization

in the prime factorizationsof c and of b are included in both products, perhapswith zero exponents. We note that fu,b):

pl'"k"0,)plinb,'b,

p:'n(oro,) ,

sincefor eachprimepi, a and b shareexactlymin(a;,6;) factorsof p;. Prime factorizationscan also be used to find the smallestinteger that is a multiple of two positive integers. The problem of finding this integer arises when fractions are added. Definition. The least common multiple of two positive integersa and D is the smallestpositiveinteger that is divisibleby a and b. The leastcommonmultiple of a and b is denotedby Io, bl. Example. We have the following least common multiples: ll5,2l l: lZ q, X l : 72, l Z , Z 0 l : 2 A ,a n d [7 , l l l : 7 7.

105,

Once the prime factorizations of a and b are known, it is easy to find p l r. a n d ,b : p i ,pur2 .. . pun,w herept,pz,...,pn I a, bl. I f a : p i ,p i , are the primes occurring in the prime-powerfactorizationsof a and b, then for an integer to be divisible by both c and D, it is necessarythat in the factorization of the integer, eachp; occurs with a power at least as large as ai and bi. Hence, [a,b], the smallestpositiveinteger divisible by both a and b is la,bl:

pl

*Grb,) Omaxb,'b,)

pf

*Gru')

where max(x, /) denotesthe larger, or maximum, of x andy. Finding the prime factorization of large integers is time-consuming. Therefore, we would prefer a method for finding the least common multiple of two integers without using the prime factorizations of these integers. We will show that we can find the least common multiple of two positiveintegersonce we know the greatest common divisor of these integers. The latter can be found via the Euclideanalgorithm. First, we prove the following lemma. Iemma 2,4. If x

and y are real numbers, then max(x,y) + min(x,y)

:x+y. then min(x,y):y and max(x,!):x, so that P r o o f .I f x ) y , and m a x ( x , y ) +m i n ( x , y ) : x * y . If x 0 I is dropped, is the conclusionthat a = b (mod rn ). If the condition (a,m): a = b (mod z) still valid? 16. Show that if n is a positive integer, then +(n-l)

il

t+2+3+

b)

13+23+33+

+

=0(modn).

(n-l)3=o(modn).

17. For which positive integers n is it true that 1 2+ 2 2 + 3 2 +

* ( n - l ) 2 = o ( m o dn ) ?

18. Give a complete system of residuesmodulo l3 consistingentirely of odd integers. 19. Show that if n = 3 (mod 4), then n cannot be the sum of the squares of two integers. 20.

il

Show that if p is prime, then the only solutions of the congruence x 2 = x ( m o d p ) a r e t h o s ei n t e g e r sx w i t h x = 0 o r I ( m o d p ) .

100

Congruences

b) 21.

Show that if p is prime and ft is a positive integer, then the only solutionsof x2 =x (mod pk) arethoseintegersx such that x E 0 or I (modpe).

Find the least positive residuesmodulo 47 of a)

232

b)

c)

247

22w

2 2 . Let

t/t1,t/t2,...,n\r be pairwise relatively prime positive integers. M : mifiz' ' ' mp and Mj : M/mi for; - 1,2,...,k. Show that M(tr*

M2a2*

Let

* Mpap

runs through a complete system of residues modulo M when a1,a2,...,a1,run through complete systemsof residuesmodulo rn1,nt2,...,r/t1, respectively.

2 3 . Explain how to find the sum z * v from the least positive residue of u * v modulo m, where u and. v are positive integers less than z . (Hint: Assume that u ( v and consider separately the cases where the least positive residue of u I v is less than a, and where it is greater than v.)

24. on a computer with word size w, multiplicertion modulo n, where n I w f2, can

be performed as outlined. Let T:IJn + %1, and t : T2 - n. For each computation, show that all the required computer arithmetic can be done without exceedingthe word size. (This method was describedby Head t67]). a)

Show that lr | < r.

b)

Show that if x and y are nonnegativeintegers less than n, then x:aT*b,

y:cT*d

where a,b,c, and d are integers such that 0 ( a ( 0 ( c < T, and 0 < d < T. c)

Letz = ad * bc (mod n), with 0 ( z ( z. Show that

d)

Let ac:eT*f where e 0 | is an integer,then the equationrh) solutions. 8.

g.

: ft has infinitely many

Which positive integers have exactly a)

two positive divisors

b)

three positive divisors

c)

four positive divisors?

What is the product of the positive divisors of a positive integer n ?

10. Let o1,h) denote the sum of the kth powers of the divisors of n, so that o1,h) : 2 dk. Note that o1h) : sfu). dln

a)

Find or(4), or(6) and o{12).

b)

Give a formula for o1(p), wherep is prime'

c)

Give a formula for o1(po), wherep is prime, and a is a positiveinteger.

d)

Show that the function op is multiplicative'

e)

Using parts (c) and (d), find a formula for o;(n), where n has prime-power factorizationn : pi'pi' . . . p:;.

11. Find all positiveintegersn such that d(n) + oQ):2n. 12. Show that no two positive integers have the same product of divisors. 13. Show that the number of pairs of positiveintegerswith least common multiple equal to the positive integer n is r(nz). 14. Let n be a positive integer. Define the sequence of integers fl1,tr2,rt3,...b! n 1 : r ( n ) a n d n 1 . , 1: r ( n * ) f o r f t : 1 , 2 , 3 , . . . . S h o w t h a t t h e r e i s a p o s i t i v e integer r such that 2 : f,r : flr1t : rlr+2: 15. Show that a positiveinteger n is compositeif and only if o(n) > n + ,/i.

180

MultiplicativeFunctions

16. Show that if n is a positiveinteger then r(n)z : )r(d)3 dln

6.2 Computer Projects Write programs to do the following: l.

Find the number of divisorsof a positive integer.

2.

Find the sum of the divisors of a positive integer.

3.

Find the integer r defined in problem 14.

6.3 Perfect Numbersand MersennePrimes Becauseof certain mystical beliefs, the ancient Greeks were interested in those integers that are equal to the sum of all their proper positive divisors. Theseintegersare called perfect numbers. Definition. If n is a positive integer and o(n) : 2n, then n is called a perfect number. E x a m p l e . S i n c eo ( 6 ) : l + 2 + 3 + 6 : 1 2 , w e s e et h a t 6 i s p e r f e c t . w e a l s on o t et h a t o ( 2 8 ) : 1 + 2 + 4 + 7 +14*28:56. sothat28 is another perfect number. The ancient Greeks knew how to find all even perfect numbers. The following theorem tells us which even positive integersare perfect. Theorem 6.9. The positiveinteger n is an even perfect number if and only if n :2m-r(2^-l) where m is a positiveinteger such that 2^-l

is prime.

Proof. First, we show that if n:2m-r(2^-l) where 2^-l is prime, then n is perfect. We note that sincezn-l is odd, we have (2m-r,2m-l) : 1. Since o is a multiplicative function, we seethat o (n ) - o (2 ^ -t)o (2 ^-l )

.

L e m m a 6 . 1 t e l l su s t h a t o ( 2 ^ - r ) : 2 ^ - l and o(2^-l):2^, assumingthat 2m-l is prime. Consequently,

s i n c ew e a r e

181

6.3 PerfectNumbersand MersennePrimes

o(n) : Q^-l)2^

:2n

,

demonstratingthat n is a perfect number. To show that the converseis truen let n be an even perfect number. Write : 1, we n :2'l wheres and t arepositiveintegersand f is odd. Since (2t,t) seefrom Lemma 6.1 that

(6.1)

o(n) : o(2':) : o(2')o(t) : (2'+t-t)o(l)

Since n is perfect, we have o (n ) : 2 n : 2 s + r1

G'D

Combining (6.1) and (6.2) showsthat (2 ' + r-1 )o(i

(6 .3 )

: 2 s + t1

Si n ce( 2s + r , 2s + t - l) : l , fro m L e mma 2 .3 w e s e eth a t 2' + 1 l o(r). Therefore, there is an integerq such that o(t) - 2'+rQ. Inserting this expressionfor o(t) into (6.3) tells us that (2 s + r_ l )2 s * rq- 2 ' * rt , and, therefore, (2'+t-l)q

(6.4)

: 1.

Hence,q I t and q # t. When we replace / by the expressionon the left-hand side of (6.4), we find that (6.5)

t +q:

( 2 s + t - t ) q+ q : 2 ' + r q : o Q ) .

We will show that q : 1. Note that if q * l, then there are at least three distinct positive divisors of t , namely 1, q, and t . This implies that oQ) 2 t + q -| 1, which contradicts(6.5). Hence,4: I and, from (6.4), we concludethat / :2s+l-1. Also, from (6.5), we seethat oQ): t + l, so that t must be prime, since its only positive divisors are I and t. Therefore, n :2 t ( 2r + l- 1) , where2 s + l -1 i s p ri me . tr From Theorem 6.9 we see that to find even perfect numbers, we must find primes of the form 2t-1. In our searchfor primes of this form, we first show that the exponentru must be Prime. Theorem 6.10. If la is a positiveinteger and2^-l

is prime, then m must be

182

MultiplicativeFunctions

pnme. Proof. Assume that m is not prime, so that m : ab where | 1 a 1 m and, | < b 1m. Then 2m-l

: 2ab-, - (Zo-l) 12a(b-D a2a(b-Dq...q1o+l) .

Since both factors on the right side of the equationare greater than I, we see that 2m-l is compositeif m is not prime. Therefore,if 2^-l is prime, then nr must also be prime. tr From Theorem6.10 we seethat to searchfor primes of the form 2^-1, we need to consideronly integersm that are prime. Integers of the form 2m-l have been studied in great depth; these integers are named after a French monk of the seventeenth century, Mersenne,who studiedtheseintegers. Definition. If m is a positiveinteger, then M^:2^-I is called the mth Mersennenumber, and, if p is prime and Mp:2p-l is also prime, then M, is called a Mersenneprime. Example. The Mersennenumber M7:27-I is prime, whereasthe Mersenne :2 0 4 7 : 2 3 .8 9i s c o m posi te. num berM n: 2rr-I It is possibleto prove various theoremsthat help decide whether Mersenne numbers are prime. One such theorem will now be given. Related results are found in the problemsof Chapter 9. Theorem 6.11. rf p is an odd prime, then any divisor of the Mersenne number Mp :2p-l is of the form 2kp + I where k is a positiveinteger. - 2p - I. From Fermat's little Proof. Let q be a prime -dividing Mp theorem,we know thatql(ze-t-t). Also, from Lemma 1.2 we know that ll\ (6.6) (T -t, 2c-t-t) : 2$t-D - f. Since q is a common divisor of zp-l and zc-t-L we know that > l . H e n c e , (p ,q -l ): Q p- t , 24- t - l) p , si ncethe onl y other possi bi l i ty, namely (p,q-l) : I, would imply from (6.6) that (Zp-t,2Q-t-l) : l. Hence p | (q-t), and, therefore, there is a positive integer m with q - | : mp. Since q is odd we see that m must be even, so that m : Zk. w h e r e k i s a p o s i t i v e i n t e g e rH . e n c eq , :mp * I - 2kp+1 . tr We can use Theorem6.1I to help decide whether Mersenne numbers are prime. We illustrate this with the following examples.

183

6.3 PerfectNumbersand MersennePrimes

8191 is prime, we only needlook Example. To decidewhetherMB:2r3-l: : 90.504.... Furthermore, from for a prime factor not exceeding lml Theorem6.11, any such prime divisor must be of the form 26k + L The only candidatesfor primesdividinB Mnless than or equal to1fTp are 53 and79. Trial divisioneasilyrules out thesecases,so that M s is prime. i s pri me,w e onl y need Exa m ple. T o dec idew h e th e rM z t:2 2 3 -r:8 3 8 8 6 0 7 prime less than or equal to by a is divisible whether M to determine zt prime this form is 47. of + first l. The form 46k the of 2896.309... ffi: A trial divisionshowsthat 8388607:47'178481, so that M4is composite. Becausethere are specialprimality tests for Mersennenumbers,it has been possibleto determine whether extremely large Mersennenumbers are prime. Following is one such primality test. This test has been used to find the largest known Mersenne primes, which are the largest known primes. The proof of this test may be found in Lenstra [7t] and Sierpifiski[351. The Lucas-LehmerTest. Let p be a prime and let Mo : 2! -l denote the pth Mersennenumber. Define a sequenceof integersrecursivelyby setting tr:4, andfork>2, r * ? rtq -2

(m o d M),

0 ( rr I Mo .

Then, M, is prime if and only if rp-1 - 0 (mod M)

.

We use an exampleto illustrate an applicationof the Lucas-Lehmertest. 4, Exa mple. c ons idert h e Me rs e n n en u m b e rM5 :2 5 - I - 3l ' Then r,: 2 ( m o d 3 1 ) , ( m o d 3 l ) , r + 2 a n d rt4 8 A2 rzz42-2:14 ( m o d3 1 ) . S i n c e r t t 0 ( m o d 3 1 ) , w e c o n c l u d et h a t M 5 : 3 1 i s 8 2- 2 : 0 prime. The Lucas-Lehmer test can be performed quite rapidly as the following corollary states. Corollary 6.1. Let p be prime and let Mp : 2p - | denotethe pth Mersenne number. It is possibleto determine whether Mo is prime using OQ3) bit operations. Proof. To determine whether Mp is prime using the Lucas-Lehmer test O(p2) requiresp - | squaringsmodulo iV* each requiring O((log M)2): bit operations. Hence, the Lucas-Lehmer test requires O Q3) bit operations.tr

184

Multiplicative Functions

Much activity has been directed toward the discoveryof Mersenneprimes, especiallysince each new Mersenne prime discoveredhas become the largest prime known, and for each ngw Mersenne prime, there is a new perfect number. At the presenttime, a total of 29 Mersenneprimes are known and these include all Mersenne primes Me with p ( 62981 and with 75000 < p < 100000. The known Mersenneprimes are listed in Table 6.3.

p

Number of decimal digits in M o

2 3 5 2 7 2 l3 6 + I1 2 t9 1'2 3 l 9a 'zz 68 9l ig 107 zf) q + t27 )q 52r 8 t ) 607 I (, 72 r279 ? 2^ lh 2203 -7s 2281 3 b 32r7 4253 4423 9689 Lbb 5z 994r I 1213 r9937 2r701 23209 44497 86243 r32049 I

I

L

9l Table 6.

I I 2 3 4 6 6 10 19 27 33 39 157 183 386 664 687 969 1281 t332 29r7 2993 3376 6002 6533 6987 I 3395 25962 3975I 5050

Date of Discovery

anclenttrmes ancienttimes ancienttimes ancienttimes Mid 15thcentury 1603 1603 1772 18 8 3 l91l l9l4 t876 t952 t952 1952 1956 1952 t957 1961 1961 I 963 I 963 1963 t97| I 978 r979 1979 1983 I983

f9t re Known Mersenne Primes.

6.3 PerfectNumbersand MersennePrimes

185

Computers were used to find the 17 largest Mersenne primes known. The discovery by high school students of the 25th and 26th Mersenne prime received much publicity, including coverageon the nightly news of a major television network. An interesting account of the search for the 27th Mersenne prime and related historical and computational information may be found in [77]. A report of the discoveryof the 28th Mersenne prime is given in [64]. It has been conjectured but has not been proved, that there are infinitely many Mersenneprimes. We have reduced the study of even perfect numbers to the study of Mersenne primes. We may ask whether there are odd perfect numbers. The answer is still unknown. It is possibleto demonstratethat if they exist, odd perfect numbers must have certain properties (see problems 1l-14, for example). Furthermore, it is known that there are no odd perfect numbers less than 10200,and it has been shown that any odd perfect number must have at least eight different prime factors. A discussionof odd perfect numbers may be found in Guy [17], and information concerningrecent results about odd perfect numbersis given by Hagis [681.

6.3 Problems l.

Find the six smallesteven perfect numbers.

2 . Show that if n is a positive integer greater than l, then the Mersenne number Mn cannot be the power of a positive integer.

3 . If n is a positive integer, then we say that n is deficient if ofu) 1 2n , and we say that n is abundant if oh) or abundant.

4.

) 2n. Every integer is either deficient, perfect,

a)

Find the six smallestabundant positive integers.

b)

Find the smallestodd abundant positive integer.

c)

Show that every prime power is deficient.

d)

Show that any divisor of a deficient or perfect number is deficient.

e)

Show that any multiple of an abundant or perfect number is abundant.

f)

Show that if n -2m-t(2^-l) , where ra is a positive integer such that 2 -l is composite, then n is abundant.

if Two positive integers m and n are called an amicable pair o(m\ : o(n) : m * n. Show that each of the following pairs of integers are amicable pairs

186

MultiplicativeFunctions

a) 220,294 b) 1 1 8 4l ,2 1 0 c) 7975A,98730. 5. a) Showthat if n is a positiveintegerwith n ) 2, suchthat3.2n-t-1,3.2n-1, and32'22n-r-1 are all prime,then2n(3'2'-t-DQ.2'-l) and2n(32.22n't-l) form an amicablepair. b)

Find three amicablepairs using part (a).

6 . An integer n is called k-perfect if o(il: 2-perfect.

kn. Note that a perfect number is

a)

Show that 120 : 23.3.5is 3-perfect.

b)

Show that 30240 : 2s32.5., is 4-perfect.

c)

- 27.34.5.7.n2.17.19 Show that 14182439040 is 5-perfect.

d)

Find all 3-perfectnumbersof the form n -2k.3.p, prime.

e)

Show that if n is 3-perfectand 3 I n, then 3n is 4-perfect.

7 . A positiveinteger n is called superperfectif oGh))

where p is an odd

: Zn.

a)

Show that 16 is superperfect.

b)

Show that if n : 2e where 2q+t-l is prime, then n is superperfect.

c)

Show that every even superperfect number is of the form n : 2q where zq+t-l is prime.

d)

Show that if n : p2 wherep is an odd prime,'then n is not superperfect.

8 . Use Theorem6.ll to determine whether the following Mersenne numbers are pnme a) M7

b) Mn 9'

c) Mn d) Mzs.

Use the Lucas-Lehmer test to determine whether the following Mersenne numbersare prime a) M3 b) M7.

10. a)

b)

c) Mn d Mn.

Show that if n is a positive integer and 2n i L is prime, then either (Hint: Use Fermat's little theorem to Qn+l) | M^ or Qn+D | (a,+D. showthat Mn(Mn+z) = O (mod 2z+l).) Use part (a) to show that Ms and My are composite.

187

6.3 Perfect Numbers and Mersenne Primes

11.

t2.

a)

Show that if n is an odd perfect number, then n : po m2 wherep is an odd I (mod4). p r i m e a n d p7 a z

b)

Use part (a) to show that if n=l(mod4).

n

is an odd perfect number, then

Show that if n - po m2 is an odd perfect number where p is prime, then n=p(mod8). that if n is an odd perfect number, then 3, 5, and 7 are not all divisors of

13. :**

1 4 . Show that if n is an odd perfect number then n has a)

at least three different prime divisors.

b)

at least four different prime divisors.

1 5 . Find all positive integers n such that the product of all divisors of n other than n is exactly n 2. (These integers are multiplicative analoguesof perfect numbers.) recursively by 1 6 . Let n be a positive integer. Define the sequenca fl1,tt2,rt3,..., n 1 : o ( n ) - n a n df l k + r : o Q )

- np fot k - 1,2,3,... tt3 :

a)

Show that if n is perfect,then n : nt : fi2:

b)

Show that if n and m are an amicablepair, then n1 : ftt, ttz- tt, tt3: t/t, is periodicwith period 2. n4: n,... and so on, f.e.,the sequencefl1,tt2,t13,...

c)

of integersgeneratedif n :12496:24'll'71. Find the sequence

It has been conjecturedthat for all is pefiodic. n 1,n2,n3,...

n, the sequence of integers

6.3 ComputerProjects Write programsto do the following: l.

Classifypositiveintegersaccordingto whether they are deficient, perfect, or abundant(seeproblem3).

2. Use Theorem6.ll to look for factorsof Mersennenumbers. 3. Determine whether Mersenne numbers are prime using the Lucas-Lehmer test. 4. Given a positive integer n, determine if the sequencedefined in problem 16 peric.ic. 5. Find amicablepairs.

Cryptology

7.1 CharacterCiphers From ancient times to the present, secret messages have been sent. Classically, the need for secret communication has occurred in diplomacy and in military affairs. Now, with electronic communication coming into widespread use, secrecy has become an important issue. Just recently, with the advent of electronic banking, secrecy has become necessary even for financial transactions. Hence, there is a great deal of interest in the techniquesof making messagesunintelligible to everyoneexcept the intended receiver. Before discussing specific secrecy systems, we present some terminology. The discipline devoted to secrecy systems is called cryptology. Cryptography is the part of cryptology that deals with the design and implementation of secrecy systems, while cryptanalysis is aimed at breaking these systems. A messagethat is to be altered into a secret form is called plaintext. A cipher is a method for altering a plaintext message into ciphertext by changing the letters of the plaintext using a transformation. The key determines the particular transformation from a set of possibletransformations that is to be used. The processof changing plaintext into ciphertext is called encryption or enciphering, while the reverse process of changing the ciphertext back to the plaintext by the intended receiver, possessingknowledge of the method for doing this, is called decryption or deciphering. This, of course, is different from the process someone other than the intended receiver uses to make the messageintelligible through cryptanalysis.

188

189

7.1 Character Ciphers

In this chapter, we present secrecy systems based on modular arithmetic. The first of these had its origin with Julius Caesar. The newest secrecy system we will discusswas invented in the late 1970's. In all thesesystemswe start by translating letters into numbers. We take as our standard alphabet the letters of English and translate them into the integers from 0 to 25, as sh o w nin T able 7. 1.

letter

A B C D E F G H I J K L M N

numerical 0 I equivalent

o

P

a

R S T

I I

V

w

X Y Z

2 3 4 5 6 7 8 9 l 0 l l t 2 l 3 t 4 l 5 l 6 t 7 l 8 l 9 20 2 l 22 23 24 25

Table7.1. The NumericalEquivalents of Letters. Of course, if we were sending messagesin Russian, Greek, Hebrew or any other languagewe would use the appropriate alphabet range of integers. Also, we may want to include punctuation marks, a symbol to indicate blanks, and perhaps the digits for representingnumbers as part of the message. However, for the sake of simplicity, we restrict ourselvesto the letters of the English alphabet. First, we discuss secrecy systems based on transforming each letter of the plaintext message into a different letter to produce the ciphertext. Such ciphers are called character or monographic ciphers, since each letter is changed individually to another letter by a substitution. Altogether, there are 26! possibleways to produce a monographic transformation. We will discuss a set that is basedon modular arithmetic. A cipher, that was used by Julius Caesar, is based on the substitution in which each letter is replaced by the letter three further down the alphabet, with the last three letters shifted to the first three letters of the alphabet. To describe this cipher using modular arithmetic, let P be the numerical equivalent of a letter in the plaintext and C the numerical equivalent of the correspondingciphertext letter. Then

C:P+3(mod26), 02.Then 7 n t-t(t_ t)# l (mo dpk). since G,p) : l, we know that (r,pk-t) : 1. consequently, from Euler's theorem,we know that vPL-2(o-D :

Therefore,there

,Q(Pk-tt

an integer d such that y o ' -' Q -t): I * d p k-t,

wherep trd, sinceby hypothesisyP'-'(P-t)* t (moApk). W e take the pth powerof both sidesof the aboveequation,to obtain, via the binomial theorem, yP'-'(P-l)

-

0 + dp*-t1o | + p@pt-r, * (|)o'Urk-t)2 +

* (dpk-t1n

| * dpk (modpo*'). Sincep I d, we can conclude that ,.P^-'(P-r) # I (mod po*t). completesthe proof by induction. tr Example. From a previous example, we know that r : 3 is a primitive root

247

8.3 The Existenceof PrimitiveRoots

: 3 is also a primitive modulo 7 and 72. Hence, Theorem 8.9 tells us that r root modulo 7k for all positive integers k. It is now time to discusswhether there are primitive roots modulo powers of Z. We first note that both 2 and 22: 4 have primitive roots, narnely 1 and 3, respectively. For higher powers of 2, the situation is different, as the following theorem shows;there are no primitive roots modulo these powers of 2. Theorem 8.10. If a is an odd integer, and if k is an integer, k ) : e 2 ' -' : a O QL )/2

3, then

1 (mo d 2 k).

We prove this result using mathematical induction. If a is an odd integer, then a : 2b t 1, where b is an integer. Hence,

proof.

a 2 : ( 2 b + 1 ) 2:

4 b 2+ 4 b * I : 4 b $ + 1 ) + 1 .

Since either b or b * 1 is even, we see that 8 | 4b (b + l), so that a2 :- I (mod 8). This is the congruenceof interestwhen k :3. Now to complete the induction argument, let us assumethat a2'-' = I (mod 2k) . Then there is an integer d such that e2'-': l+d'zk. Squaring both sides of the above equality, we obtain e 2 ' -' : | + d 2 k + r q 4 2 2 zk. This yields e2'-'= 1 (modzk+r), which completes the induction argument. n Theorem 8.10 tells us that no power of 2, other than 2 and 4, has a primitive root, since when a is an odd integer, ord2ta # OQk) , since a6Q')lz : 1 (mod 2k) . Even though there are no primitive roots modulo 2k for k > 3, there always is an element of largest possible order, namely OQ\ I 2, as the following theorem shows.

248

PrimitiveRoots

Theorem 8.11. Let k 7 3be an integer. Then o r d 2 . 5: O ( Z k ) D : 2 k - 2 . Proof. Theorem 8.10 tells us that 52'-' = I (mod 2k). for k 2 3. From Theorem 8.1, we see that ordr.S I Z*-2. Therefore, if we show that ordr.5 | 2l"-t , we can conclude that ord2.5- 2k-2. To show that ordr,S tr 2k-3, we will prove by mathematical induction that fork)3, 52,-'= | + 2k_t *

I (mod 2k).

For k : 3. we have

5:l+4(mod8). Now assumethat

52'-': l+zk-I (mod2ft). This meansthat thereis a positiveintegerd suchthat S 2 ' - ' _ ( 1+ 2 k - r ) + d Z k . Squaringboth sides,we find that 52'-': (l + 2k-t)2 + 20 + zk-t)dZk + (dzk)z so that 52,-,= 0 + 2k-r)2 : | + 2k + 22k-2 :

I + 2t (mod Zk+\ .

This completesthe induction argument and showsthat ordr'5 : O(2k)/2' tr We have now demonstratedthat all powers of odd primes possessprimitive roots, while the only powers of 2 having primitive roots are 2 and 4. Next, we determine which integers not powers of primes, i.e. those integers divisible by two or more primes, have primitive roots. We will demonstrate that the only positive integers not powers of primes possessingprimitive roots are twice

249

8.3 The Existenceof PrimitiveRoots

powers of odd primes. We first narrow down the set of positive integers we need consider with the following result. Theorem 8.12. If r is a positive integer that is not a prime power or twice a prime power, then n does not have a primitive root. Proof. Let n be a positive integer with prime-power factorization

,-p\,p'i...p';. Let us assume that the integer n has a primitive root r. This means that (r,n ) : I and or dn r :6 h ). Si n c e (r,n ) : l , w e know that (r,p' ) : l , wheneverpt is one of the prime powers occurring in the factorization of r. By Euler's theorem, we know that ro@') :

I (mod P) .

Now let U be the least common multiple of Q(p'r), OQ'il,..-,0(p';), i-e.

u : [oQ\'),aQ'il,...,0b'il1. SinceObh I U, we know that ru = t (modP,l') we seethat for i : l, 2 ,...,m . From this last congruence, ordrr:6Q) Z. Hence,the numberse(p'r'),Oe'il,..., Oe,;\ are not p air wis er elat iv e l yp ri m e u n l e s sm: I a n d n i s a pri mspow er o, * :2 and the factorization of n is n : 2p', where p is an odd prime and / is a positive integer. tr

We have now limited considerationto integers of the form n : 2p,, where p is an odd prime and r is a positive integer. We now show that all such integers have primitive roots. Theorem 8.13. rf p is an odd prime and r is a positive integer, then 2pt possesses a primitive root. In fact, if r is a primitive root modulopt, then if r is odd it is also a primitive root modulo 2pt, while if r is even, r * pt is a primitive root modulo 2pt. Proof. If r is a primitive root modulo pt , then rob') = I (modp,), and no positive exponent smaller than 6(pt) has this property. From Theorem -6.4, we note that O(zp') : 0Q) 66t7 : e(p,), so that ,6(2n') 1 (mod p') . If r is odd, then ,o(zp')= I (mod 2). Thus, by corollary 3.2, we see that rQQp';: I (mod 2p,). since no smaller power of r is congruent to I modulo 2pt , we conclude that r is a primitive root modulo 2pt . On the other hand, if r is even, then r * p ' (r + P'10{zP')

Hence,

I (mod 2)

Since r * p' = r (mod p'), we see that G * pt )QQP')

I (mod p' )

Therefore, (r + ot1oQfl: I (mod 2p'), and as no smaller power of r *pr is congruent to 1 modulo 2pt , we conclude that r * p' is a primitive root modulo 2p'. rt Example. Earlier

this section we showed that 3

a primitive root modulo

8.3 The Existenceof PrimitiveRoots

251

7t for all positive integers /. Hence, since 3 is odd, Theorem 8.13 tells us that 3 is also a primitive root modulo 2'7t for all positive integers /. For instance, 3 is a primitive root modulo 14. positive Similarly, we know that 2 is a primitive root modulo 5' for all * 5t is a integers/. Hence, since 2 + 5t is odd, Theorem 8.13 tells us that 2 primitive root modulo 2.5t for all positive integers f. For instance,2T is a primitive root modulo 50. Combining Corollary 8.3 and Theorems8.9, 8.12,8.13, we can now describe which positive integers have a primitive root. Theorem 8.14. The positive integer n possessesa primitive root if and only if fr :2,4, p', or 2pt, where p is an odd prime and / is a positive integer.

8.3 Problems l.

Which of the integers 4,10,16,22and 28 have a primitive root?

2.

Find a primitive root modulo a) b)

3.

c) d)

r72 D2.

Find a primitive root, for all positive integers k, modulo a) b)

4.

lf B2

3k lle

c) d)

l3k nk.

Find a primitive root modulo a)6c)26 18 b)

e)

338.

5.

Find all the primitive roots modulo 22.

6.

Show that there are the same number of primitive roots modulo 2pt as there are of p' , where p is an odd prime and r is a positive integer.

7.

Show that if rn has a primitive root, then the only solutions of the congruence x2 = I (mod m) are x E t I (mod z).

252

PrimitiveRoots

8.

Let n be a positive integer possessinga primitive root. Using this primitive root, prove that the product of all positive integers less than n and relatively prime to n is congruent to -l modulo n. (When n is prime, this result is Wilson's Theorem.)

9.

Show that although there are no primitive roots modulo 2& where k is an integer, k > 3, every odd integer is congruent to exactly one of the integers (-1)"50, where a:0 or I and B is an integer satisfying0 < B ( 2ft-2-1.

8.3 Computer Projects Write computer programs to do the following: l.

Find primitive roots modulo powers of odd primes.

2.

Find primitive roots modulo twice powers of odd primes.

8.4 Index Arithmetic In this section we demonstrate how primitive roots may be used to do modular arithmetic. Let r be a primitive root modulo the positive integer m (so that m is of the form describedin Theorem 8.14). From Theorem 8.3, we know that the integers r, 12, 13

form a reduced system of residuesmodulo nr. From this fact, we see that if a is an integer relatively prime to m, then there is a unique integer x with 1(x46@)suchthat r'

a (modm).

This leads to the following definition. Definition. Let m be a positive integer with primitive root r. If a is a positive i n t eger wit h ( a, m): l , th e n th e u n i q u e i n t eger x w i th I (x(d(z) and r* = a (mod m) is called the index of a to the base r modulo m. With this definition, we have a - ,ind'a (mod m ). If x is' the index of a to the base r modulo m, rhen we write x : indra, where we do not indicate the modulus m in the notation, since it is assumed"to be fixed. From the definition, we know that if a and b are integers relatively prime lo m and a = b (mod m), then ind,a : indrb. Example. Let m : 7. We have seen that 3 is a primitive root modulo 7 and

253

8 .4 l n dex A r it hm eti c

(mod7), 3 r = 3 ( m o d 7 ) , 3 2 = 2 ( m o d 7 ) , 3 3= 6 ( m o d 7 ) , 3 4 = 4 that (mo d = ( m od 7 ). I 5) . and 3 6 35= 5 Hence, modulo 7 we have i n d 3 l : 6 , i n d t2 : 2 , i n d l 3 : 1, i n d 3 4: 4 , i n d r5 : 5 , i n d r6 : 3. With a different primitive root modulo 7, we obtain a different set of indices. For instance,calculationsshow that with respectto the primitive root 5, i n d 5 l : 6 , i n d s 2: 4 , i n d s 3: 5, ind54 : 2, ind.55: l, inds6 : 3. We now develop some properties of indices. These properties are somewhat similar to those of logarithms, but instead of equalities, we have congruences mo d ulo6@) . Theorem 8.15. Let m be a positive integer with primitive root r, and let a and b be integersrelativelyprime to m. Then ( i) (ii) (iii)

ind, l = 0 (mo d Q fu )). ind,Gb) = ind,a * ind,b (mod O@)) -la. ind,a (mod 6h)) if k is a positive integer. ind,ak

Proof of G). From Euler's theorem, we know that ,6(m): I (mod z). Since r is a primitive root modulo m, no smaller positive power of r is congruentto 1 modulo rn. Hence, ind,l : 6(m) = O (mod Qfu)) . Proof of (ii). indices,

To prove this congruence, note that from the definition of ,ind'Qil :

ab (mod ,,, )

and ,ind,a*ind,b-

,ind,o

,ind,b = Ab (mOd ,, ).

Hence, ,ind,Gb) =

7ind,a

* ind,D

(mod

rn ).

Using Theorem 8.2, we concludethat in d ,(a b ) :

i n d ,a * i n d ,b (m o d 6@ )).

254

PrimitiveRoots

Proof of Gii). To prove the congruence of interest, first note that, by definition, we have -: ,ind',ar ak (mod m ) and ,k'ind'a

(rind'o)P :

=

(mod rn).

ak

Hence, ,ind,aL =

rk'

ind'o

(mod

rn ).

Using Theorem 8.2, this leads us immediately to the congruence we want, namely ind,ak

-

ft. ind,a (mod 6fuD,

a

Example. From the previous examples,we see that modulo 7, ind52: 4 and i n d 5 3 : 5 . S i n c eA Q ) : 6 , p a r t ( i i ) o f T h e o r e m8 . 1 5 t e l l su s t h a t i n d 5 6- i n d s 2 . 3 : i n d s 2t i n d 5 3: 4

t 5:9

= 3 ( m o d6 ) .

Note that this agreeswith the value previously found for ind56. From part (iii) of Theorem 8.15, we seethat ind53a= 4'inds3 = 4.5 : 20 = 2 (mod 6). Note that direct computation gives the same result, since i n d 5 3 a- i n d s Sl - i n d s4 : 2. Indices are helpful in the solution of certain types of congruences. Consider the following examples. Example. We will use indices to solve the congruence 6xr2 : I 1 (mod 17). We find that 3 is a primitive root of 17 (since 38 = -l (mod l7)). The indicesof integersto the base 3 modulo l7 are given in Table 8.1. a ind3a

I

2

3

16 14 I

4

6

7

8

9

r2 5 l 5

5

ll

l0

2

10 1l 3 7

t2

13 l4

t5

16

l3

4

6

8

9

Table8.1. Indicesto the Base3 Modulo 17. Taking the index of each side of the congruenceto the base 3 modulo 17, we obtain a congruencemodulo d(t7) : 16, namely

255

8.4 Index Arithmetic

in d 3 (6 x r2 )= i n d 3 l| :' l

(m o d 16).

Using (ii) and (iii) of Theorem 8.15, we obtain :, (mod 16). ind3( 6x r 2)- i n d 3 6* i n d 3 (x 1 2 ) 1 5 + 1 2 ' i nd3x Hence, 15+12'ind3x=7(mod16) or 12'ind3x=8(mod16). Using Corollary 3.1, upon division by 4 we find that ind3x : 2 (mod 4). Hence, ind3x :

2 , 6 , 1 0 ,o r 1 4 ( m o d 1 6 ) .

consequently, from the definition of indices,we find that x 2 3 2 , 3 6 ,3 t o o r 3 l a ( m o d 1 7 ) , (note

this that 32:- 9,36 : 15,310

17)' modulo holds congruence ( m o d t hat c o n c l u d e w e l 7 ) , 2 8, and 314:

Since

x 3 9 , 1 5 , 8 , o r 2 ( m o d1 7 ) . Since each step in the computations is reversible, there are four incongruent solutions of the original congruencemodulo l7' (mod 17). Example. We wish to find all solutionsof the congruence7'= 6 of this sides both When we take indices to the base 3 modulo 17 of congruence,we find that i n d 3 (7 ' ) :

i n d 3 6: 1 5 (m o d 16).

From part (iii) of Theorem 8.15, we obtain i n d 3 ( 7 ' ) : x ' i n d 3 7: l l x Hence.

(mod 16).

256

PrimitiveRoots

llx

:

15 (mod16).

Since 3 is an inverseof I I modulo 16, we multiply both sides of the linear congruence aboveby 3, to find that x = 3 . 1 5: 4 5 :

1 3 ( mod 16).

All stepsin this computationare reversible.Therefore, the solutionsof 7* = 6 (mod 1 7 ) are given by x = t3 (mod 16). Next, we discusscongruencesof the form xk = a (mod m), where m is a positive integer with a primitive root and (a,m) : l. First, we present a definition. Definition' lf m and k are positive integers and a is an integer relatively prime to ffi, then .we say that a is a kth power residue if * if the congruencexk = a (mod,m) has a solution. When z is an integer possessinga primitive root, the following theorem gives a useful criterion for an integer a relatively prime to m to be a kth power residue of m. Theorem 8.16. Let m be a positive integer with a primitive root. If k is a positive integer a1d o is an integer relatively prime to m, then the congruence xk = a (mod m) has a solutioriif and only-ii oQh)ld=l(modln) where d : (k,6(m)). Furthermore, if there are solutions of xk : a (mod m)' then there are exactly d incongruentsolutionsmodulo rn. Proof. Let r be a primitive root modulo the positive integer 17. We note that the congruence xk

(mod z)

holds if and only ( 8 .1 ) Now let d:

k ' i n d ,x ( k ,e (m))

i n d ,a (m o d 6@ )).

a n d y : i n d ,x , s o that x

(mod z ).

From

257

8 .4 In d ex A r it hm et ic

Theorem 3.?, we note that it d tr indra, then the linear congruence (8 .2 ) k y : i n d " o (m o d Q fu )) (8 has no solutions, and hence, there are no integers x satisfying l). If d lind'a, then there are exactly d integersy incongruentmodulo d(z) such that (8.2) holds, and hence,exactly d integersx incongruentmodulo z such rhat (8.1) holds. Since d I ind,a if and only if @@)/ilind,a

= o (mod Q(m)),

and this congruenceholds if and only if ooh)/d:1(modrz). the theorem is true. tr We note that Theorem 8.16 tells us that if p is a prime, k is a positive integer, and a is an integer relatively prime to p, then a is a kth power residue of p if and only if oQ-D/d: 1 (modp), where d : (k,p-l).

We illustrate this observationwith an example.

Example. To determine whether 5 is a sixth power residue of 17, i.e. whether the congruence x 6 = 5 (mo d 1 7 ) has a solution, we determine that 5 t6 /(6 ,1:6 ) 5 8 = -l

(m o d l 7).

Hence, 5 is not a sixth power residueof 17. A table of indices with respectto the least primitive root modulo each prime lessthan 100 is given in Table 4 of the Appendix. We now present the proof of Theorem 5.8. We state this theorem again for convenience. Theorem 5.8. If n is an odd compositepositive integer, then r passesMiller's te st for at m os t f u- l) / 4 b a s e sb w i th I < , 1 n -1 . We need the following lemma in the proof of Theorem 5.8.

258

PrimitiveRoots

Lemma 8.1. Let p be an odd prime and let e and q be positive integers. Then the number of incongruent solutions of the congruence x e - t = I ( m o dp r ) i s ( q , p r - r e - D . Proof' Let r be a primitive root of p' . By taking indiceswith respectto r, we see that x4: I (modp,) if and only if qy = 0 (mod 6e,D where y : ind'x . using Theorem3.j, we see that there are exactli e,6er)) incongruentsolutionsof gy :0 (mod|e"D. consequently,there are Q,6Q")) : (q,p'-tb-l)) incongruent solutions of xe = 1 {-oAp'). tr We now proceedwith a proof of Theorem5.g. Proof. Let n-l : 2't, wheres is a positiveinteger and,t is an odd positive integer. For n to be a strongpseudoprime to the baseD, either bt :

b2tt :

I (mod n )

-1 (mod n)

( s - l.

f o r s o m e i n t e g e r T w i t h 0( 7

bn-t=

Ineithercase,wehave

I (modn).

Let the prime-powerfactorizationof n be n : pi,pi, . . . p',,. From Lemma 8.1, we know that there are (n-r, p'/Qi-l)) : h-l,pi-l) incongruent solutionsof xn-r: I (modp7) , j :1,2,...,r. Consequently, the Chinese remaindertheoremtells us that thereare exactlv h-\,p1-l) fI solutionsof x'-l

= I (mod n ).

incongruent

j-r

To prove the theorem, we first consider the case where the prime-power flactorizationof n contains a prime power p[. with exponente* 2 2. Since

bo-D /pt : t/p't-t - t/p't < z/g (the largest possiblevalue occurswhen pj :3

and ei :2),

we seethat

259

8.4 Index Arithmetic

r

tu-r,pj-r)< fI Q;t) fI j -r ;:l

l+,r)

li-l

ll**

"+" Since

?"*f

0n-l) for n > 9 , we seethat r

u (n-l ,p,-l) (

(r -r)14.

j:r

Consequently,there are at most Q-Dla is a strong pseudoprimeto the base b. T h e o t h e r c a s et o c o n s i d e ri s w h e n n : distinct odd primes. Let

integersb, I < 6 ( n , for which n w h e r eP t , P z , . - . , Par r e

PPz"'P.

p t - | : 2 t' tr, i : 1 ,2 ,.. .,r, where s; is a positive integer and /; is an odd positive integer. We reorder the ( s, ' We note that primespr,p2,...,p,,(if necessary)so thatsr ( sz ( h-l,pi-l)

: 2*ink') (t,t,).

: (t,t;). From The number of incongruentsolutionsof x' = I (mod pi) is T solutions of incongruent 2il; are there problem 15 at the end of this section, * y''= - l ( m odp; ) w h e n O ( f ( s i -I, a n d n o sol uti onsotherw i se. H ence, i ncongruent u si n g t he Chines e r e ma i n d e r th e o re m , th e re a r e TrTz" ' 7, incongruent solutions of solutions of xt : I (mod n), and 2i' TrTz"'7, x/, = - 1 ( m od n) w h e n 0 ( 7 ( s 1 -1 . T h e re fo re,there area total of

[ ,,-' I

I

Z"'-t I

TrTz"' T, lt* > 2t'l- TrTz"' T,lt + .;; l,r-oJtL)

I

integers b with 1< D ( n-1, for which n is a strong pseudoprimeto the Uasetr. (We have used Theorem l.l to evaluatethe sum in the last formula.) Now note that

260

PrimitiveRoots

6h) : (pr-l) (pz-l)

(pr-l)

:

tiz

tr1t'*s'*

"' *s,

We will showthat

rrrz'" r,[,*ro] 2 ' ,-t

|

)

*,,r,ro,

which provesthe desired result. Because TrTz. . . 7, ( r1r, achieveour goal by showing that

(8.3)

*r,< r/4. [,*l'-t lrr',*',*'' z',-t | )

Since sr ( sz (

( s, , we seethat

tr, we can

' as, * Uf ( f^,* ''.'-t ,r',*',* f, f,r,,, 2 ' t | )' l . 2 ,- l J''

I

--

2"r-l 2"r(2, -l)

2",

:l++-l 2,-l

2"t

| 2'-l - -< l

I-

2rtr(2, -l)

2',-2 2"'(2'-l)

2r-r

From this inequality,we concludethat (s.r) is valid when r ( When r:2, w e h a v en : p p 2 w i t h p r | : 2 t r t 1 rr ( sz. If s1 ( s2, then (S.f) is againvalid, since

I

(

3.

and pz-l:2trtz,

with

''"

rt',-, I -L. I r ^ ) . . ?)/2',*',: +]/lz",z',-',) [t ['

:[+.#),,"-"

*+

W h e n s r : J 2 , w e h a v e( n - l , p r l ) : 2 ' T r and(n-l,pz-l):2tTz. Let us assume that pr ) pz. Note that T1 * t1, for if Tr: tr, then

261

8.4 Index Arithmetic

( p t - l ) I ( n - l ) , s ot h a t n : p r p z Z p z = 1 ( m o dp r - l ) , which impliesthat P2 ) Pr, a c o n tra d i c ti o n . S i n c e T1# t' 1 , we know that T r ( t r / 3 . S i m i l a r l v , l f t 1 pz then T2 # tr, so that 7"2( t2l3 . Hence, 7 ^2s, , I 2 '":t , w eh a v e T r T z4 t 1 2 / 3 , a n ds i n c el r * l/r"'* |

-,2 r, , l

: 6h)16, | < r t222"16

TtTzlr+ f lr)

proves

which

the

;

3)

t

theorem

for

this

final

case'

since

oh) /6 ( (n -r) /6 < (/,-r) /4. tr By analyzing the inequalities in the proof of Theorem 5.8, we can see that the probability that n is a strong pseudoprimeto the randomly chosenbase D, 1 < b ( n-1, is close to ll4 only for integers n with prime factorizations of t h e f o r m n : p r p 2 w i t hP r : | + 2 q 1a n d P z : I t 4 q 2 , w h e r e{ 1 a n d Q 2 a r e o d d p r i m e s , o r n : q f l z Q t w i t h P r : | + 2 q r ,P 2 : | * 2 q 2 , a n d pz: I t 2q3, wher e Q r,e z ,a n dq 3 a re d i s ti n c to d d pri mes (seeprobl em 16).

8.4 Problems l.

Write out a table of indices modulo 23 with respectto the primitive root 5.

2.

Find all the solutions of the congruences a) 3xs = I (mod 23)

3.

b) 3xta = 2 (mod 23).

Find all the solutionsof the congruences il

3' :- 2 (mod 23)

b) 13" = 5 (mod 23)'

4.

For which positive integers a is the congruence axa =

2 (mod 13) solvable?

5.

For which positive integers 6 is the congruence 8x7 :

b (mod 29) solvable?

6.

Find the solutionsof 2x = x (mod 13), using indices to the base 2 modulo 13.

7.

Find all the solutionsof x' :

8.

Show that if p is an odd prime and r is a primitive root of p, then ind,(p-|)

(p-r) /2.

x (mod 23). :

262

9.

Primitive Roots

Let p be an odd prime. Show that the congruence x4 = solution if and only if p is of the form gfr + l.

_l(modp)

has a

1 0 . Prove that there are infinitely many primes of the form 8ft*1.

(Hint: Assume that p6p2,...,pn are the only primes of this form. Let . . p)a+l . (ppz. e Show that Q must lave an odd prime factor different than j1p2,...,pn, and by problem 9, necessarilyof the form 8k+l .)

ll.

From problem 9 of Section 8.3, we know that if a is a positive integer, then there are unique integers a and B with a : 0 or I and 0 < B ( Z*-i-t such that a = (-l)" 5p (mod 2ft). Define the index system of a modulo 2k to be equal to the pair (a,B). a)

Find the index systemsof 7 and 9 modulo 16.

b)

Develop rules for the index systems modulo 2& of products and powers analogousto the rules for indices.

c)

Use the index system modulo 32 to find all solutions of j xs = I I (mod 32) and 3' = 17 (mod 32).

12. Let n : 2"p\'pj ' ' ' ph be the prime-power factorization of n. Let a be an integer relatively prime to n. Let r1,r2,...,r^ be primitive roots of pti,p'i,..., p';, respectively, and let 71 : ind", a (mod p'1), 72 : ind", a (mod ptl), (mod p'il. rc /o ( 2, let rs be a primitive root of 2t,,and let ...,1m:ind,.a : (mod ind,. a 2t). If ls 2 3,let (a,p) be the index systemof c modulo 2k, 7e (-l)'5P (mod 2t). Define the index system of a modulo n to be = so that a ( 1 o , 1 r , 7 2 , . . . , y ) i f t o ( 2 a n d ( a , 8 , 7 t , ^ 1 2 , . . . , 1i ^f )t o Z 3. a)

Show that if n is a positive integer, then every integer has a unique index system modulo n.

b)

Find the index systemsof 17 and 4l (mod lZ0) (in your computations, use 2 as a primitive root of the prime factor 5 of 120).

c)

Develop rules for the index systems modulo n of products and powers analogousto those for indices.

d)

Use an index system I lx7 : 43 (mod 60).

modulo

60

to

find

the

solutions

of

Let p be a prime, p ) 3. Show that if p =2 (mod 3) then every integer not divisible by 3 is a third-power, or cubic , residue of p, while if p : I (mod 3), an integer a isa cubic residueof p if and only i1 o@-t)/3: I (modp). Let e be a positive integer with e 7 2. il

Show that if ft is a positive integer, then every odd integer a is a kth power residue of 2" .

b)

Show that if /c is even, then an integer a isa / 7y urc pr..isely thosepairs satisfyingI ( x ( 3 and 1 ( y ( 11xl7. For a fixed integerx with 1 ( x ( 3, there are lttx/ll allowablevaluesof y. Hence, the total number of pairs satisfying I ( x < 3, 1 ( / ( 5, and llx ) 1y is 3

+ I33l7l: I * 3 + 4 : 8; 2 tt tlTl : ttt/tl + 122/71

j:1

(3,4)' th e s eeight pair s ar e (l ,l ), (2 ,D , (2 ,2 ), (2 ,3 ), (3 ,1), (3,2), (3,3) and The pairs of integers G,y) with I ( x < 3, I ( y ( 5, and llx 1 7y For a *r. pr..isely those pairs satisfying I ( y ( 5 and 1 ( x 4 7y /tt. allowable values of x. fixed integer y with I ( y ( 5, there are lly/ttl Hence, the total number of pairs satisfying I ( x < 3, I ( y ( 5, and llx ( 7y is

310

Quadratic Residues

5 j-r

ltj /ttl : Ij lrrl + [ t L l t r ] + [ 2 r / r t l+ I 2 8 l n] + [ 3 s l l1 ] :0*l

+ 1+ 2*3:7.

Thesesevenpairs are (l,2) , ( 1 , 3 ) ,( 1 , 4 ) ,( 1 , 5 ) ,( 2 , 4 ) ,( 2 , 5 ) ,a n d ( 3 , 5 ) Consequently,we seethat 1l-1

35

7-l

1 5: ) t r r j l l l + > l t j l t l l : 8 * 7 . j-r j-r

T;:5'3: Hence,

rr-l .7-l (_t)

2

2:(_l);*'

i,rrrr,r,* i, rtinl i-l

35

2lni/tl )Iti/rrl (- I )i-' (- I )r-' 3

Since Lemma

(t

'l

l#l

r,'J

g.2

t e l l s rrs .^

5t/

: ( -.1. )I it-ttr,rw"et s e e t h a t

+ L^+ | rr I that

Z,'rj/tl

: (-1;r-t

17 |

lI t ll fl r r" l | : ( - t ) [11J|.7 )

and

t-'rr-r 2

2

This establishesthe special case of the law of quadratic reciprocity when p:7andq:ll. We now prove the law of quadratic reciprocity, using the idea illustrated in the example. We consider pairs of integers (x,y) with I ( x ( Q -l) /2 and o -l such pairs. We divide t-hesepairs I ( y ( ( q - D/ 2. T h e re u r" 2 -l ; T into two groups, dependingon the relative sizesof qx and py.

Proof.

First, we note that qx I py for all of these pairs. For if qx : py, then q l p y , w h i c h i m p l i e st h a t q l p o r q l y . H o w e v e r ,s i n c e q a n d p a r e w e know d i s t inc t pr im es ,w e k n o w th a t q l p ,a n d s i n ce I ( y ( (q-i 12, that q I y. w i th I ( x ( Q-I)/z, To enum er at e th e p a i rs o f i n te g e rs (x y) -l) (q ( ( y 1 /2, and qx > py, we note that these pairs are precisely those (p-l)/2and ( ( x For each fixed value of the I (y where I 4qx/n. ( with are 1 x 4 b-1012, there Iqx/pl integers satisfying integer x, ( number of pairs of integers G,y) y qx total the Consequently, I 4 /n.

311

9.2 The Law of Quadratic Reciprocity

Q-t)t2

withl (x

( Q-D/2,andqx>

( Q-D/2,t (v

Pvis

Iqilpl'

?,

-l) 12, We now considerthe pairs of integersG,il with 1 ( x ( b 1 ( y ( (q-D 12,and qx < py . These pairs are preciselythe pairs of i n t e g l r sG , i l w i t h 1 ( y ( ( q - D / Z a n d 1 ( x 4 p y l q . H e n c e , f o r e a c h -1) 12, there are exactly fixed value of the integer y, where I ( y ( (q ( shows that the total This py x I 4 lq. lpy lql integers x satisfying (q-t)/z, ( (i,y) ( x I with b-D/2,1 (y ( nurnu..of pairselil/r.g"rt andqx < py is

j- r

Adding the numbers of pairs in these classes,and recalling that the total ' = rt ' + ,w e s e eth a t n u mb er of s uc h pair s ,,

')'' j-|

,r,,d:+'+ hilpt*'ni'' i-r

,

or using the notation of Lemma 9.2, p-l .q-l 22

T(q,p) + TQ,q) Hence,

1-11r{n'c): (-t) ,-t1rQ'il+r@,q): (- 11r(e'n) Lemma 9.2 tellsus that 1-1yr(a,r): f

lf

["'l

lp J

\

lzll4l:(-t) l . qJ l . pJ

p-l .q-r 22

."0 1-gr{o.o): [" .|

H ence

lq)

P-t.q-l

2 2

This concludesthe proof of the law of quadratic reciprocity. n The law of quadratic reciprocity has many applications. One use is to prove the validity of the following primality test for Fermat numbers. Pepin's Test. The Fermat number F^ : 22' + I is prime if and only if 3 G ' -r)1 2 : -l

(m o d F - ).

proof. We will first show that F* is prime if the congruencein the statement of the theorem holds. Assume that

312

QuadraticResidues

3G^-r)/2: -l

(mod F*).

Then, by squaring both sides,we obtain 3F.-1 = I (mod F*). From this congruence,we seethat if p is a prime dividing F*,then 3F.-l = I (modp), and hence, ordo3 | {f ^-I)

: 22'.

Consequently,ordr3 must be a power of 2. However,

ordo3tr2''-': (F^-D/2, since 3G^-t)/2 - -l (mod F*) . Hence, the only possibility is that o 1do3: 22^ : F ^ - l . Si n c e o rd o 3 : F m-t ( p - I and p F*, we see I that p : F^, and consequently,F^ must be prime. C o n v e r s e l y , i fF r : 2 2 ' * reciprocity tells us that

(e.5)

I is prime for m )

l , t h e n the law of quadratic

:[+] t*l:[+J

since F^ = | (mod 4) and F^ = 2 (m o d 3 ). Now, using Euler's criterion, we know that

(e.6)

t*l

3 G' -t)/' (-o d

F-).

From the two equationsinvolving I I I (9.5)and (s.e),we conclude that

[". j'

_ _1 (mod 3(J'._r)/2 F). This finishesthe proof. E x a m p l e .L e t m : 2 .

tr Then F2: 2 2 ' + l : 1 7 a n d aFr-t)lz _ 3 8 :

-1 (mod l7).

9.2 The Law of QuadraticReciprocity

313

By Pepin'stest, we seethat F2 : l7 is prime' : 4 2 9 4 9 6 7 2 9 7W - e n o t et h a t Let m :5. Then Fs:22' + l:232 t I -l (mod 4294967297). 3G,-D/2: 12": 32t4148364810324303 * Hence, by Pepin'stest, we seethat F5 is composite'

9.2 Problems l.

Evaluate the following Legendre symbols

[-u]

d)

a,

[ 6 4 r. J

[*]

u,[+l c,t*l 2.

f:ul

e)

l e e rJ

Iros]

l*'l

prime, then Using the law of quadratic reciprocity, show that if p is an odd

: [;] 3.

{lii

p = tl (mod 12) p = t 5 ( m o d 12 ) .

Show that if p is an odd Prime, then

[-r I :

[7J

{l

ifp=t(mod6) if p = -l (mod 6).

4 . Find a congruencedescribing all primes for which 5 is a quadratic residue' 5 . Find a congruencedescribing all primes for which 7 is a quadratic residue. (Hint: Let n be 6 . Show that there are infinitely many primes of the form 5Ic * 4'

of a positive integer and form Q : 5(tnr'\2+ 4' Show that Q has a prime divisor reciprocity quadratic of law the use do this, To n. greater than + 4 5k the form - t I to show that if a primep dividesQ, then | ? | t)l

314

Quadrati c R esi dues

7 . Use Pepin'stest to show that the following Ferntat numbersare primes a)

Fr : 5

b)

F3 - z5i

c)

F4: 65537.

8.

From Pepin'stest, concludethat 3 is a primitive root of every Fermat prime.

9.

In this problem, we give another proof of the law of quadratic reciprocity. Let p and q be distinct odd primcs. Let R be the interior of the rectanglewith vertices

o:

( o , o )A, : b / 2 , 0 , B : Q / 2 , 0 ,a n dC : b / 2 , q / D .

a)

Show that the number of lattice points (points with integer coordinates)in R i, P-l .q-l 22

b)

Show that there are no lattice points on the diagonalconnectingO and C.

c)

Show that the number of lattice points in the triangle with verticesO, A, C Q-D/2

is i-l

d)

Show that the number of lattice points in the triangle with verticesO, B, Q_r)/2 and C is j-l

e)

Concludefrom parts (a), (b), (c), and ( d ) t h a t Q-t)/2

Q-D/2

j-t

j-l

Derive the law of quadratic reciprocityusing this equationand Lemma 9.2 Computer Projects Write programsto do the following: l.

Evaluate Legendresymbols,using the law of quadratic reciprocity.

2.

Determine whether Fermat numbersare prime using Pepin'stest.

9.3 The Jacobi symbol I n t his s ec t ion ,w e d e fi n eth e J a c o b is y m b o l . Thi s symboli s a general i zati on of the Legendresymbol studied in the previoustwo sections. Jacobi symbols a r e us ef ul in t he e v a l u a ti o no f L e g e n d res y m bol sand i n the defi ni ti onof a type of ps eudop ri me . Definition. n : p' ipt i

Let n be a positive integer with prime factorization ' p; a n d l e t a b e a p o s i ti v ei n te ger rel ati vel ypri me to n. Then,

315

9 .3 The J ac obi s Y mb o l

the Jacobi symbol

:

[.]

l, ,|

t

; I

bY t' denned

I

p\'p'; " ' p';

l:[*]'t;l lh)'

Legendre S on the right-hand side of the equality are where the symbol symbols. Example. From the definition of the Jacobi symbol, we see that

: lz)'let:(-r)2(-r):-r' ['l: lzl :lil l45,11."ij

l;l

#l:[+*l:[+l[+l[+]:[+l[+l l*l -r

and

:

: '-D2 t2(-'l): [+]'[+l'[+]

When r is prime,the Jacobisymbolis the sameas the Legendresymbol' the valueof the Jacobisymbol ' However,whenn is composite, lq I Oott nor

lr)

tell us whether the congruencex2 = a (mod n) has solutions..,*. that if the congruencex2 = a (mod n) has solutions,then l*

ln)

|

- t

do know To see

(modn) has th i s, not e t hat if p i s a p ri me d i v i s o r o f n and i f x2 = a solutions, then the congruencex2 = a (mod p) also has solutions. Thus,

r I Ii | : t lp).. tl

that I

g

m ( ^ )t f -l : l. To seethat it is possible : Consequently, ' | + I II | * I ln) i-1lPi)

: | : 1 when there are no solutions to xz

a (mod n), let a : 2 and

ln )

: (-r)(-1): r. However, : are there that[+l n: t5.Nore t?l t+.| ) ^l. ,l t J t r no solutionsto x2 i 2 (mod i S), rin* x2 = 2 (mod 5) have no solutions.

the congruencesx2 = 2 (mod 3) and

We now show that the Jacobi symbol enjoys some propertiessimilar to those of the Legendresymbol.

316

QuadraticResidues

Theorem 9.5. Let n be an odd positive integer and let a and b be integers relativelyprime to n. Then

(i)

(ii)

(iii)

if a:

D (modn),then

ll: l*)

lol: l["] fql n ) ln )

I n )

r )- t | | : t _ 1 1 h - D / z' f tr ) /)

(iv)

. I Ll :1-1) (n':-r)/a ln )

Proof- In the proof of all four parts of this theorem we use the prime factorizationn : p\,p'i . . p';. Proof of (i). we knowthat if p is a rrime.,dividinqn,then a =b (modp). Hence,from Theoremg.z G\ we have : we see l* | l+ | consequentry, IDJ lp) that

: f*l"l+J" [-tL'lo)"lol" I ol'': fal :lr'l i*l f,,J lo,Jlp,) lo^,| lo,t lp^):l;j Proof of (i). From r v " ' Theorem r r r v v r w t t9.2 7t ' L (ii), w s know \ I r ' f ' we K l l u w that fq) lo, ,l

Hence.

: | , I i a I ltl

F)'

[+):l*)"[#]" l*)': [;]"l*)" " {t)" [*] l*)'l*)'' : [;]

[*]

317

9 .3 The J ac obi s Y m b o l

t+l

p

if

Theorem 9.3 tells us that

Proof of Gril.

is

prime' then

- (-11 Q-r)/2.ConsequentlY,

f-r I l-l:

l'-rl"l-r ll_

ln,|

LP,)lPrJ

:

(- ,1tJn;t\/2+

'l" l"'rll

. [-' ]" tP^)

t'(p'-t)/Z + '"

+ t^(p^-r)/2

From the prime factorization of n, we have n-

(r + Qr-l))"(l + bz-l))"'''

(t * (p^-l))''

is even. it follows that

Si n ce Q i- l)

(t + (pi-l))"

= | + tib,-t)

(mod4)

and (l + r,(pi-l))(r + r, Qi-D):

I + tiQl-t)

+ tibi-l)

( m o d4 ) .

Therefore, n = 1+ tlpr-t)

+ t2(p2-i + '''+

t^(p^-l)

( m o d4 ) '

Th i s i m pliest hat

+ Q-D/2 = tJprD12 * tz(pz-D12

+ t^(p*-D12 (mod2) .

for for (n-1) lZ wittttheexpression this congruence Combining

r'

'no*t

l+J

/)n-l

-'

rlr-

that |

| :

(-l)

2

l,r )

r) ( i i l t h e n p r i m e , p i s .If Proofof l+l

: ( - 1 ; ( r ' l - r ) /'8H e n c e '

lp)

Izl : Il" [z] L,J lp'J lp,)

+t^Qi-r\tt t+'lt : (_l),,bi_t,tts+t,gt-r)/8+ lp^)

As in the proof of (iii), we note that

n 2: ( r + ( p ? - r ) 0" + @ ? - l ) ) "" ' ( t + b T - l ) ) " .

318

QuadraticResidues

Sincepl-I = 0 (mod8), weseethat 0 + Q?-l))', = | + tie?-l) (mod64) and

( l + r , b ? - l ) ) ( l+

4 e l - t ) ) = | * t ; e ? _ D+ t , A ? t ) ( m o d 64).

Hence,

n2:t+tJp?-D+tze?-D+

+ t ^ ( p T - l ) ( m o d6 4 ) .

This implies that

( n 2 - t ) / 8 : t J p ? - D / B+ t z e ? - D / s + . . . + t * ( p 3 , _ l ) / (8m o ds ) . combiningthis congruence for (n2- l)/g with the expression for [el teils ln ) f u s t h a t l L"l' l : 1 - 1 ; ( n ' - t ) / 8 . D

ln )

We now demonstratethat the reciprocity law holds for the Jacobi symbol as well as the Legendre symbol. Theorem 9.6. Let n and m be relatively prime odd positive integers. Then m-t n-l f lf I

l r l -| l L l : lm )l n )

( _t ) ,

Proof. Let the prime factorizations of rn and n : ql' q! , . . . qo r,.w e s e eth a t

, .

n be m : pl,pl, . " p!' and

w)'"'

lr):,4 tt)':,q,s and

l*): t

( n l4/

IIl;l j-t I'J

Thus,

I

)

s r :rtrt j-t

i-t

It)"''

319

9.3 The Jacobi symbol

,sti*lt l+l[*]:,g

q'l

h)

10tu'

l

From the law of quadratic reciProcity, we know th at

[ o , - ,f n,-, 1 I

t*ltr)

lr

:(-rllrj

t-)

l

Hence,

|^) [ , I [7Jl;): We note that

r

( '

f| ff(-l)

(-l)'-'l-' \

r \ "):

/

j-l

t-l

:z",1+] ',[+] ,.a''t+] ",1+l t,p,

As we demonstratedin the proof of Theorem 9.5 (iii),

(mod2)

=*

Doif+] j-t(o)z and

5u,[+]=

n-l 2

(m od 2).

Thus,

(e.8)

r s i-t

^fr,-tl

^[Qr-tl =.-l J

i-r

+(mod2).

\

Therefore,from (g.Z) and (9.8), we can concludethat f

)f

)

l Lnl l a l : ( _ r ) I

m-l

2

n-l

2 tr

)lm )

We now develop an efficient algorithm for evaluating Jacobi symbols. Let a : and b be relatively prime positive integers with a < b. Let Ro Q and R r : D Using the division algorithm and factoring out the highest power of two dividing the remainder, we obtain

32A

Quadratic Residues

Ro:

Rflr+2t'R2,,

where s1 is a nonnegativeinteger and R2 is an odd positive integer less than R I ' When we successivelyuse the division algorithm, and factor out the highest power of two dividing remainders,we obtain Rr: *r:

Rr-r : R n -z :

Rzez+2"'R3 Rflt+2"Ra

Rn_2Qn_2 * 2t.-rRn_1 R n -tQ r-, + 2 t .-t. I ,

where s; is a nonnegativeinteger and R; is an odd positive integer less than : 2,3,...,n-l Note that the number of division, ,"qu-ir"d to reach &-r for i the final equation does not exceed the number of divisions requiied to find the greatestcommon divisor of a and b using the Euclidean algorithm. we illustrate this sequenceof equationswith the following example. E x a m p l e .L e t a : 4 0 1

andb:

lll.

Then

4 0 1: 1 1 1 . 3 + 2 2 . n lll17.6+20.9 17:9.1+23.1. Using the sequence of equations we have described, together with the properties of the Jacobi symbol, we prove the following theorem, which gives an algorithm for evaluating Jacobi symbols. Theorem 9.7. Let a and b be positive integers with a > b . Then ni-r R,-r f ^'l + " ' + s ' - r& - !a!**f, +...+R"_,-tR._r_r t 8 r z 2 2 2 l+l:(-l)'' lb ) where the integersR; and s;,,t :1,2,...,n-l

: i+l:[+]

'

, are as previouslydescribed.

Proof. From the first equation and (i), (ii) and (iv) of Theorem 9.5. we have

fglla,|-

: (-1)

321

9 .3 The J ac obi s y m b o l

we have using Theorem9.6,the reciprocitylaw for Jacobisymbols,

+ :'-')+ t#l t*l

so that f ^ I

R,-l

l+l:(-r)T LDJ

R,-l

ni-t-

[ n, I

IR,J

Similarly, using the subsequentdivisions,we find that

ry*n#i+l :,-,rT '/ lgl ^, 1R;+rJ ,| [ * n e n w e c o m b i n ea l l th e e q u al i ti es,w e obtai n the desi red

fo rT :2, 3, . . . , n- t \

for l+ I tr expression ' [b ,l The followingexampleillustratesthe useof Theorem9.7. Example. To evaluate

we use the sequenceof divisionsin the

[++], previousexampleand Theorem9.7. This tells us that

[+orl:,-,lt F*o'"lt*'

n't'.ttr!:r +*!+

+:r.

l.111 J

The following corollary describes the computational complexity of the algorithm for evaluating Jacobi symbols given in Theorem 9.7. relatively prime positive integers with a > b ' ,,be O(loezb)3) bit Then the Jacobi symbol l+ | can be evaluated using " lb) operations.

Corollary 9.1. Let a and D

rt Proof. To find lf

of O1ogzb) a sequence I uting Theorem9.7,we perform

t . DJ

divisions. To see this, note that the number of divisions does not exceed the number of divisions needed to find G,b) using the Euclidean algorithm. Thus, by Lam6's theorem we know that O (log2b) divisions are needed. Each

322

QuadraticResidues

divisioncan be doneusing o ((lo^gzD2) operations. Each pair of integers si can be found using o(logzb).bit bit operationson"" ih" appropriate fl.u.nd divisionhasbeencarriedout. consequently,o((log2D)3)bit operationsare required to find the integers R;,s7,i :1,2,"',n-t a andb. Finaily,to evaluate the exponent of -l lr.T in the expression for l+l in Theorem9.7, we usethe last threebits in the lD ) binary expansion:of Ri,i : r,2,...,,n-r and the last bit in the binary expansions of sy,,r: r,,2,...,n-r. Therefore,we use 0(lo926) additional bit operations to find I+l Sinceo((log2D)3)+ ooog2b): o(tog2,D2) the , lD ) corollarvholds. tr

9.3 Problems I.

Evaluatethe followingJacobisymbols

a, t+]

b, [*]

b, [*]

, lx)

c,[*] 'tml

2 . For which positive integers n that are relatively

to 15 does the Jacobi

symbor equar r? t*l 3 . For which positive integers n that are relatively

to 30 does the Jacobi

symbor equar r? |.+l 4 . Let a and b be relatively prime integers such that b is odd and positive and a :

(-l)'2'q

where q is odd. Show that b-l

:

(-l)--'r

+

br-l

l-''

["1 lb )

5.

Let n be an odd square-free., positive integer. Show that there is an integer a

: -t such that(a,n): I and l;,J

323

9 .3 Th e J ac obi s Y m b o l

6.

Let n be an odd square-freepositive integer' r\ w h e r et h e s u m i s t a k e n o v e r a l l k i n a r e d u c e ds e t a ) S h o wt h a t ) l + l : 0 ,

ln )

of residuesmodulon. (Hint: Use problem5') b)

,n. numberof integersin a reduc?O"ti'ofresidues From part (a), show 11"\ O : -t. - r - - to the number*itn l* I modulon suchttut I | : I" -is- equal

l'J

lrj

7 . Let a and b:ro be relatively prime odd positive integers such that A :

lOQt *

e1r1

tO:

rlQ2 I

e2r2

enfn

fn-tQn-t*

fn-l:

with where q; is a nonnegative even integol, €; : t l, r; iS a positive integer by : obtained are l. These equations ri 1 ri t, for t : 1,2,...,frj , and rn Section l0 of problem in given algorithm successivelyusing the modified division t.2.

f^'l a)

Show that the Jacobi symbol |*

f"l

Irl b)

I i, given by

l . DJ

:(-l)[

l++*++:. 2

t

2

2

*t-f'+l 2

2

)

Showthat the Jacobisymbol [+.| t, givenbv lD ) t'^l

l+ | : (-r)r' lb;

w h e r e T i s t h e n u m b e r o f i n t e g e r si , I < , (mod 4). 8.

( n, with ri-r 7 ciri = 3

Show that if a and b are odd integers and (a,b): reciprocity law holds for the Jacobi symbol: I

(

" lt

a-t b-t

b l -:l - ( - r ) ; - ; a-'b-'

' ) \ll;l-J '--'J lr;l-l l,_ [(-l)2

2

l, then the following

ira aidi (mod 22r+2). Therefore.

t2'-t : h-D/2

m

) r s Z/ a ; d i ( m o d 2 ' + t ) . i-l

329

9.4 EulerPseudoprimes

This congruenceimPlies that 12s-t-r = i

aidi (mod 2)

i-l

and (9.10)

66-r\/2 : (6rt7z:-'- :

(-t)'.*

2 o'd' (mod n). (-1)t-t

:

On the other hand, from (9.9), we have

m ^) : fI el)"'"' : ((-r)d,)., : : fr lnl ft [+.|. I n J , . : r| . p , J i _ r

.fo,o, (-1)i-t

t-l

Therefore, combining the previousequation w i th (9 .10),w e seethat

- [ql 6(n-t)/z ln)

(m o d n ).

Consequently,n is an Euler pseudoprimeto the base D' tr Although every strong pseutloprimeto the base D is an Euler pseudoprime to this base, note that not every Euler pseudoprimeto the base b is a strong pseudoprime to the base b, as the following example shows. Example. We have previously shown that the integer 1105 is an Euler pseudoprimeto the base 2. However, 1105 is not a strong pseudoprimeto the base 2 since :2552: 2(llos-l)/2

I (mod 1105),

while 2 0 t 0 s - r ) / 2: 222 7 6 :

7gl + t

1 (mod ll05).

Although an Euler pseudoprime to the base b is not always a strong pseudoprime to this base, when certain extra conditions are met, an Euler pseudoprimeto the base D is, in fact, a strong pseudoprimeto this base. The following two theoremsgive results of this kind. Theorem 9.9. If n : 3 (mod 4) and n is an Euler pseudoprime to the base b, then n is a strong pseudoprimeto the baseb.

330

Quadratic Residues

Proof. From the congruence n = 3 (mod 4), we know that n-l : 22.t where t : (n-l)/z is odd' Since n is an Euler pseudoprime to the base b, it follows that

- ql (mod bt : 6..'-t)/2 n). f ln ) r\ tbl : Drnce l- | +1, we know that either bt = l (mod n) or ln ) -l (modn). Hence,oneof the congruences b' = in the definitionof a strong pseudoprimeto the base b must hold. consequently, n is a strong pseudoprime to the baseb. tr Theorem9.10. If n is an Euler pseudoprime to the base6 and lal

: -r.

l\ n l '/

then n is a strong pseudoprimeto the base b.

Proaf. We write n-l : 2't , where / is odd and s is a positive integer. Since n is an Euler pseudoprimeto the base b, we have

br-,t: 6,.'-r)/2 fa l (modn). ln) r) B u t s i n c el 4 I : - t , w e s e et h a t

ln)

b ' r-' = -l

(m o d r).

This is one of the congruencesin the definition of a strong pseudoprime to the base b. Since n is composite,it is a strong pseudoprimeto ihe base ,. tr Using the concept of Euler pseudoprimality, we will develop a probabilistic primality test. This test was first suggestedby Solovay and Stiassen [7g]. Before presentingthe test, we give some helpful lemmata. Lemma 9.3. If n is an odd positive integer that is not a perfect sguare,then there is at least one integer b with | < b I

ft,(b ,n) :

r , a n dl 4 | : - , , ln )

where

is the Jacobi symbol.

331

9 .4 E uler P s eudop ri me s

Proof. If n is prime, the existence of such an integer b is guaranteed by Theorem 9.1. If n is composite,since n is not a perfect square,we can write n : rs wher e ( r , s ) : I a n d r: p ' , w i th p a n odd pri me and e an odd positive integer. Now let / be a quadratic nonresidue of the prime p; such a / exists by Theorem 9.1. We use the Chinese remainder theorem to find an integer b with 1 < b 1 n, (b ,n) : 1, and such that b satisfiesthe two congruences b = t (mod r) b = | (mods). Then,

fal : (ul

|,bl"-(_r),-_r,

f;J l7): tp)

that : -' r ro,,ows and : , Since : [*] [*] [*] ii] t1],', Lemma 9.4. Let n be an odd compositeinteger. Then there is at least one integerD with | < b I n, (b,n) : 1, and r\ 6 6 - D / z1 l 4 | ( m o dn ) . ln) Proof. Assume that for primeto n, that

positiveintegers not exceeding n and relatively r) 6h-t)/2 :

( e . 1l )

d). l4 | (mon ln)

Squaring both sides of this congruence tells us that r t2

b,-t :

lAl

l 3 I = ( + l ) z : I ( m o dn ) , ln )

if (b,n) : I Hence, n must be a Carmichael number. Therefore, from a rr e d i s t i n c t T h e o r e m8 . 2 1 , w e k n o w t h a t n : Q t 4 z " ' e , , whereQt,Qz,...,Q odd primes. We will now show that

332

QuadraticResidues

6 h - t ) / 2= 1 ( m o d n ) for all integers b with I ( b ( n and (b,n) :1. integer such that 6 h -r)/2 :

-l

Suppose that b is an

(mod n).

we use the chinese remainder theorem to find an integer a | 1 a { fl, (a,n): l. and

with

a=b(modq1) a : - | ( m o d Q z Q s .. . q , ) . Then, we observethat

o.r2)

o G - 1 ) / 2-

6b-D/z:

_ l ( m o dq 1 ) ,

while

(e.13)

= I (mod ezQt...Q,). o(n-r)/Z

From congruences O . l D a n d ( 9 . 1 3 ) ,w e s e et h a t o h _ t ) / 2* contradictingcongruence(q.tt).

+ 1(modn),

Hence, we must have

6 (,-t)/2= I (m o d n), for all D with I < , ( n and (b,n) - r. Consequentry, from the definition of an Euler pseudoprime,we know that

6".-t)/2:|,aj : I (modn)

l, )

for all D with I < b ( n and (b,n) : r. However, Lemma 9.3 tells us that this is impossible. Hence, the original assumption is false. There must be at l e as tone int eger6 w i th | < b 1 fl , (b ,,D : l , and 6G-D/z1

|r l4 | (modn). tr

ln ) We can now state and prove the theorem that probabilistic primality test.

the basis of the

333

9.4 Euler Pseudoprimes

Theorem 9.11. Let n be an odd composite integer. Then, the number of positive integers less then n, relatively prime to n , that are basesto which n is an Euler pseudoprime,is less than 6fu) /2. Proof. From Lemma 9.4, we know that there is an integer b with I < b 1 n, (b,n): l, and

ql (mod n). 6b-r)/2 l f lnJ

(s.rq

Now, let e1,e2,...,e^denote the positive integers less than n satisfying 1 ( a ; ( n, ( ai, n) : l , a n d

r) n), afn-rtrzlLl (mod

(e.ls)

In )

for; : 1,2,...,m. Let rr{2,...,rm be the least positive residuesof the integers bayba2,...,ba^ I for modulo n. We note that the integers rj are distinct and (ri,n): j : 1,Z,...,frt.Furthermore,

, ( n - , ) t 2 1 ( m ond) . [+]

(e.16) For, if it were true that

,e-,)/2-

[+]

(mod n),

then we would have

$a)(n-,)/2 l+l r-"0,r This would imply that,

: t+l 6h-t)/2o(n-t)/2

I r 1J

and since (9.14) holds.we would have

[+]

(mod n ),

334

QuadraticResidues

_ fqI 6."-t\/2

l, )'

c ont r adic t ing( 9 .1 4 ). S inc e aj, j :1 ,2 ,...,m , s a ti s fi e s th e congruence (9.15) w hi l e r j, j : 1, 2, . . . , n, d o e sn o t, a s (g .to ) s h o w s ,w e know thesetw o setsof i ntegers share no common elements. Hence, looking at the two sets together, we have a total of 2m distinct positive integers less than n and, relativ-elyprime to n. Since there are Qh) integers less than n that are relatively prime to /r, we -filis can conclude that 2m < qfu), so that m < proves the eh)/2. theorem. tr From Theorem 9.1l, we see that if n is an odd composite integer, when an integer b is selectedat random from the integers 1,2,,....,n-1, th; probability that n is an Euler pseudoprimeto the base 6 is less than I/2. This leads to the following probabilistic primality test. The Solovay-StrassenProbabilistic Primality Test. Let n be a positive integer. Select, at random, ft integers bpb2,...,boLorr the integers i,2,...,r-r. For each of theseintegersbj,j : 1,2,...,k,determinewhether 6Q-t)/2

t+]

(modn)

If any of these congruencesfails, then n is composite. If n is prime then all these congruences hold. If n is composite, the probability that all k congruenceshold is less than l/2k. Therefore, if n passesthis test n is ,,almost certainly prime." Since every strong pseudoprime to the base b is an Euler pseudoprime to this base, more composite integers pass the Solovay-Strassenprobabilistic primality test than the Rabin probabilistic primality test, altirough both require O(kQag2n)3) bit operations.

9.4 Problems l.

Show that the integer 561 is an Euler pseudoprimeto the base 2.

2.

Show that the integer 15841 is an Euler pseudoprime to the base 2, a strong pseudoprimeto the base 2 and a Carmichael number.

3.

Show that if n is an Euler pseudoprimeto the basesa and 6. then n is an Euler pseudoprimeto the base a6.

335

9.4 EulerPseudoprimes

4.

Show that if n is an Euler pseudoprimeto the base b, then n is also an Euler pseudoprimeto the basen-b.

5 . Show that if n= 5 (mod 8) and n is an Euler pseudoprimeto the base 2, then r is a strong pseudoprimeto the base 2. 6.

Show that if n = 5 (mod 12) and n is an Euler pseudoprimeto the base 3, then n is a strong pseudoprimeto the base 3.

7.

Find a congruencecondition that guaranteesthat an Euler pseudoprimeto the base 5 satisfying this congruencecondition is a strong pseudoprimeto the base 5.

8.

Let

the

composite positive integer

, : pl,pi, . . . ph,

where pi : | *

kr ( kz ( < k-, and where n: pseudoprimeto exactly

n

have

prime-power factorization

where for zfqi i:1,2,...,ffi, | * 2kq. Show that n is an Euler

6" II ((n-l)/2, p1-t) j-l

different basesb with l < b ( n , w h e r e

(

12

D r : 1 1/Z It t

if kr:

1,

if kj < k and a; is odd for some j otherwise.

9.4 ComputerProjects Write programsto do the following: Determine if an integer passesthe test for Euler pseudoprimesto the base b. Perform the Solovay-Strassenprobabilistic primality test.

10 Decimal Fractions and GontinuedFractions

10.1 DecimalFractions In this chapter, we will discuss rational and irrational numbers and their representationsas decimal fractions and continued fractions. we begin with definitions. Definition. The real number a is called rational are integers with b * 0. If a is not rational. then

a - a /b, where a and b say that u is irrational.

If a is a rational number then we may write a as the quotient of two integers in infinitely many ways, for if ot : a b, where o f uni b are integers with b ;t' 0, then a : ka f kD whenever fr is a nonzero integer. It is easy to see that a positive rational number may be written uniquely as the quotient of two relatively prime positive integers; when this is done we say that the rational number is in lowest terms. Example. We note that the rational number ll/Zl also see that

is in lowest terms. We

-tt/-21 - tt/2r : 22/42: 33/63: The following theorem tells us that the sum, difference, product, and quotient (when the divisor is not zero) of two rational number is again rational.

337

1O.1 DecimalFractions

Then a + 0, a - 0' a9' Theorem 10.1. Let a and B be rational numbers. and a/0 (when P+0 are rational' : alb and B : cld' where Proof. Since a and p are rational, it follows that a * O' Then' each of the e, b, c, and d are integers with b * 0 and d numbers a * B : a /b + c l d : (a d * b c)/bd' a - 0: a/b - c/d : (ad-bc)lbd' a0-b/b)'k/d)-acfbd, a/0 : b /b) lG ld) : ad lbc @*0 ' denominatcr different is rational, since it is the quotient of two integers with from zeto. D We start by The next two results show that certain numbers are irrational' considering ,/T Proposition 10.1. The number '/T is irrational' prime integers Proof. Suppose that .,,6 : a lb, where c and b are relatively with b I 0. Then, we have 2:

a2lb2,

so that 2b2 : a2. Since 2lor,problem 3l of Section2.3 tells us that2la. b2:2c2.

Let q :2c,

so that

6. H ow ever, He n c e, 21b, , and b y p ro b l e m 3 l o f Se c ti o n2 .3 ,2 al so di vi des a nd b' This a b o t h d i v i d e c a n n o t we^know that 2 since G,b)':1, B contradiction shows that .6 is irrational' it We can also use the following more general result to show that .6 irrational. * cnlxn-t * Theorem 10.2. Let o( be a root of the polynomial x' with cs * 0. integers * cp * cs where the coefficientsca, ct,...,cn-r,are Then a is either an integer or an irrational number' and b Proof. Supposethat a is rational. Then we can write ot: alb whete a

338

DecimafFractionsand ContinuedFractions

are relatively prime integers with b o. x' + c r - 1x n- l * * c p * ,0 , w e h a v e b/b),

rc,_tG/6y,-t *

Since ot is

+cJa/D

a

root

of

*ca:0.

Multiplying by bn, we find that an + cn_pn-tb +

* c p b o - r + c s b n: 0 .

Since

' '!n',*n', ^:,,;;'i-. ,,n*'u* * , u'^o!,', u"rli-" o;ui, orp x,'-::'il Since p I b and b I an , we know that p

Hence, by problem 3l of I a, Sec t ion 2. 3, w: s e e th a t p l a . H o w i v e r, si nce (a, b) : l , thi s i s a contradiction which shows that b : t 1. Consequently, if a is rational then d : * o, so that a must be an integer. tr we illustrate the use of Theorem 10.2 with the following example. Example' Let a be a positive integer that is not the mth power of an integer, so that "\/i it not an integer. ThJn x/i i, irrationat by Theorem 10.1, since

" l. L e t a b e a re a l n u mb e r, a n d ret a:Ial be the i ntegerpart of a, so that r:o--[a] i s t h e f r a c t i o n a lp a r t o f a a n d o t : a * 7 w i t h 0 < 7 < I' From Theorem 1.3, the integer a has a unique baseb expansion. We now show that the fractional part ^yalso has a unique base 6 expansion. Theorem 10.3. Let 7 be a real number with 0 ( y ( l, and let b be a positive integer, b > | . Then can be uniquely written T as r:

; ci/bi j-r

where the coefficientsc; are integers with 0 ( c; < 6-l for j : 1,2,..w ., ith the restriction that for every positive integer l/ there is an integer n with n2Nandc, lb-1.

339

1 O,1 D ec im al F r ac t i o n s

series' We will use the In the proof of Theorem 10.3, we deal with infinite geometric series' following formula for the sum of the terms of an infinite < t. Then Theorem 10.4. Lets and r be real nurnberswith lr[

V o r i: a / 0 - ' ) .

j-0

(Most calculusbookscontain a proof') For a proof of Theorem 10.4,see [62]. We can now ProveTheorem 10'3' Proof. We first let c1: IbTl , l et so th a t 0 ( c r ( b_ 1 , s i n c e0 < b 7 < b . In a d di ti on, ^ fr : b l - c r : b ^ Y- l b l l ' sothat0(?r(land ^Y:

c1 , 7l ' 1 b b

^yg for k : 2,3,..., bY We recursivelYdefine c1 and ck :

[bfr-r]

and

nlk-t:+.+' so that 0(cr follows that

(b-t,

C"t

C1

7:T* Si n ce 0 ( ln (

and 0(rt

s i n c e0 ( b z t - r 1 b ,

Ur*

l, w e s e eth a t a 4 l r/b n

)tgntO' Therefore. we can conclude that

*

< I'

Cn

n,

+^Y,

b,

< l /b n . consequentl y,

:0.

Then'

340

DecimalFractionsand ContinuedFractions

lim

7:

n , /uo.

while (10.3) j:k+t

j-k+l

:(b-l)

l lLK+l

, "u | _ t/b

: l / b k, where we have used Theorem 10.4 to evaluatethe sum on the right-hand side of the inequality. Note that equality holds in (10.3) if and only if d j - c . i: b- l f o r a l ! i w i th 7 ) t 1 t, a nd thi s occurs i f and onl y i f d j : . b- l- and c i:0 fo r i 2 k + t. H o w e v e r,such an i nstancei s excl uded by the hypothesesof the theorem. Hence, the inequality in (tO.:) is strict, and therefore, (to.z) and (10.3) contradict (to.t). ttris shows that the baseb \ expansionof a is unique. tr

341

1 O.1 Dec im al F r ac ti o n s

The unique expansion of a real number in the form ). c1/bi is called the J-t

base b expansionof this number and is denotedby kp2ca..)6. To find the base b expansion(.cp2ca..)6 of a real number 7, wo can use the recursive formula for the digits given in the proof of Theorem 10.3, namely ^ fk : b y * -t - l bl t -J ,

ck : lbt*-J , where ^Yo: ^Y,for k : 1,2,3,...

Example. Let ( . c p2 c a ..)6 b e th e b a s e8 e x p a n s ionof l /6. Then tc 1: [ 8 ' ; l : o _

1,,

l_

c2:[8';'l:2, J

_ )_ ca:[8']l-5, J

_ tca:[8'Tl:2, J

cs:[8'?t:t,

^yt:8 -l : I + T, ^y2:s -2: 2 t' + ^y3:B -5 - I T' + 2 74:8 + -2 - T' I ^ys-s +-s: T,

and so on. We see that the expansionrepeatsand hence, t/6 : (1 2 5 2 5 2 5 ..)8. We will now discussbase b expansionsof rational numbers. We will show that a number is rational if and only if its base D expansion is periodic or terminates. Definition. A base D expansion (.cp2ct..)r is said to terminate if there is a : 0. positiveinteger n such that c, - cn*l - cn+z: Example. The decimal expansionof l/8, (.125000...)ro: (.125)ro,terminates. Also, the base 6 expansionof 419, (.24000...)o- (24)6, terminates. To describethose real numbers with terminating base b expansion,we prove the following theorem.

342

DecimalFractionsand ContinuedFractions

Theorem 10.5. The real number a, 0 < q I 1, has a terminating base D expansion if and only if a is rationaland a : r/s, where 0 ( r ( s

and every

prime factor of s also divides D.

Proof. First, supposethat a has a terminating base 6 expansion, (c 1c2...c)6 .

d:

Then Q:

b' so that a is rational, and can be written with a denominator divisible only by primes dividing b. Conversely,supposethat 0 ( a (

l, and

a:

rfs .

where each prime dividing s also divides 6. Hence, there is power a of D, say bN, that is divisible by s (for instance, take N to be the largest exponent in the prime-power factorization of s). Then bNot:b*r/t:er, where sa : bN ,, and a is a positive integer since slbr. (a*a^-1...aps)6 be the baseb expansionof or. ln"n a^b^*o^-tb^-r + . . . * atb*ag

Now

let

ar/bN :

a:

6u :

d*b--N

:

(.00...a m o m - t . . . a , a s )y .

+ am_tbm-l-fl

+

*a1b|-tr+ aob-N

Hence, a has a terminating base6 expansion. D Note that every terminating base b expansion can be written as a nonterminatingbase6 expansionwith a tail-end consistingentirely of the digit b-1, since (.cp2... c^)r- (cp2... pir cm-lb-lbi...lu (.ttl l l ...)ro . T h i s i s w h y w e requi re i n Theorem i n stanc e,( 12) t o: 10.3 that for every integer N there is an integer n, such that n ) N and

343

1 O.1 Dec im al F r ac ti o n s

cn# b-l; A base

without this restrictionbaseb expansionswould not be unique. instance b expansionthat does not terminate may be periodic, for I 1 3 : ( . 3 3 3 . . .1)s' | / 6 : ( . 16 6 6 . ' . t) o '

and | /7 : (.t+ztst 142857142857..) rc' if there are Definition. A base b expansion (.cp2ca..)6 is called periodic : ' N n for cn 7 positive integers N and k such that cn11 Wedenoteby(cp2...cv1-,']]-"*1-')6theperiodicbaseb (.cp 2...c7,1"') a' For instance'we have t -( t t...cN+t-rc.nv rclr...cry+

expanslon

r/3 : (.J)_.,0 , 7 1 6: ( . 1 6r)o, and

ll7 : (.taxsz)ro. begin Note that the periodic parts of the decimal expansionsof 1/3 and l/7 the proceeds immediately, while in the decimal expansion of l/6 the digit I b base periodic periodic pirt of the expansion. We call the part of a part periodic L*punsion preceding the periodic part the pre-period, and the thi period, where we take the period to have minimal possiblelength' (.ootorzr)r. Example. The base 3 expansionof 2/45 is (0 0 1) 3and t he per io di s (O t2 l )3 .

The pre-period is

The next theorem tells us that the rational numbers are those real numbers gives with periodic or terminating base b expansions. Moreover, the theorem of rational expansions b base periods of and the lengths of the pre-period numbers. Theorem 10.6. Let b be a positive integer. Then a periodic base b expansion representsa rational number. Conversely,the base b expansionof a rational ( 1, a: rfs, number either terminates or is periodic. Furthero if 0 < a : T(J where every where r and J are relatively prime positive integers, and s prime factor af T divides 6 and (U ,b) : 1, then the period length of the base of a is ordy b, and the pre-period length is .l/, where N is the b ""punrion smaliestpositiveinteger such that TlbN.

344

DecimalFractionsand ContinuedFractions

Proof. First, suppose that the baseD expansion of a is periodic,so that

a: (.crrr...r*ffi)o c1

ct I-J-

b62 C1

C'; I-J-

b62 where we have used Theorem 10.4 to see that €l

6tc

s^_ t"^ ojo

I

,r - . _

bk-l

bk

Since a is the sum of rational numbers, Theorem l0.l rational.

tells us that a is

Conversely,supposethat 0 ( a ( l, a : r /s, where r and s are relatively prime positive integers, s : T(J , where every prime factor of T divides b, Ql,b): 1, and I/ is the smallestinteger such-that Tlb* Since Tlb*, we have aT:

(10.4)

bN, where c is a positiveinteger. Hence bNa:bN

LTUU

or

Furthermore,we can write (r0.5)

ar

c

i:n*i,

where A and C are integers with

0 < I < 6N, and (c,u):

0 < c < u.

l. (the inequalityfor A followssince0 ( bNa: +

< bN.

U which results from the inequality 0 ( a ( I when both sides are multiplied by bN) . The fact that (C,tl): I follows easily from the condition (r,s) : l. Fr om T heor em 1 .3 ,A h a s a b a s eb e x p a n s i o nA : (anan_t...epo)u.

lf U : l, then the base b expansion of a terminates as shown above. Otherwise, Iet v : ord,ub. Then,

34s

1O.1 DecimalFractions

b'#:

(10.6)

Q u+ t )c U

+t,

where/ is an integer, since b' = | (mod U). However, we also have (t

(-

+ c' * al. b')

-C+j 62

b'+:b'l]+ U

(10.7)

LA

b'

that o' where(cp2ca...)6is the baseb expansion t,so c k : l b l t -J , where To :

C

T, f o r k

(10.8)

: 1 , 2 , 3 , . . . . F r o m ( 1 0 . 7 )w e s e et h a t

(-( b' *: U\

^ y k- b ' y t -r - l bl * -J

l r , b u - t+ c 2 b ' - z+

* r"] t ru.

( T, ( l, Equatingthe fractionalparts of (10.6) and (tO.S),notingthat 0 we find that C 4 t : -

Iv

u'

ConsequentlY,we seethat

^Yv: ": t' so that from the recursivedefinition of c1,c2,...we can concludeIhzt cpau: c1, nuta periodic baseb expansion for k : 1,2,3,.,.. Hence $ c - (n-rcr-Q6. U Combining (tO.+) and (10.5), and inserting the base b expansionsof A and

9. *. huu, U'

(ro.s)

bNa :

( a n a n - 1 . . . a t a o. c p 2 . . . c v 6) .

Dividing both sidesof (10.9) by bN, we obtain a : ( . 0 0... a n a n - r . . . o p o f f i )

u,

(where we have shifted the decimal point in the base b expansion of brya N

346

D e c i ma l F ra c ti ons and C onti nued Fracti ons

spaces to the left to obtain the base b expansion of a). In this base D expansionof a, the pre-period (.00...a,an-t...ipo)a is of length N, beginning with.A/ - h*1) zeros,and the period f.ngit, ir r. We have shown that there is a base b expansionof a with a pre-period of length r/ and a period of length v. To finish the proof, we must ,t o* that we cannot regroup the base b expansion of a, so that either the pre-period has length less than ry', or the period has length less than v. To do this, suppose that

q: (.crrr...trffi)u :

C1

b

Ct

*;*

*#*(*)la.

k f t M - t + c2 b M - 2 q

+cM)(bk-t) + Gyar6k-t+ bM (bk -t)

, cM+k -;m

f cTaap)

S i n c eq . : r f s , w i t h ( r , s ) : l , w e s e et h a t s l b M $ k _ D . C o n s e q u e n t l y , TlbM uTd ul(tk-o. H e n c e , M > N , a n d v l k ( f r o m T h e o r e mg . l , s i n c e bk = I (mod tD and v : ord,ub). Therefore,'the pre-period length cannot be less than ,^/ and the period length cannot be less than v. D We can use Theorem 10.6 to determine the lengths of the pre-period and p e r iod of dec im a l e x p a n s i o n s . L e t a : r/s , 0 < a ( l , and , :2" , 5r,, , where (1,10) : l. Then, from Theorem 10.6 the pre-period has length max (s1,s2)and the period has length ord,l0. Example. Let ot:5/28. since 2g - 22.7,,Theorem10.6 tells us that the prehas length 2 and the period has length ord710 : 6. Since rylt:d: (fiasll4z), 5/28 we seethat theselengthsare correct. Note that the pre-period and period lengths of a rational numb er r f s, in lowestterms, dependsonly on the denominators, and not on the numerator /. we observe that from Theorem r 0.6, a base b expansion that is not terminating and is not periodic representsan irrational number. Example. The number with decimal expansion o r: . 1 0 1 0 0 1 0 0 0 1 0 0 0 0 . . . , consisting of a one followed by a zero, a one followed by two zeros, a one followed by three zeroes, and so on, is irrational because this decimal expansiondoes not terminate, and is not periodic.

347

1O.1 DecimalFractions

so that its decimal The number d in the above example is concocted occurring numbers expansion is clearly not periodic. To show that naturally becausewe do 10.6, Theorem such as e and 7( are irrational, we cannot use No matter numbers' these of not have explicit formulae for the decimal digits cannot still we compute, we how many decimal digits of their expansions could period the because conclude that they are irrational from ihis evidence, be longer than the number of digits we have computed'

10.1 Problems l.

Show that dE is irrational a)

by an argument similar to that given in Propositionl0'l'

b)

using Theorem 10.2.

2.

Show that :/i

3.

Show that a)

+ ..6 is irrational.

log23 is irrational.

which logob is irrational, where p is a prime and b is a positive integer is not a Power of P rational or 4 . show that the sum of two irrational numbers can be either irrational. either rational or 5. Show that the product of two irrational numbers can be irrational. b)

6.

Find the decimal expansionsof the following numbers

d) 8lrs

a) 2/5 b) slt2 c) r2113 7.

e) lllll f) 1/1001.

Find the base 8 expansionsof the following numbers

d) r16 e) rlrz f) r122.

a) rl3 b) rl4 c) rls 8.

Find the fraction, in lowest terms, representedby the following expansions

a) .rz

b) .i

c) n.

348

D e c i ma l F ra c t i ons and C onti nued Fracti ons

9'

Find the fraction, in lowest terms, representedby the following expansions

a) (.rzi, b) (.oar6 l0' Il'

For which positive integers D does the base 6 expansion of l r/zro terminate? Find the pre'period and period lengths of the decimal expansions of the following rational numbers

il 7/t2 b) tt/30 c) t/7s 12'

c) (.iT),, d) (M),6.

d) rc/23 e) B/s6 f) t/6t.

Find the pre'period and period lengths of the base 12 expansions of the following rational numbers

a) t/+ b) r/B c) 7/ro

d) s/24 e) 17h32 f) 7860.

13' Let b be a positiveinteger.Showthat the period lengthof the base6 expansion of l/m is m - I if andonlyif z is piimeand, i, primitiveroot of m. " 14. For which primesp doesthe decimalexpansion of l/p haveperiodlengthof a)l b)2 c)3

d)4 e)5 f) 6?

15. Find the baseb expansions of

a) r/(b-r)

b) r/6+D .

16. Showthat the baseD expansion of t/G-1)z;, 1.9ffirJp1;u. 17. Showthat the real numberwith base6 expansion

(otzt.,.o-tlol rr2..)t, constructed by successivelylisting the base b expansions of the integers, is irrational. 18. Show that

+.#.#.#.#

349

1 O.1 Dec im al F r ac t i o n s

r9.

one. is irrational, whenever D is a positive integer larger than integers greater than one' Let byb2,fur... !s an infinite sequence of positive as represented be can Show that every real number

,o*?.#+#;+, ( ct ( bp for k : I'2'3'"" where cs,c1,cz,c!,...are integers such that 0

20.

a)

Show that every real number has an expansion CrCtrt+

to+l!

*

zl* 3!

: are integers and 0 ( ct ( k for k where cs,c1,c2,c!,-.-

l'2'3'""

of the type show that every rational number has a terminating expansion (a). describedin Part llp is ('t,tr'-oJ" Supposethat p is a prime and the base b expansionof is p - l. show that llp of expansion b base so that the period length of the ( p, then. if z is a positive integer with I ( ln

b)

Zl.

( 2...c1sacP) 6' m /p : (.cya1...coac where k : indtm modulo P. has an even period length'

2 2. Show that if p is prime and l/p - ('ffi)6 k :2t,

f o r . , ;:r 1 , 2 , " ' , t

thenci * ci+t: b-l

whete h and' k 2 3 . The Farey series Fn of order n is the set of fractions hlk (h,k): are integers,0 ( ft < k ( n, and include 0 and I in the forms i

1, in ascendingorder' Here, we

and I respectively' For instance, the Farey I

seriesof order 4 is

3l

0l112

T ' T , T ' T ' 7 , 7 ,T a)

Find the Farey series of order 7.

b)

Show that if a/b bd - ac :1.

c)

Show that if a/b, c/d, and e/f then

and c/d

c

are successiveterms of a Farey series' then

a*e

are successiveterms of a Farey series,

7- E7'

3so

DecimalFractions and ContinuedFractions

d)

Show that if a/b ordern, then b*d

and, c/d ) n.

are successiveterms of the Farey series of

24. Let n be a positiveinteger,n ) l. Show that I not an integer. l0.l

ComputerProjects

Write computerprogramsto do the following: I' 2'

Find the base 6 expansionof a rational number, where b is a positive integer. Find the numerator and denominator of a rational number in lowesr rerms from its base b expansion.

3'

Find the pre-period and period lengths of the base D expansion of a rational number, where b is a positive integer.

4'

List the terms of the Farey series of order n where n is a positive integer (see problem 23).

10.2 Finite Continued Fractions Using the Euclidean algorithm we can express rational numbers as continued fractions. For instance, the Euclidean algorithm produces the following sequenceof equations:

62:2.23 + 2 3 : l . 1 6+ 1 6: 2 - 7 + 7:3-2 +

lG 7 2 l.

When we divideboth sidesof eachequationby the divisorof that equation,we obtain 62:r*16:,)r 23 23

I nlr6 I ?3-:t+L:t* 16 16 16/7 16 : I I Z : r + + 7 7 7/2 L

+ !. +:3 2 2' By combiningtheseequations, we find that

351

1 O.2 F init e Cont inu e d F ra c ti o n s

62 :2+ 23 :2+

1 23116 t

r '- L :

I

I

rc17 I

:2*

1+h I

:2*

1+

2++3*;

in the abovestring of equationsis a continuedfraction The final expression of 62123. expansion We now definecontinuedfunctions' of the form . A finite continuedfraction is an expression Definition I aot

atl ctz *

+-

1 a n - rt L

an an positive' The real ale real numbers with Q1,Q2,Q3',"'' where Qg,a1,a2,...,an continued fraction' the quotients of numbers ej,a2,...,Q'nare called lhe partial as,c r,..., an are all numbers real the if The continued fraction is called simple integers. we use the Because it is cumbersome to fully write out continued fractions, the above in fraction to represent the continued notation Lso;a1,e2,...,Ctn| definition. a We will now show that every finite simple continued fraction represents can number rational every that will demonstrate we rational number. Later be expressedas a finite simple continued fraction'

352

DecimalFractions and ContinuedFractions

Theorem l0'7 ' Every finite simple continued fraction represents a rational number. Proof' we will prove the theorem using mathematical induction. For n : 1 we have [ao;arl:oo+

I *aoar*l al og

which is rational. Now assume.that for the positive integer k the simple continuedfraction [ag;at,e2,...,eklis rational whlnevst as,or,...,okare integers with a r,...,ak positive. Let as,at,...,ek+tbe integers with er,...,ek+t positive. Note that [ a g . a 1 , . . . , a k +: t la g + Ia;a2,..., a1r.a1ra1l

By the induction hypothesis,[a ria2,...,ek,ek+r] is rational; hence, there are integers r and s, with s*0, such that this continued fraction equals r/s. Then l a o ; a 1 , . . . ,a k , o k + t l : a g +

I r/s

agr*S

which is again a rational number. tr We now show, using the Euclidean algorithm, that every rational number can be written as a finite simple continued fraction. Theorem 10.8. Every rational number can be expressed by u finite simple continued fraction. Proof. Letx:a/b w h e r ea a n d b a r e i n t e g e r s w i t h b > 0 . L e t r s - a and r't : b. Then the Euclidean algorithm prodr.", the following sequenceof equations:

353

1O.2 Finite ContinuedFractions

rO : r| :

r1Q1* 12 r2Q2* 13

Q 1r2 ( tt, 0(131rr,

12:

r3Qtl 14

0(ra113,

: ln-3

:

fn'ZQn-Z*

:

tnQn

fn-1Qn-1*fn

fn-Z: fn-l

0(rn-11tn-z, 0(rnlrn-t

fr-t

Writing these In the above equations 4z,Qt,.",Qn are positive integers. equations in fractional form we have L: b

lo

tt

:

tt:

.

6

13

I

q2+;:Q2.Trt

r2 rZ:

ta,

nr*;:et*

r3

ln-3

:

rn-2 ln-2: rn-l' fn-l

I

Qr*;:qt+

/1

tn-l

:

Qn-2

tn-2 L Qn-l t

,n

I

rrt^

I -L -t rn-2/rn-t : - nq- -n.-+l 4- , n - r , / r ,

;

: ,QN

rn

first equation' Substitutingthe value of r1/r2from the secondequation into the we obtain

(l 0.10)

al T:4tt

, 4z r

t ,rlry

into (10.10) Similarly, substituting the value of r2fr3 from the third equation we obtain

3 54

DecimalFractionsand ContinuedFractions

c

Qr*

b

Qz*

Q t *+rilrt Continuing in this manner, we find that

T:

I

q ' t+ Qz* Qt*

*

Qn-t

,l Qn

Hence

q n l . T h i s s h o w s t h a t e v e r yrational number can be t:rnriQz,..., written as a finite simple continuedfraction. ! We note that continued fractions for rational numbers are not unique. From the identity an :

Gn-l)

+

we seethat [ a g ; a1 , e 2 , . .e. ,n _ t , o n l: I a g ; a 1 , c t 2 ,e. .n. ,_ t , e n whenevera, )

L

Example. We have 1

#I I

: [ o ;I , l , l , 3 1: [ o ; l, l , l , 2 ,I ] .

In fact, it can be shown that every rational number can be written as a finite simple continued fraction in exactly two ways, one with an odd number of terms, the other with an even number (see problem 8 at the end of this se c t ion) . Next, we will discussthe numbers obtained from a finite continued fraction by cutting off the expressionat variousstages. Definition. The continued fractions [as;a1,o2,..., a1l, where ft is a nonnegative integer less than n, is called the kth convergenr of the continued fraction

355

1O.2 Finite ContinuedFractions

by Ct ' [ao;a1,e2,...,Qnl The kth convergentis denoted the convergentsof In our subsequentwork, we will need some properties of starting with a properties, these a continued fraction. We now develop formula for the convergents. an be real numbers,with a 1;a/;...,a, positive' Theorem 10.9. Lel ag,a1,e2,..., recursivelyby Let the sequencesP0,Pt,..., Pn and qs,qt,"', Qn be defined Qo: I Po: aO : q1: ar a s o l * l Pt and P * : o k P k - t t P*-z

Qk:

q*-z

apQt-t t

t k : I' ao;at,.' .,okl i s gi ven by fo r /c : 2, 3, . . . ,n . Then the k th c o n v e rg e n C Cp -- P*lqr' : 0 proof. we will prove this theorem using mathematical induction. For k we have Co: lael : asll : Polqo. For k : l, we seethat Cr : l a o ;a 1 l : a s + ! :

a1

aoat*l a1

:Pt Qt

Hence. the theorem is valid for k : 0 a n d k : l where Now assume that the theorem is true for the positive integer k Thismeansthat 1n 2 Crj*ro )

Cz*'

-numbered so that every odd-numberedconvergentis greater than every even convergent. tr

360

D e c i ma l F ra c ti ons and C onti nued Fracti ons

Example. Consider the finite simple continued fraction 12:3,1,1,2,41. Then the convergentsare CoC1 CzC: : C+: Cs :

2/l-2 7/3:2.3333... 9/4:2.25 16/7:2.2857... 4l/lS:2.2777... ftA /7 9 : 2 . 2784....

We seethat Co : 2 1 Cz: 2.25I Ca : 2.2777... ( Cs :2.2784... ( Cr :2.2957... ( Cr :2.3333...

10.2 Problems l'

2'

Find the rational number, expressedin lowest terms, representedby each of the following simple continued fractions

a) b) c)

IZ;ll [t;z,z] [0;5,0]

e) f) e)

[ r ;r ] [ l ;l , l ] [ I ; t , l, l ]

d)

5 , 1] [3;7,1

h)

[ l; I ,l ,l,l ].

Find the simple continued fraction expansion not terminating with the partial quotient one, of each of the following rational numbers

il

6/s

d)

b) c)

22t7 t9/29

e) f)

slsss

-4311001 873/4867.

Find the convergentsof each of the continued fractions found in problem 2 . Let up denote the kth Fibonaccci number. Find the simple continued fraction, terminating with the partial quotient of one, of u1,-,1fup,where ft is a positive lnteger. 5. Show that if the simple continued fraction expressionof the rational number a , a . ) 1 , i s [ a 6 ; a t , . . . , a kthen l, the simple continued fraction expressionof l/a is l};a o,ar,...,a k'l. 6.

S h o w t h a t i f a e * 0, then

1O.3 InfiniteContinuedFractions

P*/p*-r

361

: I o o i a * - t ., - . , a 1 , a s l

and q* / q tr-r: I'au:ar-r,"',a2,a11, convergentsof the where Ck-r: p*-t/qrr-r and C* : pt lq*,k ) l,are successive : (Hint: a*P*-1 * pp-2 to relation the Use P* continued fraction la6;a1,...,an1 * I / ( p x t / p * ) . a r s h o wt h a t p t / p * - r : of the 7 . Show that q1,) u1, for k:1,2,... where c*: p*lqr is the kth convergent and all denotesthe kth Fibonacci number' simple continued fraction las;a1,...,an1

8 . Show that every rational number has exactly two finite simple continued fraction expansions. be the simple continued fraction expansion of rls where 9 . Let lao;ar,a2,...,a211 Show that this continued fraction is symmetric, i'e. I and r)l : o s : a 2 1 t a t a n - t d 2 : a n - 2 , . .i.f, a n d o n l y i f s l ( r 2 + t ) i f n i s o d d a n d s l ( r 2 - t ) i f n is even. (Hint: Use problem 6 and Theorem 10.10). (r,s):

10.

Explain how finite continued fractions for rational numbers, with both plus and minus signs allowed, can be generated from the division algorithm given in problem 14 of section1.2'

ll.

be real numbers with a r,o2,...positiveand let x be a positive Let as,ar,a2,...,ak real number. Show that Ias;a1,.'.,ar,l1 lao;a6--.,a1,*xl if k is odd and I a s ; a 1 , . . . , a t>1 [ a o ; a 1 , . ' . , o 1 r * xi f] t i s e v e n .

10.2 Computer Projects Write programs to do the following: l.

Find the simple continued fraction expansionof a rational number

2.

Find the convergentsof a finite simple continued fraction.

10.3 InfiniteContinuedFractions . Supposethat we have an infinite sequenceof positive integersQo,Qt,ay,... How can we define the infinite continued fraction Las,at,a2,...l? To make sense of infinite continued fractions, we need a result from mathematical analysis. We state the result below, and refer the reader to a mathematical analysisbook, such as Rudin lezl, for a proof. Theorem ll.l2. Let xs,x r,x2,... be a Sequenceof real numbers Such that xo ( x r ( x z ( . . . a n d x 7 , < u fo r k : 0 ,1 ,2 ,... for somereal number u, or x o 2 x r 2 x z 7 . . . a n d x t 2 L f o r k : 0 , 1 , 2 , . . . f o r s o m er e a l n u m b e rl .

362

D e c i ma l F ra cti ons and C onti nued Fracti ons

Then the terms of the sequencexu,xr,x2,... tend to a limit x, i.e. there exists a real number x such that

14to:"' Theorem 10'12 tells us that the terms of an infinite sequencetend to a limit in two specialsituations,when the terms of the sequence are increasingand all less than an upper bound, and when the terms of the sequenceare decreasing and all are greater than a lower bound. We can now define infinite continued fractions as limits of finite continued fractions, as the following theorem shows. Theorem 10.13. Let as,e1,ct2,...be an infinite sequenceof integers with ar,Qz,... positive, and let ck : lag;a1,a2,...,e1a1Then the convergents cp tend to a limit ot.i.e

J4to:"' Before proving Theorem l0.l 3 we note that the limit a described in the statement of the theorem is called the value of the infinite simple continued fraction [as;at,o2,...1. To prove Theorem 10.13, we will show that the infinite sequenceof evennumbered convergents is increasing and has an upper bound and that the infinite sequenceof odd-numbered convergentsis decreasingand has a lower bound. We then show that the limits of these two sequences,guaranteedto exist by Theorem 10.12,are in fact equal. W e now will p ro v eT h e o re m 1 0 .1 3 . Proof. Let m be an even positive integer. From Theorem 10.1l, we seethat

cr ) ct) cs ) ca1cz1cq1

)

C^-t

1C^,

and C2i 7 Czn+t whenever 2j 4 m and 2k + | . Cs) co(czlc+(

) C z n - t ) C zn+ , 1 Czn-z 1 C2n I

and czi ) Cz**t for all positive integers j and k. we see that the hypothesesof Theorem rc.12 are satisfied for each of the two sequences C 1, C3, C2, . . and . C s ,C z ,C 4 ,.... H e n c e , th e sequenceC 1,C 3,C 5,...tends to a

363

1O.3 lnfinite Continued Fractions

a2 ' i'e' limit d1 and the sequenceCs,C2,C4,"' tends to a limit : dr )i*c"*r and : o(2'

)*c"

Using Our goal is to show that these two limits a1 and oQ are equal' Corollary 10.2 we have (-l)(z'+tl-t : lzn*t - Pzn * zt n C z n +-r C n Qzn+t

Qzn

Qzn+lQz,

Qzn+lQzn

Since e* 2 k for all positive integers /c (see problem 7 of Section 10.2), we know that I ( z n + l )Qn) ezn+rQzn and hence Czn*t - Cz,

Qzn+tQzn

tends to zero, i.e. nlim

(C z ra 1- C 2 n ) : 0 .

s 1,C 3 ,C s ,...a n d C g ,C 2 ,C 4 ,...have the S amel i mi t, si nce H e n c e,t he s equenc eC

j*

(cr,*t - cz) :

,lg

Czn*t-

,lg

cz, : o.

Therefore ayr: aq, z11dwe conclude that all the convergentstend to the limit d : (rr : dz. This finishesthe proof of the theorem' D Previously, we showed that rational numbers have finite simple continued fractions. Next, we will show that the value of any infinite simple continued fraction is irrational. Th e o r em 10. 14. Le t o s ,,o 1 ,e 2 ,...b e i n te g e rs w i th a1,Q2,...posi ti ve. Then Ia o ;ar , , a2, . . . 1is ir r ati o n a l . Proof. Let a : las;at,ctz,...land let

364

DecimalFractionsand ContinuedFractions

Cr : pr/qp : [ a o ; at , . . . , a k l denote the /cth c o n v e r g e n t o af . W h e n n is positive a integer,Theorem 10.I I shows that C2, ( a ( C z r + t , s o t h a t 0 ( a - Czn I

Czn*t - Czo .

However, from Corollary 10.2, we know th a t I

Czn*t - C2n :

' 4zn+tQzn

this meansthat Pzn

0(a-Czn:a-

4zn

a Qzn+ tQzn

and therefore, we have 0 1 a q 2 , - p z n 1 l / qzr+ t . Assume that a is rational, so that ot : e /b where a and b are integerswith b + A. Then oaoQr" b

-pzn
b, so that b/Qzr+t < I . This is a contradiction,sincethe integer aQzn- bprn cannot be between0 and I . We concludethat a is irrational. n We have demonstrated that every infinite simple continued fraction representsan irrational number. We will now show that every irrational number can be uniquely expressedby an infinite simple continuedfraction, by first constructing such a continued fraction, and then by showing that it is unique.

365

1O.3 Infinite Continued Fractions

and define the sequence Theorem f0.15. Let a: cvObe an irrational number Q0 ,Qt, Q 2, ' . . r eCuf s iv e l bYY Qk : lapl,

c r k + :l I / b t - a )

continued fo r k : 0, l, 2, . . . . Th e n a i s the value of the infinite, simple fra cti o n Lag;ar , az , - ..1 .

is an integer Proof. From the recursivedefinition given above, we see that ap that induction mathematical using show easily we can for every k. Further, : a is irrational' Next, if d0 that note first We k. every for a7, is irrational is also we assume that a1, is irrational, then we can easily see that a,p1' relation irrational, sincethe dk+r:l/(at-a*)

impliesthat (10.12)

otk:A**Ls

I qk+l

and if d;611were rational, then by Theorem10.1,a7. would also be rational' Now, since a7, is irrational andap is an integer,we know that 47, I at, and aplatlap*|,

so that 0(a1-ap t (we know that sP*-rQr is a nonzero integer sincer ls #pplqr), it follows that

372

DecimalFractionsand Continued Fractions

-x

|

sQ*

lspt-rq*l

-

': lor tl sl ,

sQ*

lqo ll

I

qrl

.l*l 2tq*

F:l

2s2

(where we have used the triangle inequality to obtain the second inequality above). Hence, we seeth a t t/2sqp I

t/2s2

Consequently, Zsqp ) 2s2, which implies that q1, ) s, contradicting the assumption. tr

10.3 Problems L

2'

Find the simple continued fractions of the following real numbers

a)

,rf2

b)

^f3

c) d)

-,/i r+.6

.

Find the first five partial quotients of the simple continued fractions of the following real numbers

a) b)

1/, 2r

c) d)

(e-l)/(e+l) (e 2 -t)/(e 2 + D .

Find the best rational approximation to zr with a denominator less than 10000. The infinite simple continued fraction expansionof the number e is

e : l 2 ; 1 , 2 , 1 , 1l , 1 4 , 61, ,1 , g , . . . 1 . a)

the first eight convergents of the continuedfractionof e

1 O.3 I nf init e Cont in u e d F ra c ti o n s

b)

373

less than Find the best rational approximation to e having a denominator 100.

expansion 5 . Let d be an irrational number with simple continued fraction -ot is Show that the simple continued fraction of o : loo;ot,a2,...f at: 1' a n d I [ a s l ; a 2 l l d v " ' l i f a 1 2 [-as-l;1,a,-l,as,a3,...lif simple 6 . Show that if p*lqx and,p1,a/q1a1 2f€ consecutive convergents of the continued fraction of an irrational number a, then

l o - p r/q rl < tl z q o '

( l /2 q l a. l o - p o * r/q o * ,1 ( Hint : F ir s t s h o wth a t l o - p r* r/q * * ,1+ l o - pol qol- l po* r/q& +- r pr,/qtl : l/q*q**t using CorollarY 10.2.) 7.

Let a be an irrational number , a ) I . Show that the kth convergent of the simple continued fraction of l/a is the reciprocal of the (k-t)th convergent of the simple continued fraction of a .

8 . Let a be an igational number, and let pllei denote the jth convergent of the simple continued fraction expansion of a. Show that at least one of any three consecutiveconvergentssatisfiesthe inequality

la- pileil < t/G/-sqil. Conclude that there are infinitely many rational numbers plq, where p and q are integers with q # O, such that

l''- plql 0. c # 0. and b not a perfect square such that

a:

(a+Jb)/c.

fur*cl)+rJb (at rcu) +t Jt I Gr + cil + r JF lI ht + cil -t.'.6 | IGt *cu) +t .,/blt(at +cu)-t ./n I lGr *cs\ (at*cu) -rtblt[r (attcD -t Gr *cl)l../T (at *cu)2-t2b

377

1 O.4 P er iodic Cont i n u e d F ra c ti o n s

i s a q u a drati ci rrati onal ' unl essthe H e n ce ,f r om Lem m a l 0 .l (ra * s )/Qa + d rational' tr G is zero, which would imply that this number is ;;;d;i";, "t fractions of quadratic In our subsequentdiscussionsof simple continued irrational' quadratic irrationals we *iil use the notion of the conjugateof a -- (a+JD lc be a quadratic irrational' Then the coniugate Definition. Let a : (a -J b )l c ' o f a , denot edby o' , i s d e fi n e db y a ' the polynomial Lemma 10.3. If the quadratic irrational d. is a root of the conjugate a', is Axz + Bx * C : 0, then the other root of this polynomial of a. Proof. From the quadratic formula, we see that Axz+Bx*C:0are

the two

roots of

_B*[EW ZA of If a is one of these roots, then a' is the other root, because the sign is reversedto obtain a' from a. tr tr4AC The following lemma tells us how to find the conjugates of arithmetic expressionsinvolvingquadratic irrationals' L e mma 10. 4. I f a' : (a ftb ffd )/c 1 irrationals,then (i)

(a1+a2)' -- al t

(ii)

(a;c.2)'

a n d ,,2 : (a2* bzJd)f cz are quadrati c

a'2

: o| - d'2 : d'td2

(iii)

(ap)'

(iv)

(c"rlc.)':

a't/o.z.

parts are easier. The proof of (iv) will be given here; the proofs of the other reader' the problems for as section this of These appear at the end Proof

of

(iv).

Note that

378

D e c i m a l F ra cti ons and C onti nued Fracti ons

G ftbr.'./Z) /r,

t

".'

v l l q )

Gr+bz,/cl)/cz _ cr(a ,+b r/7) G 2-.bz,/T) lb2)''/7

: ,, While , - t--, " "

.^lrsl---7

G;brE)/cz (or-brrE) /cz cz(arbtQ)Gr+br,/V)

_

c {a 2- b 2,/7 ) (a z+ b 2,/7 ) k z a p z -c z b ftz d ) - (czazbrczaft)fi

Hence (at/a)' : or'r/a'2. D The fundamental result about periodic simple continued fractions is Lagrange's Theorem. (Note that this theorem is different than Lagrange,s theorem on polynomial congruncesdiscussedin Chapter 8. In this chapter we do not refer to that result.) Lagrange'sTheorem. The infinite simple continued fraction of an irrational number is periodic if and only if this number is a quadratic irrational. We first prove that a periodic continued fraction represents a quadratic irrational. The converse,that the simple continued fraition of a quadratic irrational is periodic, will be proved after a special algorithm for obtaining the continued fraction of a quadratic irrational is developed. Proof. Let the simple continued fraction of a be periodic, so that a : la g;at,,e2,..,,a N -r,ffi| Now let 0 : la1s;aN+r,...,41r+ft l Then

379

1 O.4 P er iodic G on ti n u e d F ra c ti o n s

g : lal;aN*I,...,4N **,01, and from Theorem 10.9,it follows that (10.13)

^ t) -

1 P * tP* -t oq*tq*-r'

Since the where p*lq* and p1r-r/Q1r-1ata convergentsof Ia11;av"1'"''oru+kl' (tO't3) we from and simple continued f.u.tlon of p is infinite, B is irrational, have

qr,02t Qr,-r-P)0 - P*-r : a' so that p is a quadratic irrational. Now note that a : l a g ;a1 ,Q 2 ,...,Q N -r, 01, so that from Theorem 10'9 we have

'a;;:fr; ' 0pr,r-ftPN-z

Since B where pN-t/qN-1 and pr,t-zlqN-2uteconvergentsof [ao;a t.a2'"''o7'1-11' quadratic a is also a that us tells 10.2 Lemma is a q*Oruii. irrational, irrational (we know that at is irrational because it has an infinite simple continuedfraction exPansion). D To develop an algorithm for finding the simple continued fraction of a quadratic irrational, we need the following lemma' Lemma 10.5. If a is a quadratic irrational, then d. can be written as

: @+,/V)/Q, w h e r eP , Q , a n d d a r e i n t e g e f s , Q* O , d QIQ-P2) .

> O , d i s n o t a p e r f e c ts q u a r ea, n d

Proof. Since a is a quadratic irrational, Lemma 10.1 tells us that

, : (a+Jb)lc, where a,b, and c are integers, b > 0 , and c # 0 . We multiply both the numerator and denominator of this expressionfor q by Itl to obtain

380

DecimalFractionsand Continued Fractions

a.-

(wherewe haveusedthe fact that lrl: -,tr\. Now let p : alcl, clcl, e: a n dd : b c 2 . T h e np , e , a n dd a r e i n t e g e r s , l 0 s i n c e, 7 0 , d e >O (since6 > 0), d is not iperfect sinceb is not a perfectsquare,and f i n a l l ye l @ - p \ s i n c ed - p 2 : 6 r z 'lQuare oirz :;rbjoif:;T'(ilorl. n We now presentan algorithmfor findingthe sample continuedfractionsof quadraticirrationals. Theorem 10.19. Let a be a quadratic irrational, so that by Lemma 10.5there are integers Ps,Qs, and d such that

@o+,/7)/Qo , whereQ0*0,d > 0, d is not a perfectsquare, and eel @-p&). define

Recursively

dk:(ro+,/7)/Qr, C tk: [a 1 ], Pk+r:atQt-Pk, Q**r : (d-roL*t)/Q*, for k : 0,1,2,... Thena : fag;at,a2,...1. Proof. using mathematical induction, we will show that pk and e* are i n t e g e r sw i t h Q 1 ,* 0 a n d e * l @ - r p , for k:0,r,2,.... F i r s t ,n o t e t h a t t h i s assertion is true for k : 0 from the hypothesesof the theorem. Now assume that P1 and Qp are integerswith e* * 0 and e*l@_p?i. Then Pk+r:

a*Qt - Pp

is also an integer. Further,

Q*+r: @-rf *r11qo : [d-(o*Q,,-pr)2]/e* : @-rfi/Qo + (2a1,P1,-a?er). Since Qrl@-pil, by the induction hyporhesis,we see that Qpal is an integer, and since d is not a perfect square, we see that d I Pi, so that t o . Since Q*+t : @-rf*;/Qo

Q* : U-rf*1/Qo*t

381

1O.4 PeriodicContinuedFractions

we can concludethat Q1,ql@-pt*t)

. This finishesthe inductive argument.

To demonstratethat the integerses,a1,a2,...are the partial quotientsof the simple continuedfraction of a', we use Theorem 10.15. If we can show that o ( k + t:

llbr-ap),

t he n w e k n o w th a t a : fa s ;a 1 ,a 2,...1.N ote that

fork:

ap-ak:

Pk + ,/7

-ap

Af

: l^/7 - G*Qr,- P)llQ* : G/7 - pt +) lQ*

: G/V- P**')(JV+ P*+)/er,G/T+ P**r) : @-rl*)/Q*QI + Pr*r) : Q*Qr,n/Qr,G/7+ Pt*,) : Q**r/('/i + Pr,*) : lla*+r , where we have used the definingrelation for Qp* to replaced-Ppzar with that a : las;a1,e2,...f . D QtQ**r. Hence,we canconclude We illustratethe use of the algorithmgiven in Theorem10.19with the followingexample. Example. Let a : Q+J1)/2 . Using Lemma 10.5,we write

: G+.,/N)/4 wh e r e we s et P o : 6 , Q.o : 4 , a n d d : 2 8 . H e n ceoo: 2'4-6:2, (28-22)/4:6,

a1

Qr

: :

P2

:

l'6-2:4,

ot2

Og-+2)/o:2,

A2

Pr

Qz :

O1

[a] : 2, and

Q + ..E)/e, : r, IQ+,/z$/61 G+,,/Tg/2,

t

382

Decimal Fractions and Continued Fractions

P3 Qt :

4'2-!:4, Qg-+2)/2:6

d3 : o3 :

e+.,m)/6, tG+6>Jil:r,

P4 Qq

: -

l'6-4:2, (28-22)/6:4,

d4 a4

: :

e+rFZ$/q, t7+.'-z$/il:

Ps Qs

-

l'4-2:2, Q8-22)/4:6,

a5 a5

: :

e+r/-Z$/6, t ( z + , , / N ) / 6 :1 l ,

andso,with repetition, sincepr:

t,

p5 and

er: es. Hence,we seethat : G + . n ) / 2 I 2 ; 1 , 4 , 1 , 1 , r , 41,,r. ,.I . : I2;1,4,1,11.

We now finish the proof of Lagrange'sTheoremby showingthat the simple continuedfractionexpansion of a quadraticirrationalis periodic. Proof (continued). Let a be a quadraticirrational,so that by Lemma 10.5 we can write a as o : (po + .,8) /eo . Furthermore, by Theorem10.19we haveo: dk ap Pwr Q*r

: : : :

lao;ar,ez,...l where

(r1, + ,,/7)/Q* , [apl, atQ*-Pk*t, Q -rf *1 /Qo*r,

fork: Since a

: Ias;a' "")'lrl,o; ]:ffi _ll;l Ijl "_

that * q*-).

Taking conjugates of both sides of this equation, and using Lemma 10.4, see that

(ro.r+)

o' : (pr,-p'* * p*-) /(qt,-p'n * q * - ) .

When we solve (tO.t4) for ol1,, ws find that

383

1O.4 Periodic Continued Fractions

,

dk:

( - P*-zI -ex-,l" tr- | qk^ t ,

p*t t

,*t l to a as k tends to Note that the convergents p*-z/Q1r-2 and p*-rlqrr-t tend infinity, so that

| , - P*-z la. Q*-z I

t fr' -

P*-t

I

Q*-t

Since tends to 1. Hence, there is an integer N such that a ' * 1 0 f o r k > N . o ' t > - 0 for k > l, we have Pp

otk-Otk :

+ Jd Q*

Po-Jd Q*

Zfi r0. Qr

sothatQ*> 0fork>N. SinceQ*Qrr*,- d - P?*r, we seethat for k 2 ly', 0t ( Q*Q**r-- d P?*t < d . Alsofork>N,wehave Pl*, (d:

Pl*t-Q*Qx*r,

sothat - ,/7 I P*+r < -,/7. - -,[d < P*+r N , we see that there are only a finite number of possiblevalues for the pair of integers Px,Qx for k > N . Since there are infinitely many integers k with k > N,therearetwointegersi andT suchthatPi:Pi andQi:Qi : with i < j . Hence, from the defining relation for cu;., we see that o(i di

Hence conseque "t'*:;:;,";:"',i: ,-,,i:"',oi,*,'lo,ol,.;:,,':,.:,: i:i-,,, :

I a g ; al , o 2 , . . . , a i - 1 , Q i i,+o 1 , . . . ,ia- t l .

This shows that a has a periodic simple continued fraction.

D

384

DecimalFractionsand ContinuedFractions

Next, we investigate those periodic simple continued fractions that are purely periodic, i.e. those without a pre_period. Definition. The continued fraction [as;at,ez,...f is called purely periodic if t h e r ei s a n i n t e g e rn s u c h t h a t a 1 r : e n t k , f o r k : 0 , 1 , 2 , . . . , s ot h a t

lag;at,az,...l:Iffi. Example' The continued fraction tl;jl: [2;2,41: JA is not.

(t+.1:) /2 is purely periodic while

The next definition and theorem describe those quadratic irrationals with purely periodic simple continued fractions. Definition. A quadratic irrational at if called reduced if -l ( a' ( 0, w h e rea ' i s th e c o n j u g a teo f a .

a )

I

and

Theorem 10.20. The simple continuedfraction of the quadratic irrational a is purely periodic iI-and only if a is reduced. Further, if a is reduced and a: l,as;at,e2,...,enl then the continuedfraction of - l/oi i, to;o,,_ffi Proof. First, assume that a is a reduced quadratic irrational. Recall from Theorem 10.15 that the partial fractions of the simple continuedfraction of a are given by ek :

fork:

where ato: d

lapl, otk+t :

l/@tr-o*),

We see that l/qt+t:ek-ak,

and taking conjugates,using Lemma 10.4, we see that

(ro. rs)

l/a'*+t:

c , ' k- a 1 r .

we can prove, by mathematical induction, that - I ( a1 ( 0 for k:0,1,2,.... F i r s t , n o t e t h a t s i n c e c . 0 : a i s r e d u c e d ,- l l a o < 0 . N o w a ssum et hat - r 1 a ' 1 ,< 0 . T h e n , s i n c ea * 2 1 for k :0,1,2,-... (note that a o 2 I s i n c ea > 1 ) , w e s e ef r o m ( t O . t 5 ) t h a t l / o t t+ r < - 1 , so that -l

1 a'k+t < 0 . Hence, -l

< a) 10

for /c :

38s

1 O.4 P er iodic Conti n u e d F ra c ti o n s

Next. note that from ( t o . t 5 ) w e h a v e d'k:a**lla'*+t and since -l

1 a'* < 0 , it follows that -l 1a**lfa'1ra1

t

I , s o t h a t - l < s 7 ' : - l / p < 0 . Hence, a is a reduced quadratic irrational. Furthermore, note that since fi :

-l/ot,. it follows that

387

10.4 PeriodicContinuedFractions

tr

-l/o':ffiol'

fraction of '/D , We now find the form of the periodic simple continued Although \6 is not where D is a positive integer that is not a perfect square' -l -,/D and 0, the quadratic is not between reduced, since its conjug-ate

.6-ii

r.*,o*r"i6-t;

l,/Dl - '[5 ' doeslie r.duced,sinceits conjugate,

that the between-1 and 0. Therefore,from Theorem 10.20, we know initialpartial the continuedfractionor [.lill +.,/D is purely periodic. Since is quotient of the simple continued fraction of tJD | + "/D w h e r ea o : I . . / D l ' w e c a nw r i t e if faf + ,/Dl:21,/Dl:2a0,

I,/DI+-,/D:tml-

: I 2 ao ; at , Q2 , . . . ,na, 2 Qg , al , . . . , Q rl'

Subtracting ao : ,/6

from both sidesof this equality, we find that ./ D : l a g ;a3 a 2 ,...,2 a g 1 0,...1 ,a2 ,...2a ,,a

:log;orro'zmol. To obtain even more information about the partial quotients of the continued fraction of ,/D, we note that from Theorem 10.20, the simple be obtained from that continued fraction expansionof -l /$'IDl "/D) can period, that so ..lD the reversing by + for t.,6l ,

r/G/D-t.D1):tffi. But also note that

-t-6-l:lo;orprGol,

6

so that by taking reciprocals,we find that

| / G/D - t.D-l) - tor;orGrl

-

for the simple continued Therefore,when we equatethese two expressions we obtain t.D]) , fractionof llG/D Al:

QnrQ2:

Cln-ys...;On: Ol,

so that the periodic part of the continued fraction for ..lD is symmetricfrom the first to the penultimate term. In conclusion, we see that the simple continued fraction of 16

..ld:loo;ffi.

has the form

388

Decimal Fractions and Continued Fractions

We illustrate this with some examples. Example. Note that

8-

[ 4 ; l, 3 , 1, 8 ]

.16l

ts,ffii.rol

,,/Te :

,Fqe -

1 6 ;,l 2 , 1, 1 , 2 , 6 , 2, l, ,12 , 1, l 2 l [ 8 ; 1 , 2l ,,I , 5 , 4 , 1 5 , 1 , 2 ,I16 ,l

-,/ri:

tq;ml,

and

where each continued fraction has a pre-period of rength l and a period ending with twice the first partial quotient which is symmetric from the first to the next to the last term. The simple continued fraction expansionsof ,E fo, positive integers d such that d is not a perfect square and d < 100 can be found in Table 5 of the Appendix.

10.4 Problems l.

Find the simplecontinuedfractionsof a)

Jt

b) c)

d)

,/41

Jr r Jzt

e) r)

6 ,/-gq.

2 . Find the simple continued fractions of

il

o+,fi /z

b) Qq+,81)lt c) (tt-.E)t. 3 . Find the quadratic irrational with simple continued fraction expansion

il [z;t,5] b) tz;rSI c) t2JJI. 4.

il

Letd

,,/N

beapositive

isla:Tdl.

Show that the simple continued fraction of

389

1 O.4 P er iodic Cont i n u e d F ra c ti o n s

fractionsoi tffit't'fZgg' and b) Uggrrt (a) to find the simplecontinued J22r0. 5.

Let d be a integer,d 2 2' a)

Show that the simple continued fraction of ,/F

b)

show that the simple continued fraction of JFd

c)

Ugparts

is [d-l ;@l' is [d- t;zla-zl.

(a) and (b) to find the simple continued fractions of rfg9' tffg'

,lnz. and..G60' 6.

a)

of Shory lhat if d ,l un int"g.t, d > 3 , then the simple continued fraction

,tm b)

of Show that if d is a positive integer, then the simple continued fraction

'/fu. c) 7.

i s[ d - 1 ' l H , l 2 d - 2 1 .

rsld;c$71. -l,ft-gt , anO

Find the simple continued fraction expansionsof ,/6,.6f

Let d be an odd positive integer' a)

that

Show

the

simple

continued

fraction

of

JF+

continued

fraction

of

J d2-q

is

ld;ffil,ird>l' b)

that

Show

thr __qgple

la-lM,zd-zi,\f

d>3.

, where d is a positive integer, : a is a nonnegativeinteger. *here a2+l has period length one if and only if d

8 . Show that the simple continued fraction of Ji

, where d is a positive integer, : where a and b are integers, + b a2 if d only if and has period length two . b l \ a > a n d l , b

9 . Show that the simple continued fraction of Jd

6,1: (ar+brJrl)lct

10. prove that if

and a2-- (a2*urJd)/c,

^re quadratic

irrationals, then a)

(a1*42)'

:

c , ' t*

o''2

b)

(a1-a2)'

:

d'r -

d2

c)

(c''c.z)'

:

ot't'or2.

1 1 . Which of the following quadratic irrationals have purely periodic continued fractions

a) b)

l+.6 2 + ,/-B

c) 4+',m

c) d) e)

(tt - ,/-toltg e + ,f?l)/z (tz + -'.ft-g)l:t

12. Supposethat a : G+JF)/c, where 4,b, and c are integers,b ) 0, and b is noi u perfecl square. Show that is a reduced quatratic irrational if and only if 3 let a1,:2ak_t I a*_z Show that if p : (tar + l)2 * 2a1,-1* r, where / is a nonnegativeinteger, then rD has a period of length k + l.)

15' Let k be a iF:r. Let Dk - (3k+t)2 + 3 lgsitiu: continued fraction of JOp has a period of length 6ft.

Show that the simple

10.4 Computer Projects Write computer programs to do the following: 1'

Find the quadratic irrational that is the value of a periodic simple continued fraction.

2'

Find the periodic simple continued fraction expansionof a quadratic irrational.

11 some NonlinearDiophantine Equations

11.1 PythagoreanTriPles The Pythagoreantheorem tells us that the sum of the squaresof the lengths of the legs of a right triangle equals the square of the length of the hypothenrur.. Conversely, any triangle for which the sum of the squares of the lengths of the two shortest sides equals the square of the third side is a right triangle. Consequently,to find all right triangles with integral side lengths, we need to find all triples of positive integ ers x ,y ,z satisfying the diophantine equation

(rr.t) positive Triples of Pythagorean triPles.

x2+!2:22

integers

satisfying

this

equation

are

called

Example. The triples 3,4,5; 6,8,10; and 5,12,,13are Pythagorean triples b e ca us e32 + 42 : 5 ' .6 2 + 8 2 : 1 0 2 ,a n d 5 2 + 1 22: 132. Unlike most nonlinear diophantine equations, it is possible to explicitly Before developing the result describe all the integral solutions of (ll.l). definition. we a need triples, Pythagorean all describing : l. Definition. A Pythagoreantriple x,!,2 is calledprimitive if (x,y,z) Example. The Pythagoreantriptes 3,4,5 and 5,I2,I3 are primitive' whereas 391

392

S o m e N onl i near D i ophanti ne E quati ons

the Pythagoreantriple 6,g,10 is not. Let x,!,2 be a pythagorean triple with (x,y,z) : d . Then, there are " i -r' r,,r1,21): int eger s x r , t,z r w i th x : d x i ,y : d yt,, J i r, l. Furthermore, because ""A x2+y2:22, we have G/d)2+(y/il2:(z/d)2, s o t hat

x?+y?:r?. Hence, xt,!t,21 is a primitive pythagoreantriple, and the original triple x,!,2 is simply an integral multiple of this primitive pytgagoreantriple. Also, note that any integral multiple of a primitive (or for that matter any) Pythagoreantriple is again a pythagorean triple. If x1 ])t,zt is a primitive Pythagoreantriple, then we have

x?+ y?: r?,, and hence. @x)2+(dyr)r:(dz)2, so that dx 1,dy1,dz1 is a Pythagoreantriple. Consequently, all Pythagorean triples can be found by forming integral multiples of primitive Pythagorean triples. To find all primitive pythago*un triples, we need some lemmata. The first lemma tells us that any two integers of a primitive Pythagoreantriple are relatively prime. Lemma 11.1. If x,!,z is G,y) : (x ,z) : (y,z) : l.

a

primitive

Pythagorean

triple,

then

Proof. suppose x ,! ,z is a primitive pythagorean triple and (x ,y) > l. Then, ther e is a pr im e p s u c h th a ,tp ^ l (x y ), s o th at p I x andp y. S i ncep x I I a n d p l . - y ,* . k n o w t h a t p | ( r ' + y ' ) : 2 2 . B e c a u s p e l;r,'*..un conclude that p I z (using problem 32 of Section 3.2). This is a contradiction since (x ,y ,z) : l. Therefore, (x g) : l. In a similar manner we can easilv show that ( x , z ) : ( y ,z ) : l . D

393

1 1 .1 P y t hagor ean T ri Pl e s

integers of a primitive Next, we establish a lemma about the parity of the PythagoreantriPle. then x is even and y Lemma 11.2. If x,y,z is a primitive Pythagoreantriple, is odd or x is odd and Y is even' 1l '1, we know Proof. Let x ,!,z be a Primitive Pythagoreantriple. By Lemma x and y cannot that (x ,y\ : 1, so that x and y cannot both be even. Also (from 2 problem of Section 2'1) both be odd. If x and Y were both odd, then we would have ) x - = v z = I (mo d 4 ), so that 22:x2*y2

= 2(mod4).

x is even This is impossible (again from problem 2 of Section2.1). Therefore, and y is odd, or vice versa. E The final lemma that we need is a consequenceof the fundamental theorem of arithmetic. It tells us that two relatively prime integers that multiply together to give a square must both be squares' (r,s) : I and Lemma 11.3. If r,s, and t are positive integers such that : m2 and s : n2. ; : t2, then there are integersz and n such that r Proof. If r :1 ,upptr. that r ) lbe

or s : l, then the lemma is obviously true, so we may I and s ) 1. Let the prime-power factorizationsof r,,s, and

,:p1,pi2... p:", p:" s : p:,i\ p:,it and

t : ql' ql'

quo'.

Since (r,s ) : l, the primes occurring in the factorizations of r and s are distinct. Since rs : t2, we have

pi'pi'

pi"pi,+ipi,n pl,': q?"q'ru'

qiur'

From the fundamental theorem of arithmetic, the prime-powers occurring on

394

S o m e N onl i near D i ophanti ne E quati ons

the two sides of the above equation are the same. Hence, eachpi must be equal to Qi for some j with matching exponents, so that a; : 2bi. consequently,every exponenta; is even,and therefore ai/2 is an integer. we seethat r - m2 and , : 12, where m and n arethe integers a./2 a-/z m : pt' P2'

a/2

Pu"

a nd

n : pi,r('pi,C'

a/2

Pr"

!

We can now prove the desired result that describes all Pythagoreantriples.

primitive

Theorem ll.l. The positive integers x,l,z form a primitive pythagorean triple, with y even,if and only if there are relatively prime positiveintegers 172 and n, |/t ) n, with m odd and n even or m even and,n odd, such that x : m2-n2

'r7-'#ir' Prot{. Let x ,y ,z be a primitive Pythagoreantriple. Lemma I 1.2 tells us that x is odd and y is even, or vice versa. Since we have assumed that y is even, x and z are both odd. Hence, z*x and z-x are both even,so that there are p os it iv eint eger sr a n d s w i th r : (z + i /2 a n d s : (z-i l /2. S i n c ex 2 + y 2 : 2 2 , w e h a v ey 2 :

Ir)'

z2-x2:

(z*x)G-x).

Hence.

lz+x] f ,-"1

lr): I , .lt ' J:" w e n o t et h a t ( r , s ) : 1 . T o s e et h i s , l e t ( r , s ) : d . S i n c ed l , a n d d l s , dlG+s)z and,dl(r-s):x. T h i s m e a n st h a t d l ( * , r ) : 1 , sothat d :1. Using Lemma I 1.3, we see that there are integers la and n such that r : m 2 and,s : n 2 . W ri ti n g x ,y ,a n d z i n te r msof m andn w e have x:r-.s:m2-n2.

y:rM:rffi:2mn.

395

1 1 . 1 PY t hagor ean Tri P l e s

z:r*s:m2+n2. and n must also - xalso that (m ,n) : 1, since any common divisor of m we see (x,y,z) : l ' that : know w e * ' + r' , a n d Oi "i O" : m 2- n2' ,y :2 m n , a n d z then x y ' were' if they We also note that rn and n cannot both be odd, for : (x,y l ' Since ,z) and z would all be even, contradicting the condition n is odd, and is even (m,n) : I and m and n cannot both be odd, we seem has the triple or vice versa. This shows that every primitive Pythagorean appropriate form. To seethat everYtriPle x : m2-n2 y:2mn :2m2*n2, : 1, are positive integers, m ) n, (m,n) where m and n that note first m * n (mod 2), forms a primitive Pythagoreantriple, x 2 + y 2 : ( m 2 - n 2 ) 2+ ( 2 m n ) 2 : (ma -2 m2 n 2 + n 4 )* 4m2n2 : ^ 4 * 2 m 2 n 2t n a : (m2+n2)2 : 22.

and

To see that these values of x,y, and z are mutually relatively .prime, assume t h a t ( x , y , z ) : d ) ! . T h e n , t h e r e i s a p r i m e p - s u c h t h a t p l ^ ( x , y , z ) ^ .W e note that p * 2, since x is odd (becausex: m2-n2 where mz and n2 have o fp o rit " par it y ) . A l s o , n o te th a t b e c a u s ep I,x and p l t, p I G+ i :2m2 H e n c e p I m a n d p I n , c o n t r a d i c t i n gt h e f a c t t h a t a n ' dp l i t - ; : 2 n 2 . (* ,i ) : 1. T her efo re , (r,y ,z ) : l , a n d x o y ,z i s a pri mi ti ve P ythagorean triple. This concludesthe proof. D The following example illustrates the use of Theorem I I .l to produce PythagoreantriPles. so that (m,n): and n:2, Example. Let m:5 us that I .1 tells 1 m ) n. Hence, Theorem x:m2-n2:52-22:21 Y:2mn:2'5'2:20 z:m2+n2:52+22:29 is a primitive Pythagoreantriple.

I , f f i * n ( m o d2 ) , a n d

396

S o m e N o nl i near D i ophanti ne E quati ons

We list the primitive pythagorean triples generated using Theorem I l.l with rn : < 6 in T abl e I l .l .

m

n

2 3 4 4 5 5 6 6

I 2 I 3 2 4 I 5

x :

m2-n2

y:2mn

3 5 15 7 2l 9 35 1l

t : m2+n2

4 t2 8 24 20 40

5 l3 l7 25 29 4l 37 6t

r2 60

Table 11.1. Some Primitive pythagoreanTriples.

I l.l l.

Problems Find all il

primitive Pythagoreantriples x,l,z

b)

Pythagoreantriples x,!,2 with z < 40.

2 . Show that if x,!,2 divisibleby 3.

with z

< 40.

is a primitive pythagorean triple, then either x or y is

3 . Show that if x ,!,z is a Pythagorean triple, then exactly one of x,y and,z is , divisibleby 5.

4 . Show that if x,l,z is a Pythagorean triple, then at least one of x,y, and z is divisible by 4.

5 . Show that every positive integer greater than three is part of at least one Pythagoreantriple.

6 . L e t x l - 3 ,l t : recursivelv bv

4,zt:

5, and let

for n :2,3,4, ..., be defined

397

11.2 Fermat'sLast Theorem

xntl- 3xn*Zzn*l !n+r-3xn*2zo*2 zn+t-4xn*3zn*2' Show that xnln,zn is a Pythagoreantriple' + l, thenx,l,Z isoneof

7.

S h o w t h a t i f x , ! , 2 i s a P y t h a g o r e a tnr i p l e w i t h y : x the Pythagorean triples given in problem 6'

g.

: Find all solutions in positive integers of the diophantine equation x2 I 2y2 t2'

g.

Find all solutions in positive integers of the diophantine equation x2 * 3y2:

10. Find all solutions in w2+xzry':t'. I l.

positive integers of

the

t2-

diophantine equation

Find all Pythagorean triples containing the integer 12.

1 2 . Find formulae for the integers of all Pythagoreantriples x,l,z with z y*l 1 3 . Find formulae for the integers of all Pythagoreantriples x,l,z with z y * 2'

1 4 . Show that the number of Pythagorean triples x,-y,z (with x2 + y2 : z2) with a fixed integer x is (rk2)-l)/2if

x is odd, and (r!2l4-1)/2

if x is even.

1 5 . Find all solutions in positive integers of the diophantine equation *' * py' : 22, wherep isaprime. 11.1 Computer Projects Write programs to do the following: with xy,and

z lessthan a given bound.

l.

Find all Pythagoreantriples xJ,z

2.

Find all Pythagorean triples containing a given integer'

ll.2

Fermat's Last Theorem

In the previous section, we showed that the diophantine equation x2 + y2 : z2 has infinitely many solutionsin nonzerointegersx, !, z . What happens when we replace the exponent two in this equation with an integer grrut.. than two? Next to the discussionof the equation xz + y2 : z2 in his copy of the works of Diophantus, Fermat wrote in the margin: "However, it is impossibleto write a cube as the sum of two cubes, a fourth power as the sum of two fourth powers and in general any power the sum of two similar powers. For this I have discovereda truly wonderful proof, but the margin is too small to contain it."

398

So me N onl i near D i ophanti ne E quati ons

Since Fermat made this statement many people have searchedfor a proof of this assertion without success. Even trrouitr no ,or...t proof has yet been discovered,the foilowing conjecture is knowi as Fermat,s rasttheorem. Fermat's Last Theorem.

The diophantine equation x'+ln:zn

has no solutionsin nonzerointegersx, r, z when n is an integer with n D 3. Currently' we know that Fermat's last theorem is true for all positive integers n wit h 3 ( n < 1 2 5 0 0 0 . In th i s s e c ti o n ,w e wi l l show that the speci alcaseof Fermat's last theorem with n: 4 is true. That is, we will ,ho* that the diophantineequation xa+!4:24 has no solutionsin nonzerointegersx, !, z. Note that if we could also show that the diophantineequations xP + YP:7P has no solutionsin nonzero integersx,!,2 wheneverp is an odd prime, then we would know that Fermat's last theorem is true (seeproblem 2 at the end of this section). The proof we will give of the special case of n - 4 uses the method of infnite descent devised by Fermat. This method is an offshoot of the well-ordering property, and shows that a diophantine equation has no solutions by showing that for every solution there is a "smaller', solution. contradicting the well-ordering property. Using the method of infinite descent we will show that the diophantine equationxa + !4 : 22. has no solutionsin nonzerointegersx, !, and z. This is strongerthan showingthat Fermat's last theorem is true for n: 4, because a n y s o l u t i o no f x a + y 4 : t a : ( 2 2 ) 2g i v e sa s o l u t i o no f x a * v a : 2 2 . Theorem 11.2. The diophantine equation

**',ro,r: t' hasnosolutions in nonzer" ,",.*1, Proof. Assume that the above equation has a solution in nonzero integers x,l,z. Since we may replaceany number of the variableswith their negatives

399

1 1 .2 F er m at ' s Las t T h e o re m

we may assumethat x,Y,z are without changing the validity of the equation' positiveintegers' : 1' To see this, let (x,Y) : d. Then We may also supposethat (x,y) (x v Yt) : 1 ' w h e re x 1 and y 1 itro Positiveintegers' x : dx 1 and y = dY ,, w i th since xa + Y4 : '2 ' vte have (dx)4+(dYr)4:22, so that

d a ( x f + Y f ): ' 2 ' 2'2' we know t h a t d ' I t . Hence do | ,', and, by problem 32 of Section Thus' integer' positive Therefore, z : d'r r, where z 1is a da(xf + yf): (d2tr)': dor?, so that

xf+yl:t?. Th i s giv esa s olut io no f x a + y a : with (xr,yr) : 1.

: l r' z : zr ' 2 i n p o s i ti v ei n tegersx : xt' !

z2'where t h a t x : x , , l : 1 0 , z : z . ' i s a . s o l u t i o no f x a + y 4 : So, suppose : that there (xe,-/o) will show 1 ' We xo, lo, andzsare positiveintegerswith : 1' (xr' yl ) : : w i th zt x r,! l t, z: i s a not hers olut ioni n p o s i ti v ei n te g e rsx su ch t hat 21 1 z s . S i n c ex d + y t : z l , w e h a v e

G i l z + ( y & ) 2 :z E , we have so that x&, y&, ,o is a Pythagoreantriple. Furthermore, p I xs y&' then p and I l-fi, r&> - i, ro. if p is a prime suchthat p I x3 is a zs (xq,lrq): *3,yE, Hence, l. the fact that contradicting ;;';'l'ro, afe there that know we 11.1, prim-itiveiythagoreantriple, and by Theorem(mod (z 2) and rl ' ,n), m # positiveintegersz andn with x& : m2-n2 !& : Zmn zo: m2+n2, yfr the even where we have interchangedx62 andyfr, if necessary'to make integerof this Pair.

400

So me N onl i near D i ophanti ne E quati ons

From the equationfor xfr, we seethat x&+n2:m2. Since (m,n) : l, it foilows that x,s,n,m is a primitive pythagorean tripre. Again using Theorem I I .1, we seethat there are fositive integersr and s with (r,s) : l, r # s (mod 2). and ro : ,2-s2 n:2rs m - r2+s2. Si nc e m is odd a n d (m,n ) : l , w e k n o w that (m,2d : l . W e note that b e c aus ey & : ( 2 d m, L e mma l l .3 te l l s u s th at there are posi ti vei ntegersz1 andw with m:t? a n d 2 n : w 2 . S i n c ew i s e v e n ,w : 2 v w h e r ev i s a positiveinteger,so that v2: n/2:

rs.

si nc e ( r , s ) : I , L e m m a 1 1 .3 te l l s u s th a t th ere are posi ti vei ntegersx1 erd y1 s uc h t hat r : x l a n d s : y ? . N o te th a t si nce (r,s) : l , i t easi ryfol ow s th at ( x l, - y r ) : l. H e n c e .

x{+yf:

-2

zl

where x t,! t,z 1 ?re positive integers with (r r,y1) : l. zt I 26, because

Moreover, we have

zr(zf:m2 o /-

and

, : 1[(s 2Jd

+ t-./7)- (s - t',17)]> o.

t' 2 y1, by the Th i s m eanst hat s,/ i s a p o s i ti v es o l u ti o n ,s o th a t s 2 x1,and the contradicts this But choice of x1,y1 as the smallest-positivesolution' some for xpy1, be must X,I' Therefore inequality s * f ../7 < xr * ytfi. choice of /c. tr To illustrate the use of Theorem I1.6, we have the following example' positive solution of Example. From a previous example we know that the least l3y': I is xt:649, -Pr: 180' Hence' all t h e d i o p h a n t i n ee q u a t i o nx 2 positive solutions are given by xt, yp where

x* * yr,./n : (649+ tgo\[Lte . For instance,we have

408

Some NonlinearDiophantineEquations

x z * y 2,8

: 842361+ 233640.,/l t

H e n c e x 2 : 8 4 2 3 6 1 , y 2 : 233640 is the least positive solution of x 2 - l 3 y 2 : l , o t h e rt h a n X 1 - 6 4 9 ,y ' : 1 8 0 .

ll.3 l'

2'

3'

Problems Find all the solutionsof each of the foilowing diophantine equations a)

x2+ 3y2:4

b)

x 2 + 5 y 2: 7

c)

2 x 2+ 7 y 2 : 3 0 .

Find all the solutionsof each of the following diophantine equations a)

x'-y':B

b)

x2 - 4y2: 40

c)

4xz - 9/2 : loo.

For which of the following values of n x2 - 3ly' : n havea solution

a)l b) - 1 c)2 4.

does the diophantine equation

d ) -3 d4 f) -s ?

Find the least positive solution of the diophantine equations a) b)

x2 - 29y2: -1 x2 - 29yz: 1.

5.

Find the three smallest positive solutions of x2-37y2:1.

6.

For each of the following values of d determine whether the diophantine equationx2 - drz : -l has solutions

il2 b)3 c)6 d) 7.

13

e) f)

tj 3l

e) h)

4r s0.

the diophantine equation

The least positive solution of the diophantine equation xz - 6lyz : 1 is xt:1766319049, lt2261i398A. Find the least positive solution other than x t,l t.

409

1 1 .3 P ell' s E quat i o n

8.

S!g* that if pr/qt is a converggntof the simple continued fraction expansionof

9.

Show that if d is a positive integer divisible by a prime of the form 4ft * 3, then the diophantineequationx2 - dy': -l has no solutions.

Jd thenlp?- dq?l < | + zJd.

Let d and n be positive integers.

I l.

il

Show that if r,s is a solution of the diophantine equation x2 - dyz : I and X,Y is a solution of the diophantine equation x2 - dy' : , then Xr + dYs, Xs t Yr is alsoa solutionof x2 - dy': r.

b)

Show that the diophantine equation x2 - dyz: infinitelv many solutions.

n either has no solutions,or

Find those right triangles having legs with lengths that are consecutiveintegers. (Hint: use Theorem 11.1 to write the lengths of the legs as x -.r2 - 12 and y :2st, where s and t are positiveintegerssuch that (s,t) : l, s ) / and s and t have opposite parity. Then x-y:il implies that (s - r)2- 2t2: +1.)

12. Show that each of the following diophantine equationshas no solutions a)

xa-2ya:1

b)

x4-2y2--1.

11.3 Computer Projects Write programs to do the following: 1.

Find those integers n with lrl < Ji x2 - dyz: rz has no solutions.

2.

Find the least positive solutions of the diophantine equations x2 - dy': x 2 - d y 2- - 1 .

3.

Find the solutionsof Pell's equation from the least positive solution (see Theorem I 1.6).

such that the diophantine equation

I and

Appendix

412 Appendix Tabfe 1. FactorTable. The leastprimefac1o1,of .::h.odd positiveintegerlessthan 10000and not divisibleby five is givenin the table. ThJinitial digits of tile integeiare listedto the sideand the lastdigit is at the top of the column. primes are indicatedwith a dash..

1379 0 I 2 3 4 ) 6 7 8 9 t0

1379 3

3-

33-

3-

3 7 3-

337 3-

3 7^ 33

n 3- 3 7 t2 ll 33 13 7-14 3 ll 3 15 33 l6 7--13 t7 33l8 3 11 3 l9 20 3 7 3 ll 2l 373 22 1 3 23 3324 313 3 25 - l l 7 26 3327 33 28 717 29 33 13 30 7 33 - ll 3l 32 3t7 3 7 33 33 34 l l 7 - 35 33-

4A 4l 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 6l 62 63 64 65 66 67 68 69 70 7l 72 73 74 75

-131180 3 7 38l 373 82 19_ 83 3384 lt 33 85 7 86 3 ll 3 87 13 33 88 -r7 7_ 89 3390 7 3lI 3 9r 1 72 3 92 313 3 7 93 33 94 19 7-13 95 3396 33 97 7ll-19 98 3399 33 100 13l0l 3 7 3t7 t02 373 103 _ ll rc4 33105 3 2 3 3 106 l l - - 7 107 3 - 3 r 3 r0 8 3 1 7 3 109 -19 7ll0 323 3lll 7 33 lt2 t7 - 1l ll3 33 7 rt4 33 ll5

1379 3 ll 3 319 3

1379 t20 t2l

r22 3 7 329373 23-3 - 3 lr 13 33 7 3 19 329 t7 33 - ll 7 313 37 33 -23 - 13 3 8 37 3t 33 7 -tl 3323 33 7t71933313 3 t73 7 337 3 1t 329 313 23 33 7 33lt 33 19- 73 ll 3 t7 7 331 3 1 31 9

123 t24 125 r26 t27 128 t29 130 131

r32 r33 134 135 136 r37 r38 139 140 t4l

r42 r43

317 3 7 --23 3333 17fi29_ 3 7 313373 3 11 9 3333 7 3 13 3 33 rr 31 7 13 317 319 7 323 3 -29-37 33 7 319 3 13 7tt323 317 313 3 7--33ll 33 3 13 7 313 37 3

144 145 t46 147 148 r49 33150 1 9 3 1 1 3 r5l -1737 7 rs2 3 - 3 1 1 153 329 3 r54 2 3 - 7 t55 3 - 3 -

413

Appendix Table 1. (Continued).

r379 19 33 7-t333t7 33 7-3333 723-ll 331 317 t3 33 ll 3 7 323 41 373 19313 329 3r7 3 - - ll 7 3337 33 r77341 329 7 33 133ll 3 7 313 3 723t7 r82 3 - 3 3 1 r83 3 ll 3 184 7 1 9 - 4 3 185 3 1 7 3 l r 186 33 t87 188 3 7 3189 3 1 3 7 3 r90 - l t - 2 3 l9l 33r9 36 37 38 39 160 r6l t62 r63 t64 r65 r66 r67 r6 8 r69 170 17l 172 173 174 175 176 177 178 179 180 l8l

76 77 78 79

2m 201 202 203 204 205 206 207 208 209 210 2tl 2t2 2r3 2t4 2t5

2r6 217 2t8 2r9 220 22r 222 223 224 225 226 227 228 229 230 231

7t33r9 33 ll 3713-17 3 7 33 343 7 -3r9 313 323 3 7-tl29 33t9 33t 3 3 7 3u373 2 9 t3 3 ll 3 33 --19 7 33t7 3 ll 3 l34t 7 337 3rl 7 313 3 3r--47 33 7 317 3 23 7-33r3 337 3 731-3343 33 29--ll 3 7 337 3

1379

1379

1379

3 7 3 lt 3 7 -29 118 3 ll9 3 ll 7 329 3 240 241 - 1 9 - 4 1 242 33 7 243 l l 3 3 244 7 -31 245 3 ll 3 246 2 3 3 3 247 7 --37 248 313 319 249 4 7 3 l l 3 250 4 t - 2 3 1 3 251 3 7 3 ll 252 37 3 253 - t 7 4 3 254 33?5S 33 256 t 3 l t 1 7 7 257 331 32s8 2 9 3 1 3 3 259 723 260 3r9 326r 7 33 262 - 4 3 3 7 r 1 263 33 7 264 1 9 3 3 265 l l 7 - 266 33t7 267 33 268 7--269 33270 3 7 3 3 27r ll ll6

rt7

3-

3 7 31 1t 9 3 7 33 37 3753 329 37 3rt 3 19--17 33 7 33 747t9 313 343 33 7rl-13 334t 33 2 3 3 7- 2 9 3 7 3l7 37 3 29s t 3 - - l l 296 33297 313 3 298 t t t 9 2 9 7 299 341 3300 331 3 301 - 2 3 7 302 33 13 7 3303 3 304 - t 7 1 1305 343 3 7 306 33 307 3 7 7 1 7 308 33309 1 1 3 1 9 3 3r0 729133ll 3 l1 3 156 t57 158 159 280 281 282 283 284 285 286 287 288 289 290 291 292 293 294

-

414

Appendix Table 1. (Continued).

l3

192 r93 r94 195 r96 r97 198 r99 320 321 322 323 324 325 326 327 328 329 330 331 332 333 334 335 336 337 338 339 340 341 342 343 344 345 346 347

7 9

t7 341 3 --13 7 329 3319 3 3713 7tl 337 33 1 1-

3-

3-

13 33 - l1 7 353 34r 7 317 3 313 3 7 329 3 r 7 7 1 9l l 337 333

7-3r3-

3347 3 13-_-17 3 7 337 3 - - ll 3l 3r7 3343 3 l94r7 3313 ll 323 3 47- 719 3 ll 3 7 33 323 3 7

1379 232 233 234 235 236 237 238 239

:oo 361 362 363 364 365 366 367 368 369 370

37r 372 373 374 375 376 377 378 379 380 381 382 383 384 385 386 387

n23t3t7 3333 -137 3r7 323 33 73313 33 23-7 3319 33 ll - 7 4l 313 37 319 3 -13 329 3 7 33 7tt_ 347 361 33 7-37319 323 ll 313 3 -53 3 7 319373 t7--29 3331 37 3ll 3 --43 7 3311 23 33 717 3353 7 33

1379 272 3 7 3273 37 3 274 - 1 34 t 275 33 3l 276 l l 3 3 277 1 7 4 7 - 7 278 3 11 3 279 33 _19 400 40r 33402 33 403 2 9 3 7t t 7 404 313 3405 33 406 3 t t 7 7 1 3 40'7 3 3408 7 361 3 409 l74r0 3ll 3 7 4tr 323 3 412 t 3 7 - 413 33414 4 1 3 1 l 3 4t5 7--416 323 311 4t7 4 3 3 3 4 1 8 37 47 53 59 419 3 7 313 420 37 3 421 - l t 422 341 3423 319 3 424 - - 3 1 7 42s 33426 3t7 3 427 7tl

1379 3t2 353 3 3r3 ; 13-43 3r4 3 7 347 3 1 5 23 3 7 3 316 2 9 317 3 19 3 ll 318 33

3le- * 3 r z 37

440 44r 442 443 444 445 446 447 448 449 450 451 452 453 454 455 456 457 458 4s9 460 461 462 463 464 465 466 467

3 7 3lr 3 7 3 1943 3ll 323 33 -6t7 334r 17 3 11 3 767 33 ll 7 33 1333 7 23 313 3 19 7*329 347 33 71723t9 3 - 3 13 33 43-t7lI 3 7 331 373 tt4t-3333 59-13 7 33-

415 Appendix Table 1. (Continued).

468 3 1 3 4 3 3 33-1113428 348 5 9 3 1 l 3 3 8 8 469 - 1 3 7 3 7 3 3 7 429 7 3r7 3 389 713349 317 3470 3 430 1 11 35 9 3 l 390 4 7 3 't 353 3 3 31 3 ll 350 471 3 ' , l 319 431 7 -3 39r 3351 293 4',72 29 3432 3 3 392 7 1 3 352 3',l 3 473 7 6 1 433 3 33r 393 333s3 3 4 7 3 1 1 474 343 3434 7 --rl 3 394 3354 7 6 7 475 3 3 3 5 9 3 3 7 435 1 9 39s 355 5 31 1 3 1 9 3 1 1 476 7 l r 1 7 436 3 3 5 6 3 7 3 4 3 396 t 7 3 3 3 1 7 1 3 477 3 2 9 3 437 2 3 4 1 3 7 3 397 t r 2 9 357 --4 3 8 1 3 3 4 1 3 478 7 3 7 31 73 7 398 358 3- _ 3 2 3 5 3 479_ 439 13373 3 5 9 399 -Tt1 3359 3 3 1 7 600 7t qt*r :-Tl 560 3- 5zo 480 :tn7tl13 601 3 3 4 1 3 1 5 6 1 3 1 7 313 6 1 521 481 t7 3r9 3602 7-1713 3 562 3 7 3 1 1 522 2 3 3 482 3 3 7 3603 3 1 3 3 4 3 563 3 7 3 523 483 --23 7 604 3 3 564 3 7 329 484 47 29 37 13 524 373 3605 565 3 2 3 3 4 3 525 5 9 3 7 3 485 3 3 1 1 606 3 7 3 3 3 1 3 526 - 1 9 2 3 r r 566 486 1 3 5 9 607 5 3 3 7 3 s67 3- 37 527 487 - 1 l 3 7 3608 3 1 7 3 568 1 3- 1 1s28 319 3488 37 3 - 7 s69 3 4 1 609 3489 6 7 3 5 9 3 s29 1 1 6 7 r 7 3141 6 1 0 3 313 570 33530 490 1 3 - 7 329 3 2 9 6 1 1 7 571 531 47 313 3 317 3491 3 11 3 -r 612 3 t 7 3 5 9 572 7 7 3 t 7 3 1 3 3 532 '7 492 1'1 - 1 l 533 3 1 9 573 1 1 3 - ' t - 3 6 1 3 3493 3 l1 3 614 3 5',74 7 33 7 534 3494 3 4 7 3 6 1 5 1 3 3 1 1 5 3 1 1 2 3 3 ) t ) 3 535 3495 731 6 r 6r6 3 3 7 3 7 576 7 3 3 3 1 s36 496 1 1 7 337 3617 3 1 3 537 4 1 3 1 9 3 5',77 2 9 2 3 s 3 3497 '7 3 323 6 1 8 7 1 7 3 3 578 7 3 538 498 1 7 3 1 1 4 1 6 1 9 3 1 1 3 579 3 3s39 7-19499 3 7 3620 JI | 3 580 540 1 1 3 33s00 3 3 621 1 1 3 581 3 7--3 2 9 3 54r ' 71 3 5 01 622 J 5 582 311 361 l l 4 7 542 502 323 3r7 623 719133 583 3543 3 7 3503 I

4 ^ -

^ a

416 Appendix Table 1. (Continued).

504 7 r 3 7 3 544 13_ 584 3- 3624 7 9 3 3 505 - 3 1 1 3 _ 545 3 7 3 5 3 585 3 _ 3 625 713-tr 506 3 6 1 3 3 7 546 4 3 3 7 3 586 - l l 626 33507 I r 3 3 547 - 1 3 587 3 7 3627 3 3 508 - 1 3 7 548 33 ll 588 373 628 l l 6 l - 1 9 509 3 lt 3 _ 549 1 7 3 2 3 3 589 4 3 7 1- 1 7 629 3 7 3510 33 550 7 590 33 1 9 630 373 5ll 19- 7551 337 3591 2 3 3 6 1 3 6 3 1 - 5 9 - t l 512 3 4 7 3 2 3 552 33 592 3 t - - 7 632 335 1 3 7 3 l l 3 553 - l l 7 2 9 593 317 3_ 633 1 3 33 514 5 3 3 7 - 1 9 s54 3 2 3 3 3 1 594 1 3 3 1 9 3 634 t 7 - t t 7 515 33 7 555 7 33 595 1 1- 7 s 9 635 33516 1 3 3 3 556 6 7 - 1 9 596 3 6 7 3 4 7 636 33 517 731_ s57 33 7 597 7 3 4 3 3 637 2 3 - 7 518 37r 3s58 3 3 7 3 598 - 3 1 - 5 3 638 313 3_ 519 29 33 559 729tl 599 3 1 3 3 7 639 7 33 640 3 7 1 94 3 t 3 680 33 1 l 720 1 9 3 3 760 l l - 7 64r 3 1 l 3 7 6 8 1 7 3 t 7 3 721 7761 323 319 642 33 682 t 9 722 3 31 3 762 329 3 643 s 9 7 4 1 4 7 683 33 7 723 7 33 763 1 3 1 7 7 644 317 3684 3 4 r 3 724 1 3 - - l l 764 33645 3 l l 3 685 1 3 7 - 1 9 725 33 7 765 7 3 t 3 3 646 7 2 3 2 9 - _ 686 33726 5 3 3 1 3 3 766 4 7 7 9 1 1 647 3 3 ll 687 3 1 3 3 727 il 7 1929 767 33 7 648 3 1 3 3 688 7 -7t83 728 33 3 7 768 33 649 - 4 3 7 3 6 7 689 361 3129 2 3 3 3 769 7 4 3 _ 650 3 7 3 2 3 690 5 7 3 3 730 767-770 33 13 651 1 7 3 7 3 69r - 3 1 - l t 73r 3 7 t 3 1 3 771 l l 3 3 652 - l l 6 l _ 692 3 7 3 1 3 732 3 t 7 3 772 7--59 6s3 3 4 7 3 1 3 693 2 9 3 7 3 733 - - l t 4 l 773 311 371 654 3 1 3 3 694 l l 5 3 734 3 7 3774 361 3 655 - - 7 9 7 695 317 3735 3 7 3 775 2 3 656 33696 33 736 1 7 3 7 5 3 776 3 7 317 657 33 697 - 1 9 - 7 737 3 7 3 3 4 7 777 t 9 3 7 3 2 9 658 7 1 1 698 33 2 9 738 1 l 3 8 3 3 778 3 1 4 3 1 3 659 319 3699 33 739 1 9 - 1 3 7 779 3 - 3 1l

417 Ap p e ndix Table 1. (Continued)'

1379

1379 660 661 662 663 664 665 666 667 668 669 670 671 672 673 674 675 676 677 678 679 800 801 802 803 804 805 806 807 808 809 810 811 812 813 814 815

- 4 7 7 4 3 '740 3 1 1 3 3 1 3 3741 333 742 4 1 1 3 1 1 7 7 3343 3743 703 7 9 1 3 3 1 7 3ll 3 3 7 744 3704 3 745 - 2 9 705 1 l 3 3r7 3 7 746 7 3 7 2 3 706 3 747 3 1 3 3 11 3 707 7-708 7 3 3 1 9 3 748 359 37 4 1 4 7 3 r 749 709 3 750 1 3 3 7r0 3 - 3 13 3 11 3 751 7 rl -73 7tl 33752 7t2 3 3 7 3 1 1 753 t 7 3 713 7t4 3 7 3 7 3 754 - 1 9 755 3 7 3 715 - 2 3 1 7 37 3 3 1 3 3 6 7 756 7t6 3 757 6 7 - - 1 l 717 7 r 3 337 1 8 4 3 r r - 7 758 371 3 3 2 3 759 37t9 840 3 1 3 7 3 880 1 3 - - 2 3 881 3 7 3 841 t 3 4 7 1 9 37 3 882 5842 3 l l 3 883 - l l 843 7 884 3 3 7 3 844 2 3 - 885 5 3 3 1 7 3 3 l l 3 7 9 845 7 3 - 3 886 846 319 313 847 4 3 3 7 7 6 r 887 3 r 7 3 1 3 888 8 3 3 - 3 848 7 3 2 9 3 889 t 7 - 7 l l 849 329 3s9 8 5 0 - 1 1 4 7 6 7 890 7 337 3 7 8 9 1 3 851 33 - 3 892 1 1- 7 9 852 3 7 3893 8s3 1 9 7 - 323 3 3 - 3 8 3 894 854 7 r7-29 71317 3 3 1 3 4 r 8 5 5 r 7 3 4 3 3 895

3 7 3111713337 3 7 3 19 329 71761 33359 3 7-lr341 3337 3 193 7 3ll 3 7 3 53--23 3 ll 317 43 329 3 --67 7 313 33 1l 3 713 353 33 31371237 329 31l 3 13 3 83- 73 ll 3 7 341 3 -59 3 7 33 ll 3 7 -23 3 1l 3 47 379 3

700 701 '702

a a

1379

1379 780 781 782 783 784 785 786 787 788 789 790 791 792 793 794 795 796 797 798 799 920 92r 922 923 924 925 926 927 928 929 930 931 932 933 934 935

29 331 3 7313- 7 3341 3t7 3 - 11 7 47 329 33 7 33 7 313 353 3 7 -rl 341 3* 3 89 37 --r7

3r3 3* 373 3 19-3113 3 7 379 23373 6l-1119 3361 3r3 3 -23 - l1 3 7 337 3 i l 1 9- 4 7 3s9 313 3 73 3--37 7 3r7 371 34r 3 -67 7 319 33 7 313347 3 7

418

Appendix Table 1. (Continued).

1379 816 33 817 -1113818 3 7 319 819 373 820 5 9 1 3 2 9 _ 821 343 3_ 822 319 3 823 7 824 3373 825 3 7 3 2 3 3 826 l 1 7827 33r7 828 7 33 _43 829 830 319 3 7 831 33 832 5 3 7 t t 833 3 13 3 31 834 t 9 3 1 7 3 835 7-6r13 836 33837 i l 3 3 838 1 78 3 839 3 7 3 3 7 960 3 13 3 961 7-59962 33963 323 3 964 3 l - l l 965 3 7 313 966 37 3 967 t 9 t 7 968 323 3 969 l 1 3 3

1379 856 7-1311 857 3323 858 331 3 859 l l 1 3 860 3 7 386r 7 9 3 7 3 862 3 7 - 863 389 353 864 33 865 4 t t 7 t t 7 866 33867 1 3 3 3 868 - 1 9 7 869 33870 7 33 871 3 1 - 2 3 872 311 3 7 873 33 874 7-13 875 3319 876 3 ll 3 877 73167878 33 1l 879 5 9 3 1 9 3 970 8 9 3 11 8 7 971 3 ll 3 972 37t 3 973 t974 33975 7 3rt 3 976 4 3 t 3 977 329 3 7 978 33 979 19741 a n 4

1379 896 897 898 899 900 901 902 903 904 905 906 907 908 909 910 9 tl 912 913 9r4 915 916 917 9r8 919 980 981 982 983 984 985 986 987 988 989

1379

3-

3936 l l 3 1 7 3 3 4 7 3 937 7 _83 7 1 3 1 1 8 9 938 3ll 341 317 3939 33 33 940 7 -2397 7l 29 94r 333 7 3942 3 ll 3 l r 3 7 3 943 83944 3 7 3 ll 3 ll 3 945 t 3 3 7 3 13 3-17 3 946 4 7 4 3 2 9 7 947 333 3 1 3 6 1 948 1 9 3 5 3 3 3 l l 3 949 - l l - 7 19- 7950 3 13 337 3 3 1 3 1 l 951 3 31 3 7 33 952 - 8 9 7 t 3 2 3 - - 1 3 953 333 4 t 3 7 954 7 33 33 955 - 4 1 1 9 l 1 7 8 9 s 3 956 373 3 7 33 6 7 957 t 7 3 6 1 3 33 958 1 1 7 - 4 3 72917959 353 329 33 1 7 990 33 33 991 r r 2 3 4 7 7 7tt3r992 3333993 319 3 1 3 3 4 3 3 994 - 6 1 7 -59 99s 337 323 3 7 3 7 r 996 7 33 373 997 1 3 - u t 7 4t - - ll 998 367 3 7 3 1 3 3 1 9 999 9 7 3 1 3 3

Reprinted with permission from u. Dudley, Elementary Number Theory, Second Edition, copyrighto 1969 and l97g by w. H. Freeman and company. All rights reserved.

419

Appendix Table 2. Values of Some Arithmetic Functions'

I 2 3 4 5 6 'l I 9 l0 ll t2 l3 l4 l5 l6 t'l l8 l9 2A 2l 22 23 24 25 26 2'I 28 29 30 3l 32 33 34 35 36 5I

38 39 40 4l 42 43 44 45 46 4"1 48 49

I I 2 2 4 2 6 4 6 4 l0 4 t2 6 I 8 l6 6 l8 8 t2 l0 22 8 20 t2 l8 t2 28 I 30 l6 20 l6 24 t2 36 l8 24 l6 40 t2 42 20 24 22 46 l6 42

I 2 2

I 3 4

J

2 4 2 4 3 4 2 6 2 4 4 5 2 6 2 6 4 4 2 8 3 4 4 6 2 8 2 6 4 4 4 9 2 4 4 8 2 8 2 6 6 4 2 l0 3

6 t2 I l5 l3 l8 t2 28 t4 24 24 3l l8 39 20 42 32 36 24 60 3l 42 40 56 30 72 32 63 48 54 48 9l 38 60 56 90 42 96 44 84 78 72 48 124 57

420 A ppendi x Table 2. (Continued).

50 5l 52 53 54 55 56 57 58 59 60

6r 62 63 64 65 66 67 68 69 7A 7l 72 73 74 75 76 11

78 79 80 8l 82 83 84 85 86 87 88 89 90 9l 92 93 94 95 96 9',| 98 99 100

20 32 24 52 l8 40 24 36 28 58 l6 60 30 36 32 48 20 66 32 44 24 70 24 72 36 40 36 60 24 78 32 54 40 82 24 64 42 56 40 88 24 72 44 60 46 72 32 96 42 60 40

6 4 6 2 8 4 8 4 4 2 t2 2 4 6 7 4 8 2 6 4 8 2 t2 2 4 6 6 4 8 2 t0 5 4 2 t2 4 4 4 8 2 t2 4 6 4 4 4 t2 2 6 6 9

93 72 98 54 120 72 120 80 90 60 168 62 96 104 127 84 144 68 r26 96 t44 72 r95 74

n4 t24 140 96 168 80 186 t2r

r26 84 224 108 t32 120 180 90 234

n2 r68 128 t44 t20 252 98 t7l r56 217

421 Appendix Table 3. PrimitiveRootsModulo Primes prime p, p < 1000is givenin the table' The leastprimitive root r modulop for each

2 3 5 7 1l l3 t7 l9 23 29 31 3',1 4l 43 47 53 59 6l 67 7I 73 79 83 89 97 l0l 103 107 109 113 127 131 r37 139 t49 l5l 157 163

r67 r73 179 l8l

1 2 2 3 2 2 3 2 5 2 3 2 6 3 5 2 2 2 2 7 5 3 2 3 5 2 5 2 6 3 3 2 3 2 2 6 5 2 5 2 2 2

l9l 193 r97 199 2tl 223 227 229 233 239 241 251 257 263 269 271 277

28r 283 293 307 311 313 317 331 33',1 347 349

3s3 359 367 373 379 383 389 397 401 409 419 421 43r 433

l9 5 2 t 2 3 2 6 3 7 7 6 3 5 2 6 5 3 3 2 5 T7 l0 2 3 10 2 2 3 7 6 2 2 5 2 5 3 21 2 2 7 5

439 443 449 457 46r 463 467 479 487 49r 499 s03 s09 521 523 541 547 5)/

563 569 57r 577 587 593 599 601 607 613 617 6r9 63r 641 643 647 653 659 601 6 73 677 683 691 701

r5 2 3 l3 2 J

2 13 n J

2 1

) 2 3 2 2 2 2 2 3 3 5 2 3 7 7 3 2 3 2 3 3 ll 5 2 2

z 5 2 5 3 2

709 719 727 733 739 743 75r 751 76r 769 773 787 797 809 811 82r 823 827 829 839 853 857 859 863 877 881 883 887 907 9ll 919 929 937 94r 947 953 967 97r 977 983 991 997

2 ll 5 6 3 5 3 2 6 ll 2 2 2 3 3 2 J

2 2 ll 2 3 I

5 2 3 2 5 2 l7 7 3 5 2 2 3 5 6 3 5 6 7

422

Appendix Table 4. Indices Numbers

p

I r( I to

I l: lt2 l: l! l8 )1

22

29 28 3 r 30 3 i 36 4l 43 47 53 59 6l 67 7l 73 79 83 89 97

40 42 46 52 58 60 66 70 72 78 82 88 96

'il;il^ilrrl 'ilil,Y,l trlfr|JIl,li Numbers

p

t 7 1 8 l 1 9 20 2 l 22 23 24 25 26 27 28 29 30 3 l 32 33 l9 23 29 3l 37 4l 43 47 53 59 6l 67 7l 73 79 83 89 97

l0 7 2l 7 7 33 38 t6 l0 40 47 64 49 2l 2l 56 6 89

el

r z lr s

5 l l I e 24 z6i 4 8 1 7 1 3 5 25 1 6I e 34 z s l r c 37 1 2 l 4 s 37 3 s l ' 3 7 49 43138 8 t 3 l i 2 6 24 1 3 ll 0 17 5 8 11 6 40 2 0 1 6 2 17 6 1 3 2 70 6 3 1 4 7 29 r 8 1 3 5 t4 7 8 1 8 1 69

I

l3 t7 29 22 t4 36 6 3l t0 55 62 27 39 54 80 82 5

lt 26 t7 3l 29 t5 25 7 26 l6 60 37 63 72 25 t2 24

20 27 l5 36 t6 5 39 l5 57 28 l5 46 26 60 57 77

8 l3 29 l3 40 28 20 53 9 42 44 30 l3 I 75 49 76

l6 l0 l0 4 8 2 42 t2 44 30 56 2 46 54 52 2

l9 5 t2 l7 l7 29 25 46 4l 20 45 67 38 78 39 59

Indices tl

r s lr + l rlrol I 6l34l2l s rr j 3 Il s Il 4 r

l5 t4 23 ll t 4 l 2 2 l 3 s 39 sl116146 l3 3 4 1 2 0 1 2 857 29 s t l 2 s l 4 4 55 aI rr I oa 60 18l4el35 l5 3l6llll 67 s2lt0l12 l8 3 l 2 s l s e 87 l8l 3l13 9

nlsrlrs

9 28 34 3 33 49 59 47 ll ll 56 38 3l 46

5 20 l0 l8 9 3l 44 27 f 23 5 t7 5 2l 5 32 30 57 40 6 l 20 69 5 t4 80 85 74 60

Reprintedwith permissionfrom J. V. Uspenskyand M. A. Heaslet,Elementary Number Theory, McGraw-Hill Book Company.Copyright O 1939.

423

Appendix Table 4. (Continued).

Numbers p

3'l I 4l r9 43 23 47 34 s3 l l 59 4 l 6 l 48 67 65 7 l 55 7 8 29 79 25 83 5 7 89 22 97 27

l9 2l l8 33 9 24 ll 38 29 34 37 35 63 32

t8 2 l4 30 36 44 l4 l4 64 28 l0 64 34 t6

32 35 4 42 l 7 30 3 8 55 39 39 27 22 l l 2A 22 64 70 t 9 36 20 48 ll 5l 9l l9

20 22 9 50 9

6 33 3l 4l 3'l 46 58 65 65 35 67 24 95

I n dices

I

6 l2l I t s l 2 4 1 3 | 4314 l ')) | 8 l 29 4s132 1 4 l l l 33I 27148 2s 5 4 1 5 6 431r i I 34 l 8 s 3 1 6 3 e l 6 r l 27 46 2 5 1 3 3 481431 l 0

25 t l + t 5 r I 7 r I l 3 74 7 5 1 5 8 4 e l76164 30 4 0 1 8 1 7 t l 26 1 7 2 e l28172 30 2 l l l 0 8 5 1 3 9 4 l 5 8 1 45

23 40 16 58 29 2l 54 30 6l 73 l5

23 20 50 9 3l 59 23 54 84

2l 54 l0 43 50 38 l7 76 65 l4

23 36 38 46 2 66 28 l6 74 62

63

64

65

44

Numbers p

53 59 6l 67 7l '73 79 83 89 97

50 5 l

52

43 27 r3 32 45 5 3 3l 5t 62 5 l 0 27 50 22 5 5 46 7 68 36 63

76 47 42 2l 5l 3 42 79 55 93

53

54

))

56

)t

5 8 5 9 60

2l 52 26 t9 57 65 ll 4l 37

30 32 49 42 68 33 37 36 55

29 36 45 4 43 t5 13 75 47

6l

62

IncLices 3l 37 8 59 56 52 59 5 3 5 l 78 l 9 66 l 0 \) 8 7

22 33 57 23 53 '7'7

35 t9 52 l4 26

3 l 30 36 56 J 66 5 23 3l 7l 34 l 9 43 l 5 67 43

48 69 I 1 7 58 l l e 45 1 6 0 66 l 3 e 69 1 4 7 64 t 8 0

3 5 6 34 5 3 36 67 45 48 60 5 5 24 1 8 70 6 22 5 83 8 75 t 2 26

Numbers

p

67 7l 78 79 83 89 97

66

67

68

33 63 69 73 t5 13 94

47 50 48 45 56 57

6l

69

4l 52 29 2'7 5 8 50 38 58 6l 5l JI

't0

'tl

7 2 7 3 74 7 5

35 42 4l 36 79 66

44 5l 33 62 lt

36 t4 65 50 50

't6

II

7 8 79 80

8l

Inclices 44 69 20 28

23 4'l z l 44 27 5 3 29 72

40 49 67 53

I

43 39 32 68 1 4 3 3 l 42 77 40 1 4 2 46 4 2 l J J t 3 0 4 l 88

Numbers

p 82

83

84

85

86

87

88

83 4 l 89 3 7 97 23

6l t7

26 76 '73 90

45 38

60 83

44 92

89

90

91

92

93

94

95

96

82

48

I n d lces

s 4 l ' 7 e 1 5 61 4 9

20122

Appendix Table 4. (Continued).

Indices

p 2

I

3

5

4

2l rl | | 21 41 3l rl

6

7

8

e{t0llt

| I

I rl ' l! 2. 2l 3l 3'' 41 43 4'l 53 59 6l 67 7l 73 79 83 89 97

l3

t4

l5

t6

ll

3l 21 6l 41 5l I 2l 4l8l slrol I 2l 4l 8l 3l aln

lI l

l2

Numbers

7l3l6l 'l I ul el slrol zl I 3l elrolnl slrs trlrolr+l al tl 4 t2 2l al r l 21 4l sl rol rgl 7) t4l elrs;rzlrslrrl 5 l 2 l t o l + l z o l r l 1 7 l1 6 ll l I t l z z l r s l zrrl o l r z l s l lsl I 21 4l altol :l 6l rz z+ rs I rs t vl I n z a 1 ! ! I I I 3 l e l 2 7 1 r o l z e l r c lt 7 2 0 2 e z s n a l z + l r o lz t I z s I I : o I z sI 2 1 4 l a l r o l t z l z t l t 7 | 3 4I I I n sI : o I z : I s I o l r e l r r l z s l z t l : B l z eIl t ol 3 t l 2 s I I z+el zlr + lztl glrsl l28l 3 l s l z t l l s l z e l + r l 3 7 l rrez l 3r 2 o r o + l r z s l z sI r r I r +l z z l u l n |l 28s|; 4 0Il r zII r : |I r e| + l Il zr ct ll +z zt ll nz t Il 2 l 4 l 8 l t 6 l : z l r r l z 2 l 4 4 lr s r z : + r s | I m I t I Al zal z l + l s l r o l r z l s i t0 z0 40I 2t | +z | | | | ! I soI +r I zt I +eI z l 4 l a l r e l : z l : l 6 l t 2 1z + l+ t I t s I zs q I r sI : o I r t I z z l 2 l 4 l s l r o l t z l e q lr 5 s l + r l u l r s l o l r s l : o l s l r o i T l 4 e l s q l s r l s r l z l 1 4I 1 2 1+ 2 1+ s : r + l z t l s l z s l s z l + r l s t l s l r i l 2 l1l o l s o|l: r | e l + s l so+lIszo: Il ras II 3 l e l z t l z l e l r s l , 1 1+ l r z l x l z t I s l z + l z zlsslrol z l 4 l s l r o l t z l o a l 15| t I t+ | zaI seII zqI ssI t: I ee| +qI 3 l e l 2 7l s r l o s l r z l i t l 6 4 l t + l + z l y l z z I e oI z oI o oI z l sl2sl28l43l2tl s l o l o l l o l s : I z r I u l z g l+ s| + e l: e I Indices

p t7

l8

1 9 l0 23 l 5 29 2 l 3 l 22 37 l 8 4 1 26 43 26 47 3 8 53 J 59 3 3 6 l 44 67 20 7 l 62 7 3 20 79 48 83 l 5 89 6 9? 83

I 6 l3 4 36 33 35 2 6

I

27 40 8 27 65 30 t8 2t

t 9 20 2 l 22 23 24 25 26 27 28 29 30 3r 32 33 I 26 l2 35 34 l9 l0 t2 t4 54 t3 56 62 37 60 54 38

t2 23 5 33 40 l4 3 24 28 47 26 37 l8 32 37 73 93

l4 t7 l5 29 35 42 l5 48 56 33

I

)

s2 46 t7 t7 74 4l 77

I

t4 21 5 40 28 43 53 5 37 38 t2 5l 65 34 94

l0 II

) 30 34 46 33 47 l0 7 53 60 74 47 I3 82

20 2 l0 t6 t6 42 l3 35 20 l4 t6 8 64 ll 39 22

I Numbers l l 22 6 l8 I 20 3 | 6 l t 2 l 2 4 l l t4 2 Irzltrlzz 9 ) l5 | 2l 6l18 II 22 1 6 3 3 1 2 4 1 2 36 6 26 52 s l l 4 e l 4 s 3 7 l l 22 4 4 l 2 e l s 8 57 40 l 9 3 8 l 1 5 l 3 0 60 l 28 56 4 s l 2 3 l 4 6 25 l 4l 3 2 t l 5 1 3 5 1 32 40 54 24 34 23 6 e l 4 e l 6 8 l 46 22 44 s | 'o I zoI 40 28 84 t + 1 + + l + t l40 t 3 65 z + l t s l u l 79

I r sI r I lr:l tlzr

srl:el:+l

22 l 3 37 33 I 3 39 2 1 42 )5 5l 59 5 7 50 3 3 ll 6 47 1 6 59 t 9 80 't7 3l 4 35 1

t4 t'l 39 35

3r 43 53 66 42 57 7l t2 78

425 Ap p e n dix Table 4. (Continued).

Indices

p 42

3 5 1 3 6 37 3 8 39 40 4 l

34 17 28 4 l 20 43 3 l 47 34 53 9 5 9 27 6 l 45 6',1 65 7l l0 73 35 79 l 3 83 59 89 36 97 2

19 I 3 8 23 zl t 29 4 l 8 36 54 49 29 5 8 63 59 70 & 29 72 39 3 8 3 5 70 l 9 57 l 0 50

4 4 i | 4 5 i . 4 64',1 48 49

4t

I 8 17 6 38 l9 49 35 t2 48 26 3l 68 86

15 20 20 l9 39 55 5l 22 68 35 57 82 56

I

8 30 23 38 37 3 l3 2l 78 62 26 42

I 24 9 46 l7 l3 6 20 32 76 4l 78 l6

Numbers

29 45 39 34 26 t2 69 l4 70 82 56 80

I 37 25 9 52 24 57 70 52 8l 79 12

M 50 l8 43 48 44 58 77 79 59 60

rl

32llel I 4 7 l 4 r l 2 e 5 l 0 20 3 6 1 1 3 1 2 652 45 3 t 2 5 l 5 0 l 3 e t'l 34 2 e l 5 8 l 4 e 3 l 1 6 2l 5 i 2 4 l | 2 6 1 4 067 1 4 3I 1 7 7 1 1 6 3 1 2 342 1 6 41 2 8 ? 3 l 6 r l 2 s t) :l611 4 3 ? 5 1 6 7 1 5 1l 9 1 3 8| t 1 6 8 8 1 8 6 1 8 062 l 8 1 2 4 e l 4 5 l 3 1 58 t 9 6 1 9 2 1

lndices

p

5 3 40 59 3 6r l4 67 47 'll 48 1 3 61 19 50 8 3 69 89 72 97 72

27 6 28 27 52 43 7l 55 38 69

I t2 56 54 9 69 55 27 25 54

48 4l t5 l5 46 zl I 54 25 7 5 47 76 89

24 5l 4l 63 53

l5 42 60 25 )) 31

30 23 53 33 56 t4 l 7 34