Steps into Analytic Number Theory A Problem-Based Introduction 9783030650766, 9783030650773

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Problem Books in Mathematics

Paul Pollack Akash Singha Roy

Steps into Analytic Number Theory A Problem-Based Introduction

Problem Books in Mathematics Series Editor: Peter Winkler Department of Mathematics Dartmouth College Hanover, NH USA

More information about this series at http://www.springer.com/series/714

Paul Pollack • Akash Singha Roy

Steps into Analytic Number Theory A Problem-Based Introduction

Paul Pollack Department of Mathematics University of Georgia Athens, GA, USA

Akash Singha Roy Chennai Mathematical Institute Chennai, Tamil Nadu, India

ISSN 0941-3502 ISSN 2197-8506 (electronic) Problem Books in Mathematics ISBN 978-3-030-65076-6 ISBN 978-3-030-65077-3 (eBook) https://doi.org/10.1007/978-3-030-65077-3 Mathematics Subject Classification: 11-XX, 11-01, 97D50 © Springer Nature Switzerland AG 2021 This work is subject to copyright. All rights are solely and exclusively licensed by the Publisher, whether the whole or part of the material is concerned, specifically the rights of translation, reprinting, reuse of illustrations, recitation, broadcasting, reproduction on microfilms or in any other physical way, and transmission or information storage and retrieval, electronic adaptation, computer software, or by similar or dissimilar methodology now known or hereafter developed. The use of general descriptive names, registered names, trademarks, service marks, etc. in this publication does not imply, even in the absence of a specific statement, that such names are exempt from the relevant protective laws and regulations and therefore free for general use. The publisher, the authors, and the editors are safe to assume that the advice and information in this book are believed to be true and accurate at the date of publication. Neither the publisher nor the authors or the editors give a warranty, expressed or implied, with respect to the material contained herein or for any errors or omissions that may have been made. The publisher remains neutral with regard to jurisdictional claims in published maps and institutional affiliations. This Springer imprint is published by the registered company Springer Nature Switzerland AG The registered company address is: Gewerbestrasse 11, 6330 Cham, Switzerland

For our friends and colleagues at the Ross Mathematics Program.

Preface

These are notes from a 5-week course that I (P.P.) taught as part of the 2019 Ross/Asia Mathematics Program. The course and camp were held from July 7 through August 9 at the Jiangsu Aviation Technical College in Zhenjiang, Jiangsu, China. Clientele ranged widely in age and mathematical experience; most participants were high school students, but some were undergraduates and others were mathematics graduate students. The course was nontraditional in format. In lieu of lectures, participants received 15 problem sets (or “Steps”) over the course of the summer, one every two to three days. Class meetings (held each weekday, and sometimes also on weekends) were entirely devoted to presentations of solutions, sometimes by students and sometimes by the instructor. Participants were expected to have a strong background in mathematical problem solving (e.g., from training for mathematics contests) but not assumed to possess advanced subject-matter knowledge. As such, most of the solutions require nothing more than elementary number theory and a solid grasp of calculus. The chief exception is the proof of Dirichlet’s theorem on primes in progressions for a general modulus, where we use (without proof) certain facts from complex variables. Even there, someone familiar with the real-variables side of things will find the necessary results easy enough to swallow. It is not at all obvious to the uninitiated that analysis has something of value to offer arithmetic. I attempted in this course to marshal the most convincing examples available for this surprising thesis. This explains the somewhat atypical emphasis throughout on concrete, number-theoretic problems, in contrast to a systematic development of analytic tools. Our primary themes are the value distribution of arithmetic functions (e.g., Hardy and Ramanujan’s result on the typical number of prime factors of an integer and Erd˝os’s multiplication table theorem), the distribution of prime numbers (Chebyshev’s results, Dirichlet’s theorem, Brun’s theorem on twin primes), and the distribution of squares and nonsquares modulo p (e.g., Vinogradov’s upper bound on the least positive nonsquare mod p). Of course, in 5 weeks one can only cover so much; somewhat regrettably, these notes do not include a proof of the Prime Number Theorem.

vii

viii

Preface

Apart from the addition of a “problem track” on the values of ζ (s) at positive even integers, the problem sets are mostly unchanged from what students received in Summer 2019. What is new are the solution sets, which have been prepared by myself and Akash Singha Roy (a 2019 Ross Program counselor and enthusiastic participant in the original course). Our intent with the solution sets was to provide enough detail that novices will find the text useful for self-study. Several of the problems come attached to remarks indicating directions for interested students to explore further. Akash and I would like to conclude by thanking all of the student participants in the original course, as well as the “powers that be” behind the Ross program: Tim All, Jim Fowler, Dan Shapiro, and Jerry Xiao. Athens, GA, USA

Paul Pollack

Notation

Most of our notation and conventions will already be familiar to anyone who has taken a course in elementary number theory. Here are some possible exceptions: When we write “log”, we always intend the natural logarithm. The set of positive integers is denoted Z+ . For us, the letter p always denotes a prime, whether or not this is mentioned explicitly. We use Zm for the integers mod m and write Um for the group of units mod m. The residue class of the integer a, modulo m, is denoted “a mod m”. Finally, sums over integers are to be understood as taken only over positive integers, unless explicitly indicated otherwise. For example, “ n≤x ” means a sum over positive integers n ≤ x.

ix

Contents

Step #1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Practice with Big O notation. Estimating the partial sums of the harmonic series. Definition of the Euler-Mascheroni constant.

1

Step #2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Asymptotic estimates related to ζ (s). Many proofs that there are infinitely many primes. The Principle of Inclusion-Exclusion enters the picture.

5

Step #3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . The Euler product representation, logarithm, and reciprocal of ζ (s). The Jordan–Bonferroni inequalities. Upper and lower bounds for the divisor-counting function τ (n).

9

Step #4 . . . . . . . . . . . . . . . . . . . . ......................................................... A first pass at estimating p≤x, p prime 1/p. Average order of the Euler totient function. Counting using roots of unity. Extracting coefficients from a convergent Dirichlet series.

13

Step #5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . The quadratic  Gauss sum associated to the prime p. An improved lower bound on p≤x 1/p. The average order of σ (n). The identity theorem for Dirichlet series. Wallis’s product formula for π .

17

Step #6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . An inequality of Pólya–Vinogradov for Legendre symbol sums. Legendre’s formulation of the sieve of Eratosthenes. 0% of the positive integers are prime. Abel’s summation formula. ζ (2) = π 2 /6.

21

Step #7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Upper and lower bounds for the least common multiple of 1, 2, . . . , N . A first attempt to count twin primes. A curious identity for ζ (2k).

25

xi

xii

Contents

Step #8 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Gaps between squares, and between nonsquares, modulo p. Chebyshev’s upper and lower bounds for π(x). Dirichlet’s theorem on primes in progressions mod 3 and mod 4.

29

Step #9 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . The probability a random integer is squarefree. Brun’s  upper bound on the count of twin primes and the convergence of p: p,p+2 prime 1/p. √ Existence of a nonsquare modulo p that is < p, for all large p.

33

Step #10 . . . . . . . . . . . . . . ............................................................. Mertens’ estimate for p≤x 1/p. How many prime factors an integer has, on average. A recursive formula for ζ (2k). First musings on primitive prime factors of 2n − 1.

37

Step #11 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Vinogradov’s upper bound on the smallest nonsquare modulo a prime. The number of prime factors of a typical integer. The number of divisors of a typical integer and a sharp upper bound on τ (n). Bang’s theorem on primitive prime divisors of 2n − 1.

41

Step #12 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Erd˝os’s multiplication table theorem. Bertrand’s postulate: There is always a prime between n and 2n. Two consequences of the Prime Number Theorem.

45

Special Step A: Dirichlet’s Theorem for m = 8 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

47

Special Step B: Dirichlet’s Theorem for m =  (Odd Prime) . . . . . . . . . . . . . . .

53

Special Step C: Dirichlet’s Theorem in the General Case . . . . . . . . . . . . . . . . . . .

59

Solutions to Step #1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

63

Solutions to Step #2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

71

Solutions to Step #3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

79

Solutions to Step #4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

89

Solutions to Step #5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

97

Solutions to Step #6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 105 Solutions to Step #7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 115 Solutions to Step #8 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 123 Solutions to Step #9 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 133 Solutions to Step #10 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 139 Solutions to Step #11 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 151 Solutions to Step #12 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 159

Contents

xiii

Solutions to Special Step A . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 167 Solutions to Special Step B . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 173 Solutions to Special Step C . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 185 Epilogue . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 193 Suggestions for Further Reading . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 197

Step #1

Do not laugh at notations; invent them, they are powerful. In fact, mathematics is, to a large extent, invention of better notations. Richard Feynman

Hello to Big-Oh If f and g are complex-valued functions, we say “f is big-Oh of g”, and write f = O(g), to mean that there is a constant C ≥ 0 such that |f | ≤ C|g| for all indicated (or implied) values of the variables. We refer to C as the “implied constant”. For instance, x = O(x 2 )

on

[1, ∞),

with C = 1 an acceptable implied constant,

while x = O(x 2 ) on [0, 1]. As a more complicated example, 1 log(1 + x) = x − x 2 + O(x 3 ) 2

on [−9/10, 9/10],

meaning: there is a function E(x) with log(1 + x) = x − 12 x 2 + E(x) on [−9/10, 9/10] with E(x) = O(x 3 ) on [−9/10, 9/10]. You can prove this using the Maclaurin series for log(1 + x). (Really; try it!) © Springer Nature Switzerland AG 2021 P. Pollack, A. Singha Roy, Steps into Analytic Number Theory, Problem Books in Mathematics, https://doi.org/10.1007/978-3-030-65077-3_1

1

2

Step #1

1.1 Basic Properties (a) For any constant c, we have c · O(g) = O(g). Note. Interpret this to mean: “If f = O(g), then c · f = O(g).” Parts (b)–(e) should be interpreted similarly. (b) O(g) · O(h) = O(gh), (c) O(f ) + O(g) = O(|f | + |g|), (d) If f = O(g) then O(f ) + O(g) = O(g), (e) If f = O(g) and g = O(h), then f = O(h). 1.2 Prove: log(1 + x) = x + O(x 2 ) for all x ≥ 0. Is the same estimate true on (−0.99, ∞)? on (−1, ∞)? 1.3 We say that f (x) = O(g(x)) “as x → ∞” or “for all large x” if ∃ x0 such that f (x) = O(g(x)) on (x0 , ∞). Prove: If limx→∞ g(x) = 0, then as x → ∞, 1 = 1 + O(g(x)), 1 + O(g(x))

eO(g(x)) = 1 + O(g(x)), and

log(1 + O(g(x))) = O(g(x)).

Note. Interpret the first claimed equation to mean that if f (x) = O(g(x)) as x → ∞, then 1/(1 + f (x)) = 1 + O(g(x)), as x → ∞. Similarly for the others. 1.4 As x → ∞,     1 x 1 e . 1+ +O =e− x 2x x2 1.5 If f and g are positive-valued, then (f + g)2 ≤ 2(f 2 + g 2 ). More generally, for any real κ > 0, we have (f +g)κ = Oκ (f κ +g κ ). Here and elsewhere, a subscripted parameter indicates that you are allowed to choose your implied constant to depend on this parameter.

Asymptotic Analysis 1.6 For n ∈ Z+ , define an =

1 − n

 n

n+1

dt . t

Interpret an as an area and explain, from this geometric perspective, how to see that  ∞ n=1 an converges.

Step #1

3

1.7 There is a real number γ (the “Euler–Mascheroni constant”) such that for all positive integers N, 0≥

 1   1 − log(N + 1) + γ ≥ − . n N +1

n≤N

1.8 For all real x ≥ 1:

1 = log x + γ + O(1/x). n n≤x

Ingenuity 1.9 (NEWMAN) Let a1 = 1, and let an+1 = an + √ 2n + O(n−1/2 log n), as n → ∞.

1 an ,

for all n ∈ Z+ . Then an =

Step #2

Mathematicians have tried in vain to this day to discover some order in the sequence of prime numbers, and we have reason to believe that it is a mystery into which the human mind will never penetrate. Leonhard Euler

Asymptotic Analysis n+1 If f is strictly decreasing on [n, n + 1], then f (n) > n f (t) dt > f (n + 1) (draw a picture!). If f is strictly increasing, then the inequalities reverse. Use these observations to establish the following estimates. ∞

2.10 For s > 1:

 s 1 < . n−s < s−1 s−1 n=1

2.11 For s > 1 and x ≥ 1:

 n>x

2.12 For x ≥ 1: not x?

n−s < x −s +

1 s x 1−s ≤ x 1−s . s−1 s−1

log x! = x log x − x + O(log (ex)).

Why do we write ex and

© Springer Nature Switzerland AG 2021 P. Pollack, A. Singha Roy, Steps into Analytic Number Theory, Problem Books in Mathematics, https://doi.org/10.1007/978-3-030-65077-3_2

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6

Step #2

Infinitely Many Primes Prove each statement and deduce the infinitude of primes. 2.13 (STIELTJES) If p1 , . . . , pk is any finite list of distinct primes, with product P , and ab is any factorization of P into positive integers, then a + b has a prime factor not among p1 , . . . , pk . n

2.14 (GOLDBACH) The “Fermat numbers” 22 + 1, for n = 0, 1, 2, 3, . . . , are pairwise relatively prime. 2.15 (PEROTT) For some constant c > 0, and each N ∈ Z+ , the count of squarefree integers in [1, N] is >N−



N/m2 ≥ cN.

m≥2

Thus, there are infinitely many squarefree integers. 2.16 (RAMANUJAN, PILLAI, ENNOLA, RUBINSTEIN) Let P = {p1 , . . . , pk } be a set of k primes, where k < ∞. For each x ≥ 1, the number of integers in [1, x] divisible by no primes outside of P coincides with the number of nonnegative integer solutions e1 , . . . , ek to the inequality e1 log p1 + · · · + ek log pk ≤ log x.

(*)

The number of such solutions is (log x)k + OP ((log (ex))k−1 ).

k k! i=1 log pi Hint. Here is a way to start on the upper bound. To each nonnegative integer solution e1 , . . . , ek of (*), associate the 1 × 1 × · · · × 1 (hyper)cube in Rk having (e1 , . . . , ek ) as its “leftmost” corner. Show that all of these cubes sit inside the k-dimensional (hyper)tetrahedron defined by ‘x1 log p1 + · · · + xk log pk ≤ log (xp1 · · · pk ), all xi ≥ 0’. What is the volume of that tetrahedron? How does this volume compare to the number of cubes? It might help to first assume that k = 2 and draw some pictures.

Step #2

7

Combinatorial Methods 2.17 For all n ∈ Z+ , and all 0 ≤ r ≤ n:         n n−1 n n = (−1)r . − + · · · + (−1)r r r 0 1 2.18 For a finite set A, and subsets A1 , . . . , Ak of A, state and prove the “inclusion-exclusion formula” for |A\(A1 ∪A2 ∪· · ·∪Ak )|. Why is it called “inclusion– exclusion”? 2.19 (LEGENDRE) √ π(x) − π( x) + 1  x + = x − p1 √ p1 ≤ x

 √ p1 1:

log ζ (s) =

 1  1 = + O(1). ks kp ps p p k≥1

3.24 For 1 < s < 2:

1  1 1 + O(1). It follows (why?) that = log s p s−1 p p p

diverges. (EULER) 3.25 Find a sequence {c(n)}∞ n=1 with the property that ζ (s)

∞  c(n) n=1

ns

=1

© Springer Nature Switzerland AG 2021 P. Pollack, A. Singha Roy, Steps into Analytic Number Theory, Problem Books in Mathematics, https://doi.org/10.1007/978-3-030-65077-3_3

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10

Step #3

(for all s > 1), and describe c(n) in terms of the prime factorization of n. (We will see later that there is a unique such sequence {c(n)}.)

Combinatorial Methods 3.26 (JORDAN, BONFERRONI) If one halts the inclusion-exclusion formula after an inclusion, one always overshoots (in the sense of obtaining an estimate at least as large as correct). If one stops after an exclusion, one always undershoots.  3.27 Let A be a set of positive integers. If a∈A a1 converges, then A contains 0% of the positive integers, in the sense that lim

 n≤x, n∈A

x→∞

1

 n≤x

 1 = 0.

 3.28 Let A be a set of positive integers for which a∈A a1 diverges. List the elements of A: a1 < a2 < a3 < . . . . Then there are infinitely many m for which am < m(log m)1.01 . It follows that there are arbitrarily large values of x for which 

1 > x/(log x)1.01 .

n≤x, n∈A

Can you think of other functions that can replace x/(log x)1.01 here?

Arithmetic Functions and the Anatomy of Integers 3.29  Suppose that f, g, h are arithmetic functions related by an identity f (n) = g(d)h(n/d), valid for all n ∈ Z+ . Explain why d|n



f (n) =

n≤x

3.30 For x ≥ 1:



g(a)

a≤x

 n≤x

 b≤x/a

h(b) =

 b≤x

τ (n) = x log x + O(x).

≈ log x divisors “on average”.)

h(b)



g(a).

a≤x/b

(Thus, a number n ≤ x has

Step #3

11

3.31 Large Values of the Divisor Function (a) The numbers n = 2k all satisfy τ (n) > log n. (b) For every real A, there are infinitely many n ∈ Z+ with τ (n) > (log n)A . 3.32 For all n ∈ Z+ :

τ (n) ≤ 2n1/2 .

Ingenuity 3.33 For every N ∈ Z+ , there is a d ∈ Z+ for which the following holds: There are at least N primes p for which p + d is also prime.

Step #4

What did the analytic number theorist say when they were drowning? Log-log-log-log-log. Anonymous

Variations on a Theme of Euler 4.34 (a) For all x > 0, and every  ∈ (0, 1):  1  x  1  1 1 ≤ = x ≤ x  log + O(x  ). 1+ p p p  p p≤x p≤x p≤x (b) For all sufficiently large values of x: 1 ≤ log log x + 2 log log log x. p p≤x Hint. Use (a) with  = of ?)

1 log x·log log x .

(But how did we come up with this choice

4.35 (a) For all x > 0, and every  ∈ (0, 1):

© Springer Nature Switzerland AG 2021 P. Pollack, A. Singha Roy, Steps into Analytic Number Theory, Problem Books in Mathematics, https://doi.org/10.1007/978-3-030-65077-3_4

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14

Step #4

 1 1  1 1  1 − ≥ log + O(1). ≥ − p  n>x n1+ p1+ p>x p1+ p≤x p (b) For all sufficiently large values of x: 1 ≥ log log x − 2 log log log x. p p≤x  1 From Problems 4.34 and 4.35, we conclude that as x → ∞: p≤x p = log log x + O(log log log x). Later we will prove sharper estimates for this sum.

Arithmetic Functions and the Anatomy of Integers 4.36 Recall that Euler’s φ-function satisfies φ(n) = n

  μ(d)

 1 1− =n . p d p|n

d|n

Here μ(n) is the Möbius function, which appeared as the solution sequence c(n) in Problem 3.25. Deduce from Problem 3.29 that for x ≥ 1: 

φ(n) =

n≤x

1 2  μ(a) x + O(x log (ex)). 2 a≤x a 2

4.37 (DIRICHLET, MERTENS) For x ≥ 1: 

φ(n) =

n≤x

1 x 2 + O(x log (ex)). 2ζ (2)

4.38 (DIRICHLET) A lattice point is chosen uniformly at random from the square (0, N] × (0, N], where N ∈ Z+ . As N → ∞, the probability its coordinates are 1 relatively prime tends to ζ (2) .

Computing with Roots of Unity 4.39 Let m ∈ Z+ . For a ∈ Z:  1  2π ika/m 1 e = m 0 k mod m

if a ≡ 0 (mod m), otherwise.

Step #4

15

Here the sum on k is taken over any set of integer representatives of Zm . 4.40 (Counting Square Roots mod m) Let m ∈ Z+ . For n ∈ Z: #{a mod m : a 2 ≡ n (mod m)} =

1  2π ikn/m  −2π ika 2 /m e e . m k mod m

a mod m

Dirichlet Series  a(n) By now we have seen multiple expressions of the form ∞ n=1 ns , where the a(n) are complex numbers. These are known as “Dirichlet series”.  a(n) 4.41 Suppose that ∞ n=1 ns is a Dirichlet series that converges for some real number s  = s0 . Then for some real number C, we have |a(n)| ≤ Cns0 for all a(n) n. Hence, ∞ n=1 ns converges absolutely for every s > s0 + 1. Furthermore, for + every m ∈ Z : lim ms

s→∞

∞  a(n) = a(m). ns n=m

Mathematical Masterpieces: The Identity as Art Form 4.42 (GOLDBACH) Find the sum of the infinite series 1 1 1 1 1 1 1 1 1 1 + + + + + + + + + + ... 3 7 8 15 24 26 31 35 48 63 whose denominators, increased by 1, are the distinct numbers of the form nm with n, m ≥ 2 (the perfect powers).

Step #5

I have had my results for a long time: but I do not yet know how I am to arrive at them. Carl Friedrich Gauss

Distribution of Squares mod p Let p be an odd prime. 5.43 (GAUSS) The “Gauss sum” associated to p is 

G=

e2π ia

2 /p

.

a mod p

Show that for k ∈ Z, p  k:

 a mod p

2π ika 2 /p

e

  k G. = p

  Here pk is the Legendre symbol: 0 when p | k, and otherwise 1 or −1, according to whether or not k is a square mod p. 5.44 For n ∈ Z:   G  2π ikn/p −k . #{a mod p : a ≡ n (mod p)} = 1 + e p p 2

k mod p

Deduce:

    n G  2π ikn/p −k = . e p p p k mod p

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Step #5

5.45 Prove: G · G = p.(Here the bar denotes complex conjugation.) Deduce that G is a square root of −1 p p. √ √ [Thus, G = ± p when p ≡ 1 (mod 4) and G = ±i p when p ≡ 3 (mod 4). Gauss worked for years to determine

which sign to take, eventually proving that the + sign is always correct.]

  Hint. Start from the expression for pn proved in Problem 5.44. Take the modulus squared of both sides and sum on n mod p.

Variations on a Theme of Euler Below, we write ω(n) for the number of distinct prime factors of n and we use (n) for the number of prime factors of n, counted with multiplicity. For example, ω(45) = 2, while (45) = 3. Equivalently, ω(n) =



(n) =

1,



1.

pk |n

p|n

5.46 For every nonnegative integer k, and real x ≥ 1:  n≤x n squarefree ω(n)=k

1 1 ≤ n k!



1 p p≤x

k .

5.47 For x > 1: exp

 p≤x

1 p

 ≥

 n≤x n squarefree

1 . n

Also: ζ (2)

 n≤x n squarefree

1 1 ≥ > log x. n n≤x n

Deduce: 1 > log log x − 1. p p≤x This improves the lower bound of Problem 4.35.

Step #5

19

Arithmetic Functions and the Anatomy of Integers 5.48 (DIRICHLET) For x ≥ 1:



σ (n) =

n≤x

5.49 For all n ∈ Z+ :

1 ζ (2)x 2 + O(x log (ex)). 2

2ω(n) ≤ τ (n) ≤ 2(n) .

Dirichlet Series 5.50 (KALMÁR) A “multiplicative composition” of n is a representation of n as a product of integers > 1, where order matters. We let g(n) denote the number of multiplicative compositions of n. For instance, g(1) = 1 (the empty composition has all parts > 1), while g(6) = 3 (for 2 · 3, 3 · 2, 6). n g(n)

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 1 1 1 2 1 3 1 4 2 3 1 8 1 3 3 8 1 8 1 8

Let ρ = 1.72864 . . . be the solution in (1, ∞) to ζ (ρ) = 2. Prove: For all s > ρ, ∞  g(n) n=1

ns

=

1 . 2 − ζ (s)

 ∞ b(n) a(n) 5.51 If ∞ n=1 ns and n=1 ns converge and are equal for all large real numbers s, then each a(n) = b(n). (This implies the uniqueness of the sequence {c(n)} in Problem 3.25.)

Mathematical Masterpieces: The Identity as Art Form 5.52 For every nonnegative integer n, 

π/2

sin2n x dx =

0



π/2 0

sin2n+1 x dx =

π 1 · 3 · · · (2n − 1) · , 2 2 · 4 · · · (2n)

while

2 · 4 · · · (2n) . 3 · 5 · · · (2n + 1)

Here, as usual, empty products are to be understood to equal 1.

20

Step #5

5.53 (WALLIS) Show that as n → ∞, π/2 0

π/2 0

sin2n x dx

sin2n+1 x dx

 ∞ 

2k π 2k . Conclude that = · 2 2k − 1 2k + 1 k=1

→ 1.

Step #6

The elegance of a theorem is directly proportional to the number of ideas you can see in it and inversely proportional to the effort it takes to see them. George Pólya

Distribution of Squares mod p Let p be an odd prime. 6.54 Let k ∈ Z, p  k. For all N ∈ Z+ :     2π ikn/p   e =  n≤N

   π kN  sin p  1   ≤ .  πk   πk  sin p  sin p 

6.55 (PÓLYA–VINOGRADOV) For all N ∈ Z+ :     (p−1)/2  n  √  1 √  ≤ < p log p. p  p  n n≤N

n=1

Hint. Show (i) | sin(π θ )| ≥ 2|θ | when |θ | ≤ 12 , (ii)

1 n


y}. For x, y ≥ 1, π(x, y) = x −

  x x + − p1 p1 p2 p ≤y p

0: The count of n ≤ x belonging to any given residue class mod m lies within 1 of x/m. 7.71 Set π2 (x, y) = #{n ≤ x : p | n(n + 2) ⇒ p > y}. For squarefree d, let Ad := #{n ≤ x : n(n + 2) ≡ 0 (mod d)}. Then for x, y ≥ 1, π2 (x, y) = A1 − =







Ap1 +

p1 ≤y

p1 0. For all sufficiently large x: π(x) ≤ (log 4 + )

8.79 For x > 0:

log LCM(x) =

 p≤x

 log p

x . log x

log x log p

≤ π(x) log x.

8.80 (CHEBYSHEV) Fix  > 0. For all sufficiently large x:

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Step #8

π(x) ≥ (log 2 − )

x . log x

8.81 Fix K > 2. For all sufficiently large x, there is a prime in the interval (x, Kx].

8.82 For x > 0: x! = m≤x LCM(x/m).

Variations on a Theme of Euler 8.83 Let χ : Z → {−1, 0, 1} be defined by ⎧ ⎪ ⎪ ⎨ 1 if n ≡ 1 (mod 3), χ (n) = −1 if n ≡ −1 (mod 3), ⎪ ⎪ ⎩ 0 if 3 | n.

Put

L(s) =

∞  χ (n)

ns

n=1

=1−

1 1 1 1 1 1 1 + s − s + s − s + s − s + .... s 2 4 5 7 8 10 11

Show that for s > 1:

1 (a) L(s) = . χ (p) p 1 − ps (b) 1 ≥ L(s) ≥ 1/2. As a consequence, log L(s) = O(1).   1 1 (c) − = O(1). s p ps p≡1 (mod 3)

p≡−1 (mod 3)

(d) For 1 < s < 2, and either choice of ± sign:  p≡±1 (mod 3)

1 1 1 + O(1). = log ps 2 s−1

8.84 Both residue classes 1 mod 3 and −1 mod 3 contain infinitely many primes. In fact, the sum of the reciprocals of the primes up to x in either progression is 1 2 log log x + O(log log log x), as x → ∞. 8.85 The conclusion of Problem 8.84 also holds for the residue classes 1 mod 4 and −1 mod 4.

Step #8

31

Ingenuity 8.86 (V.-A. LEBESGUE) Let p be an odd prime. For each positive integer , let S = #{(x1 , . . . , x ) ∈ (Zp ) : x12 + · · · + x2 = 1}. By writing S =

1 p





e2π ik(x1 +···+x −1)/p 2

2

x1 ,...,x mod p k mod p

and using your evaluation of the Gauss sums, find a simple, closed form expression for S , for each .

Step #9

When I encountered Brun’s sieve for the first time, I was reminded of the legend that Alexander the Great cut with his sword the intricate knot of Phrygian King Gordius, and proceeded to Asia. In fact, in my mind Brun is mightier than the great king, for he cut the enigmatic knot that had survived 2100 years without any sign of wear. Yoichi Motohashi

Arithmetic Functions and the Anatomy of Integers 9.87 For n ∈ Z+ : which d 2 | n.

|μ(n)| =



d 2 |n μ(d).

The sum here is over all d ∈ Z+ for

9.88 (GEGENBAUER) For x ≥ 1:  n≤x

|μ(n)| =

 √ d≤ x

μ(d)

x  1 x + O(x 1/2 ). = ζ (2) d2

Hence, a “random” integer is squarefree with probability ζ (2)−1 . Can you generalize to kth-power-free integers?

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Step #9

Combinatorial Methods We recall some of the setup of Step #7. Let π2 (x, y) = #{n ≤ x : p | n(n + 2) ⇒ p > y}. Let x, y ≥ 2, and let P be the product of the primes not exceeding y. You have proved (Problem 7.73) that for every nonnegative even integer k, π2 (x, y) ≤

  2 1 1− x + xE1 + E2 , 2 p 2k

d|P ω(d)≤k

of d. Assume in the problems below that k = k(x) and y = y(x) are parameters depending on x, both of which tend to infinity with x. (We wait until Problem 9.91 to pin down k, y explicitly.) 9.89 For large enough x: E1 ≤



  2ω(d)  2ω(d) 2e 1+ ≤ eω(d)−k = e−k ≤ e−k (log y)6 . d d p p≤y

d|P ω(d)>k

d|P

9.90 For large enough x: Every d | P with ω(d) ≤ k satisfies d ≤ y k , and so E2 ≤ y k max 2ω(d) ≤ y 2k . d≤y k

1

9.91 Now choose y = x 100 log log x and k = 10 log log x. Deduce: 1

π2 (x, x 100 log log x ) = O(x(log log x)2 /(log x)2 ), as x → ∞. Conclude: π2 (x) = O(x(log log x)2 /(log x)2 ), as x → ∞. 9.92 (BRUN) If P is the set of primes p for which p + 2 is also prime, then  1 p∈P p < ∞.

Step #9

35

Distribution of Squares mod p 9.93 (WHYBURN) Let p be an odd prime. The reduced fractions a/b, with 0 < √ a, b < p, are all distinct in Zp . Deduce: For large p, these fractions represent more than 50% of the nonzero elements of Zp . 9.94 For all large primes p, the least positive nonsquare modulo p is smaller than √ p.

Mathematical Masterpieces: The Identity as Art Form 

1 . More generally, given m ∈ Z+ , determine a simple −1 n>1  1 closed form expression for . 2 n − m2 +

9.95 Evaluate

n2

n∈Z n=m

Ingenuity 9.96 For integers n > 1, writeP (n) for the largest prime factor of n. For which real numbers λ does the series n>1 nλ P1(n) converge ?

Step #10

A mathematician is a conjurer who gives away [their] secrets. John H. Conway

Variations on a Theme of Euler 10.97 (a) Using Legendre’s formula for the prime factorization of x!, deduce that for x > 0: log x! = x

(b) For x ≥ 1:

 log p + O(x). p p≤x

 log p = log x + O(1). p p≤x

 10.98 (MERTENS) Write p≤x logp p = log x + E(x), so that E(x) = O(1) by Problem 10.97(b). Show that for x ≥ 2:   1 1 = log log x + C + O , p log x p≤x where the constant C is given by C = 1 − log log 2 +

∞ 2

E(t) t (log t)2

dt.

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Step #10

Arithmetic Functions and the Anatomy of Integers 10.99

√ (a) The proportion of numbers in [1, x] possessing a prime factor > x tends to log 2, as x → ∞. (b) If we would like the proportion of numbers in [1, x] with a prime factor > y to tend to 12 , as x → ∞, how could we choose y = y(x)? 10.100 Mean values of ω(n) and ω(n)2  (a) As x → ∞: ω(n) = x log log x + O(x). (Thus, ω(n) ≈ log log x “on n≤x

average” over n ≤ x.)   (b) ω(n)2 = x n≤x

(c) As x → ∞:

p,q prime pq≤x



1 + O(x log log x). pq

Hint: ω(n)2 =



p,q prime 1. p,q|n

ω(n)2 = x(log log x)2 + O(x log log x).

n≤x

 10.101 For x > 0: n≤x ((n) − ω(n)) = O(x). Thus, (n) and ω(n) differ by a bounded amount on average. 10.102 For all large integers n: n/φ(n) = O(log log n). Hint: First show

−1 = O(1). p|n, p>log n (1 − 1/p)

Order of 2 mod   Let n > 1, and let Bn = gcd

n Bn

 2n − 1 : p|n . 2n/p − 1

2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 3 7 5 31 3 127 17 73 11 23 · 89 13 8191 43 151 257 131,071 3 · 19

10.103 Certainly if  is a prime for which o(2 mod ) = n, then  | Bn . (Make sure you see why!) We now investigate the converse. Suppose  | Bn , with  prime. Since  | 2n − 1, we know o(2 mod ) | n. Assuming that o(2 mod ) < n, show: (a) (b) (c) (d) (e)

o(2 mod ) divides n/p for some prime p dividing n, The unique prime p satisfying (a) is p = . As a consequence,  | n. If q is a prime dividing n with q = , then q divides o(2 mod ).  is the largest prime factor of n. 2  Bn .

Step #10

39

10.104 If there does not exist a prime number  for which o(2 mod ) = n, call n “repellent”. Show that if n is repellent, then Bn = 1 or Bn is the largest prime factor of n. Do you have a guess as to which n are repellent?

Mathematical Masterpieces: The Identity as Art Form 10.105 (EULER) For each integer k > 1: k−1  i=1

Thus, ζ (4) =

  1 ζ (2k). ζ (2i)ζ (2k − 2i) = k + 2

π4 , and in general, ζ (2k) ∈ π 2k Q. 90

Ingenuity 10.106 Recall that a “perfect number” is a positive integer n for which n =  d|n, d 0, the number of odd perfect numbers in [1, x] is ≤ x 1/2 .

Step #11

There are many things you can do with problems besides solving them. First you must define them, pose them. But then of course you can also refine them, depose them, or repose them, even dissolve them! David Hawkins

Distribution of Squares mod p 1 11.107 Fix  ∈ (0, 50 ). For all large primes p, fewer than 45% of the integers in 1

[1, p 2 + ] have a prime factor > p1/3 .

11.108 For all large primes p, the smallest positive nonsquare modulo p is smaller than p1/3 . 11.109 (VINOGRADOV) The conclusion of Problem 11.108 holds with 1 . by any real number > 2√ e

1 3

replaced

Arithmetic Functions and the Anatomy of Integers 11.110 (TURÁN) As x → ∞:

 (ω(n) − log log x)2 = O(x log log x). n≤x

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Step #11

11.111 (HARDY–RAMANUJAN) For large x, we have ω(n) ≈ log log x for “most” values of n ≤ x. Precisely: Fix  > 0. Then the proportion of n ≤ x for which |ω(n) − log log x| >  log log x tends to 0, as x → ∞. 11.112 (HARDY–RAMANUJAN) The conclusion of Problem 11.111 holds with (n) replacing ω(n). 11.113 (HARDY–RAMANUJAN) For large x, we have τ (n) ≈ (log x)log 2 for “most” values of n ≤ x. Precisely: Fix  > 0. Then the proportion of n ≤ x for which τ (n) ∈ / ((log x)log 2− , (log x)log 2+ ) tends to 0, as x → ∞. How does (log x)log 2 compare with the average of τ (n), of size ≈ log x, determined in Problem 3.30? 11.114 (Maximal Size of τ (n); WIGERT) Given a large positive integer n, factor n = AB, where A is the largest divisor of n composed of primes not exceeding log n/(log log n)3 . Prove: 2

(a) τ (A) ≤ 2O(log n/(log logn) ) .  n·log log log n (b) (B) ≤ logloglogn n + O log(log , and so 2 log n) 

τ (B) ≤ 2

log n log log n +O

 log n·log log log n (log log n)2

.

(c) For each  > 0 and all sufficiently large n: τ (n) ≤ 2(1+) log n/ log log n . (d) For each  > 0, there are infinitely many n such that τ (n) ≥ 2(1−) log n/ log log n . In fact, that inequality holds for n = LCM(x), once x is sufficiently large. Hint. τ (LCM(x)) ≥ 2π(x) . Find a way to use the result of Exercise 8.79. Note. Since 2(1+) log n/ log log n = n(1+) log 2/ log log n , and 1/ log log n → 0, part (c) has as a weak consequence that τ (n) = O (n ) for all  > 0.

Order of 2 mod  Recall the definition of the “cyclotomic polynomials” n (T ): 1 (T ) = T − 1 and successive values of n (T ) are determined by the relation

d|n

d (T ) = T n − 1,

for all n ∈ Z+ .

(*)

Step #11

43

Here are the first several values of n (T ): n n (T )

1 T −1

2 T +1

3 T2 +T +1

4 T2 +1

5 T4 +T3 +T2 +T +1

6 T2 −T +1

A priori, each n (T ) ∈ Q(T ), but an induction argument shows that each n (T ) ∈ Z[T ]. From (*) and Möbius inversion, n (T ) =

(T d − 1)μ(n/d) . d|n

11.115 For each n > 1, the integer n (2) divides the number Bn defined in Step #10.

11.116 For all n ∈ Z+ : n (2) = 2φ(n) d|n (1 − 1/2d )μ(n/d) > 13 2φ(n) . 11.117 (BANG) For every n > 6, there is a prime  with o(2 mod ) = n.

Step #12

I like to think of mathematicians as forming a nation of our own without distinctions of geographical origin, race, creed, sex, age or even time. . . all dedicated to the most beautiful of the arts and sciences. Julia Robinson

Arithmetic Functions and the Anatomy of Integers For N ∈ Z+ , let M(N ) = {m · n : 1 ≤ m, n ≤ N } be the collection of numbers appearing in the N × N multiplication table, and put M(N) = #M(N ). N M(N ) M(N )/N 2

1 2 4 8 16 32 64 128 256 512 1024 1 3 9 30 97 354 1263 4695 17,668 67,765 260,095 1.00 0.75 0.56 0.47 0.38 0.35 0.31 0.29 0.27 0.26 0.25

12.118 Let  > 0. If m, n ∈ Z+ are selected uniformly at random from [1, N ], independently, then the probability that (mn) is more than  log log N away from 2 log log N tends to 0, as N → ∞. 12.119 Let  > 0. If n ∈ Z+ is selected uniformly at random from [1, N 2 ], then the probability that (n) is more than  log log N away from log log N tends to 0, as N → ∞. M(N) ˝ ) lim = 0. 12.120 (ERD OS N →∞ N 2 © Springer Nature Switzerland AG 2021 P. Pollack, A. Singha Roy, Steps into Analytic Number Theory, Problem Books in Mathematics, https://doi.org/10.1007/978-3-030-65077-3_12

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Step #12

Distribution of Prime Numbers 12.121 (RAMANUJAN) For n ∈ Z+ : (a) (b) (c) (d) (e)

  2n 4n LCM(2n) · LCM(2n/3) · LCM(2n/5) · · · = ≥ . n LCM(n) · LCM(n/2) · LCM(n/3) · · · 2n n LCM(2n) 4 LCM(2n/3) ≥ . LCM(n) 2n 4n/3 LCM(2n) ≥ . LCM(n) 2n LCM(2n) is squarefree, with every prime divisor belonging to (n, 2n] or at LCM(n) √ most 2n. There is always a prime between n and 2n—meaning, in the interval (n, 2n].

12.122 (PILLAI) For every n > 1, there is a prime congruent to 1 mod n that is < 2n . At the end of the nineteenth century, Hadamard and de la Vallée Poussin (independently) established the “Prime Number Theorem” (PNT): π(x) → 1, x/ log x

as x → ∞.

Assume the PNT for the next two problems. 12.123 For every finite sequence d1 , . . . , dn ∈ {0, 1, 2, . . . , 9}, with d1 = 0, there is a prime number whose leftmost n decimal digits are d1 , . . . , dn . 12.124



p≤x p 2 x→∞ x / log x

(a) Show that lim

(b) (ALLADI) Let A(n) =



=

1 . 2

p|n p.

 Determine lim

x→∞

1≤n≤x A(n) . x 2 / log x

Ingenuity 12.125 (PENNEY–POMERANCE) Let Sk (n) = 1k + 2k + 3k + · · · + nk . It is wellknown that S3 (n) = S1 (n) · S1 (n) for all n ∈ Z+ (both sides = ( 12 n(n + 1))2 ). Classify all (nonempty) multisets of positive integers {a1 , . . . , ak }, {b1 , . . . , b } for which

Sai (n) = Sbj (n) for all n ∈ Z+ . i

j

Special Step A: Dirichlet’s Theorem for m = 8

Dirichlet alone, not I, nor Cauchy, nor Gauss knows what a completely rigorous mathematical proof is. Rather we learn it first from him. When Gauss says that he proved something, it seems to me very probable, when Cauchy says it, you can wager as much pro as con; when Dirichlet says it, it is certain. Carl Gustav Jacob Jacobi

In the 1830s, Dirichlet proved that there are infinitely many primes p ≡ a (mod m) whenever a and m are relatively prime. Our goal on this set is to show, by Dirichlet’s method, that there are infinitely many primes in each of the residue classes 1 mod 8, 3 mod 8, 5 mod 8, and 7 mod 8. This proof of Dirichlet’s theorem for m = 8 occupies a middle ground between the relatively simple arguments of Step #8 for the cases m = 3 and m = 4, and the more intricate arguments of the next two Special Steps, where Dirichlet’s theorem is proved in general.

Characters of U8 If G is a finite commutative group, by a “character” of G we mean a homomorphism χ : G → {z ∈ C : |z| = 1} (the target set here forming a group under multiplication). We use the symbol χ0 to denote the “trivial character” mapping all

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Special Step A

Table 13.1 Character table of U3

1 mod 3 2 mod 3

χ0 1 1

χ 1 −1

of G to 1. For example, when G = U3 there is precisely one nontrivial character χ , and the table of characters along with their values is very simple (shown at right, below) (Table 13.1). 13.126 Show that if χ is a character of U8 , then χ (g)2 = 1 for all g ∈ U8 . Use this to show that U8 has exactly 3 nontrivial characters, with character table as follows:

1 mod 8 3 mod 8 5 mod 8 7 mod 8

χ0 1 1 1 1

χ1 1 −1 1 −1

χ2 1 −1 −1 1

χ3 1 1 −1 −1

13.127 Calculate the sum of each row of the table. Conclude that the indicator function I1 mod 8 : U8 → {0, 1} corresponding to the element 1 mod 8 ∈ U8 is given by I1 mod 8 = 14 (χ0 + χ1 + χ2 + χ3 ). 13.128 Let a be an odd integer. Using that n mod 8 = a mod 8 ⇐⇒ (n mod 8)(a mod 8)−1 = 1 mod 8, deduce that Ia mod 8 =

1 χ (a mod 8)−1 χ , 4 χ

where the sum on χ is over all the characters of U8 . Conclude: 1 1 (χ0 + χ1 + χ2 + χ3 ), I3 mod 8 = (χ0 − χ1 − χ2 + χ3 ), 4 4 1 1 I5 mod 8 = (χ0 + χ1 − χ2 − χ3 ), I7 mod 8 = (χ0 − χ1 + χ2 − χ3 ). 4 4

I1 mod 8 =

Dirichlet L-Functions for m = 8 Let m be a positive integer. If χ is a character of Um , we associate to χ the function on Z which takes the value χ (n mod m) at integers n coprime to m, and the value 0 at all other integers. We use the same notation for both the original function on Um and this “lift” to a function on Z; this will not cause any confusion.

Special Step A

49

The functions on Z that arise this way are called the “Dirichlet characters” modulo m. By construction, each Dirichlet character χ satisfies χ (ab) = χ (a)χ (b)

for all a, b ∈ Z.

For each Dirichlet character χ , we define the “Dirichlet L-function” associated to χ by L(s, χ ) =

∞  χ (n) n=1

ns

.

The unique nontrivial character χ mod 3, along with its associated L-function, appeared already in Problem 8.84 (there L(s, χ ) was denoted L(s)). 13.129 Let χ be any Dirichlet character modulo a positive integer m. Show that the series defining L(s, χ ) is absolutely convergent in the complex half-plane (s) > 1. Deduce that for (s) > 1, L(s, χ ) =

p

1 1−

χ (p) ps

.

This generalizes Problem 8.84(a). For the rest of the problem set, we retreat from general Dirichlet characters to the safety of m = 8. 13.130 Let χ be a Dirichlet character mod 8. Show that for all s > 1, the   (pk ) double series p k≥1 χkp ks converges absolutely (so that the terms may be safely reordered), and ⎛ ⎞   χ (pk ) ⎠ = L(s, χ ). exp ⎝ ks kp p k≥1

Deduce that L(s, χ ) is positive and that log L(s, χ ) =



χ (pk ) p prime kpks . k≥1

13.131 Let χ be a nontrivial Dirichlet character mod 8. You will show in this problem that the nonvanishing of L(1, χ ) implies that log L(s, χ ) is bounded for s near 1. Set  Aχ (t) = χ (n). n≤t

50

Special Step A

(a) Prove that 

χ (n)n−s = x −s Aχ (x) + s



x 1

n≤x

Aχ (t) dt t s+1

for all x ≥ 1. (b) By examining the column sums of the character table, show that |Aχ (t)| ≤ 2 for all t ≥ 0. (c) Deduce from (a), (b) that the series defining L(s, χ ) converges for all s > 0, to ∞ Aχ (t) s 1 t s+1 dt. (d) Use these integral representations to prove that the series defining L(s, χ ) converges uniformly for s ≥ , for any fixed  > 0. Since the summands are continuous functions of s, a standard theorem in real analysis guarantees that L(s, χ ) is continuous for s > 0. (e) Deduce: If L(1, χ ) = 0, then log L(s, χ ) is defined and continuous for all s ≥ 1. Conclude that log L(s, χ ) = O(1) for 1 < s < 2.

Reduction to the Nonvanishing of L(1, χ) for χ = χ0 It follows from Problem 13.128 that for every odd integer a, and all s > 1,  p≡a (mod 8)

 1 1 = · Ia mod 8 (p mod 8) s p ps p odd



 1 = ps p odd

  χ (p) 1 1 −1 χ (a) χ (p) = χ (a)−1 , 4 χ 4 χ ps p

where the sums on χ are over all Dirichlet characters χ mod 8.  13.132 Estimating p χ (p)p−s (a) For each Dirichlet character χ mod 8:  χ (p) p

ps

= log L(s, χ ) + O(1)

for 1 < s < 2.

(b) L(s, χ0 ) = (1 − 2−s )ζ (s) when s > 1. Hence:  χ0 (p) p

ps

= log

1 + O(1) s−1

for 1 < s < 2.

Special Step A

51

(c) Suppose χ = χ0 and L(1, χ ) = 0. Then  χ (p) p

ps

= O(1)

for 1 < s < 2.

13.133 Assume L(1, χ ) = 0 for all χ = χ0 . Then for every odd a, and all s with 1 < s < 2,  p≡a (mod 8)

1 1 1 + O(1). = log ps 4 s−1

Consequently, there are infinitely many primes p ≡ a (mod 8).

Nonvanishing of L(1, χ) for χ = χ0 13.134 Write out the series for L(1, χ ), for each nontrivial character χ mod 8. Use familiar facts about alternating series to prove that each such series converges to a positive real number.

Techniques of Generalization 13.135 Use the ideas of this problem set to prove Dirichlet’s theorem when m = 12.

Special Step B: Dirichlet’s Theorem for m =  (Odd Prime)

In this sequence of problems you will carry out a proof of Dirichlet’s theorem when m = , an odd prime. The general outline of the proof is the same as for m = 8. Since the characters of U are no longer necessarily real-valued (see, for instance, Table 14.1 for the case  = 5), we will appeal to results in complex analysis rather than real analysis. Having to use complex analysis as opposed to real is more a technical inconvenience than a significant obstacle. We encounter a more serious difficulty when it comes to proving the required nonvanishing of L(1, χ ) for nontrivial χ . For this we follow a clever argument suggested by Sarvadaman Chowla and Louis Mordell.

Characters of U 14.136 Every character of U maps U into the group of ( − 1)th complex roots of unity. 14.137 Recall that U is cyclic, and fix a generator g. For every complex ( − 1)th root of unity ω, there is a unique character χ of U with χ (g) = ω. Thus, there are  − 1 distinct characters of U . 14.138 Show that every row of the character table of U sums to 0, except the row headed by 1 mod , which sums to  − 1. Deduce that for every a ∈ Z not divisible by , the indicator function of a mod  ∈ U is given by Ia mod  =

1  χ (a mod )−1 χ . −1 χ

© Springer Nature Switzerland AG 2021 P. Pollack, A. Singha Roy, Steps into Analytic Number Theory, Problem Books in Mathematics, https://doi.org/10.1007/978-3-030-65077-3_14

53

54

Special Step B

Table 14.1 Character table of U5

1 mod 5 2 mod 5 3 mod 5 4 mod 5

χ0 1 1 1 1

χ1 1 i −i −1

χ2 1 −1 −1 1

χ3 1 −i i −1

The L-Functions L(s, χ) 14.139 Let χ be a Dirichlet character modulo . The same argument you supplied on the last problem set shows that, for all complex s with real part larger than 1, the   (pk ) double series p k≥1 χkp ks converges absolutely and ⎛ ⎞   χ (pk ) ⎠ = L(s, χ ). exp ⎝ ks kp p k≥1

 k (The key fact is that, even in the complex world, exp( k≥1 zk ) = |z| ≤ 1, z = 1.) We now define (!) “Log L(s, χ )” by the equation Log L(s, χ ) =

1 1−z

whenever

 χ (pk ) . kpks

p prime k≥1

Verify that the right-hand series in this definition converges uniformly on compact subsets of (s) > 1, under any ordering of the terms. Since each summand is analytic, a well-known theorem in complex variables guarantees that Log L(s, χ ) is analytic for (s) > 1. 14.140 (a) Show that every column in the character table for U sums to 0, except the one headed by χ0 . (b) Deduce that if χ is a Dirichlet character mod  with χ = χ0 , then  | n≤t χ (n)| <  for all t.  χ (n) (c) Show that ∞ n=1 ns converges uniformly on compact subsets of (s) > 0. Hence, L(s, χ ) is analytic for (s) > 0. 14.141 Let χ be a nontrivial Dirichlet character mod , and suppose that L(1, χ ) = 0. Since L(s, χ ) is analytic and nonzero in a neighborhood of s = 1, a standard theorem in complex variables implies that there is some analytic logarithm of L(s, χ ) in a neighborhood of s = 1. Deduce that our particular logarithm function, Log L(s, χ ) (which is initially defined only when (s) > 1), analytically continues to a neighborhood of s = 1.

Special Step B

55

Conclude (assuming L(1, χ ) = 0): Log L(s, χ ) = Oχ (1) for 1 < s < 2.

Reduction to the Nonvanishing of L(1, χ) for χ = χ0 It follows from Problem 14.138 that for every integer a not divisible by , and all s > 1,  p≡a (mod )

 χ (p) 1 1  −1 = χ (a) , ps −1 χ ps p

where the sums on χ are over all Dirichlet characters χ mod .  χ0 (p)  1  1 1 14.142 For 1 < s < 2: + O (1). = − = log s s s p p p s−1 p p 14.143 Estimating

p|



p

χ (p)p−s

(a) For each Dirichlet character χ mod :  χ (p) p

ps

= Log L(s, χ ) + Oχ (1)

for 1 < s < 2.

(b) Suppose χ = χ0 and L(1, χ ) = 0. Then  χ (p) p

ps

= Oχ (1)

for 1 < s < 2.

14.144 Assume L(1, χ ) = 0 for all χ = χ0 . Then for every integer a not divisible by , and all s with 1 < s < 2,  p≡a (mod )

1 1 1 log + O (1). = s p −1 s−1

Consequently, there are infinitely many primes p ≡ a (mod ).

Nonvanishing of L(1, χ) for χ = χ0 14.145 Suppose that L(1, χ ) vanishes for two distinct nontrivial Dirichlet characters mod , say χ1 and χ2 .

56

Special Step B

L(s, χ2 ) L(s, χ1 ) and remain bounded as s ↓ 1, as does s−1 s−1 1 (s − 1)L(s, χ0 ). Deduce: L(s, χ ) remains bounded as s ↓ 1. s−1 χ   1 Log L(s, χ ) = ( − 1) ≥ 0. (b) For s > 1: ks kp k χ p ≡1 (mod )

Hence, L(s, χ ) ≥ 1, contrary to the conclusion of (a). Thus, there can be (a) The quotients

χ

at most one Dirichlet character χ = χ0 for which L(1, χ ) = 0. 14.146 Now consider more closely the situation where there is some (necessarily unique!) nontrivial Dirichlet character χ mod  with L(1, χ ) = 0. Show that in this case: (a) χ assumes only real values, (b) χ coincides with the Legendre symbol mod :

χ (a) =

  a for all a. 

Let G be the Gauss sum associated to  (as on Problem 5.43), and let   ∞    n 1 · )= . L = L(1,   n n=1

We show in the remaining two exercises that L = 0, completing the proof of Dirichlet’s theorem when m = . 14.147 Set A=

−(a)

  1 − e2π ia/ . 0 x0 . Since g(x) → 0 as x → ∞, we can also find an x1 > x0 with C|g(x)| < 12 for x > x1 . Then |f (x)| < 12 for x > x1 . We will show that 1+f1 (x) = 1 + O(g(x)), ef (x) = 1 + O(g(x)), and log(1 + f (x)) = O(g(x)) on (x1 , ∞). Turning to the first claim, we observe that on the interval (x1 , ∞), 1 1 =1− f (x) = 1 + O(f (x)). 1 + f (x) 1 + f (x) Here the final equality comes from noting that | 1+f1 (x) | < 2 for x > x1 . We know that f (x) = O(g(x)) for x > x1 , and so the O(f (x)) term in the last display may be replaced with O(g(x)), by Problem 1.1(e). The other claims are similar. Indeed, for x > x1 ,

Solutions to Step #1

65



e

f (x)

f (x) f (x)2 + + ... =1+ 1+ 2! 3!

 f (x) = 1 + O(f (x)),

since   2 2 1/2   1 + f (x) + f (x) + . . .  ≤ 1 + 1/2 + (1/2) + · · · = e − 1 .   2! 3! 2! 3! 1/2 Also,   f (x) f (x)2 f (x)3 log(1 + f (x)) = 1 − + − . . . f (x) = O(f (x)), 2 3 4 since   2 3 2 3   1 − f (x) + f (x) − f (x) + . . .  ≤ 1 + 1/2 + (1/2) + (1/2) + . . .   2 3 4 2 3 4 ≤ 1 + (1/2) + (1/2)2 + (1/2)3 + · · · = 2. Since we can replace O(f (x)) with O(g(x)), these estimates for log(1 + f (x)) and ef (x) imply the second and third claims. 1.4 We begin by noting that for small t, say |t| ≤ 12 , we have 1 log(1 + t) = t − t 2 + O(t 3 ) 2

(1.1)

et = 1 + t + O(t 2 ).

(1.2)

and

Both (1.1) and (1.2) are straightforward to prove by manipulations with Taylor series, analogous to those seen in previous solutions. Taking t = 1/x in (1.1), we see that for large x,      1 1 1 1 =x − 2 +O x log 1 + x x 2x x3   1 1 . =1− +O 2x x2 Therefore,      1 x 1 1+ = exp x log 1 + x x

66

Solutions to Step #1

   1 1 +O = exp 1 − 2x x2    1 1 . = e · exp − +O 2x x2 1 Taking t as − 2x + O(1/x 2 ) in (1.2) reveals that

     1 1 1 1 exp − +O + O = 1 − . 2 2x 2x x x2 (Here the O(t 2 ) term in (1.2) has been absorbed into the O(1/x 2 ) error.) Putting this back into the preceding display gives     1 x 1 e 1+ , +O =e− x 2x x2 which was precisely the claim. Remark. In this problem and the two preceding it, we can avoid bounding tails of series by invoking Lagrange’s remainder formula for Maclaurin series approximations: If f is (n + 1)-times differentiable on the closed interval from 0 to x, where x = 0, then f (x) = f (0) + f  (0)x +

f  (0) 2 f (n) (0) n f (n+1) (c) n+1 x + ··· + x + x 2! n! (n + 1)!

for some c strictly between 0 and x. As one example: For any nonzero t ∈ (−1, 1], there is some c 1 1 1 8 3 between 0 and t with log(1 + t) = t − 12 t 2 + 13 (1+c) 3 t . This implies (1.1) with 3 (1−1/2)3 = 3 as one acceptable value of the implied constant.

1.5 Since 2(f 2 + g 2 ) − (f + g)2 = (f − g)2 ≥ 0, the first statement of the problem is clear. To see that (f + g)κ = Oκ (f κ + g κ ), introduce the function h whose value at each input is the maximum of the values of f and g. Then (f + g)κ ≤ (2h)κ = 2κ hκ ≤ 2κ (f κ + g κ ). This proves the claimed O-estimate, with an implied constant of 2κ . n+1 1 n+1 1 1.6 Observe that an = n n dx − n x dx is the area of the region An (say) between the graphs of y = 1/n (a horizontal segment) and y = 1/x for n ≤ x ≤ n + 1. Figure 16.1 shows A1 , A2 , A3 , A4 . We can imagine translating each An to the left, to land inside the square [1, 2]×[0, 1] of area 1. For every positive integer N, the translates of A1 , . . . , AN are disjoint, and so 1≥

N  n=1

area(An ) =

N  n=1

an .

Solutions to Step #1

67

Fig. 16.1 The regions A1 , A2 , A3 , A4 in the solution to Problem 1.6

1

2

∞

Thus, the partial sums of the infinite series sequence. Therefore, ∞ n=1 an converges.

n=1 an

3

4

5

form a bounded, increasing

1.7 By Problem 1.6, it makes sense to define γ =

∞ 

an .

n=1

Put γN =

N

n=1 an ,

and notice that

 1  1  N +1 dt − = − log(N + 1). γN = n t n 1 n≤N

n≤N

Rearranging,  1 = log(N + 1) + γN n

n≤N

= log(N + 1) + γ − (γ − γN ). The proof will be complete if we show that 0 ≤ γ − γN ≤

1 . N +1

The ∞first inequality ∞is clear, as each an ≥ 0. To see the second, notice that γ − γN = a = n n=N +1 n=N +1 area(An ), in the notation of the last solution. The regions

68

Solutions to Step #1

AN +1 , AN +2 , . . . can be translated left, without overlap, into [N + 1, N + 2] × [0, N 1+1 ]. So the sum of their areas cannot exceed N 1+1 . Remark. It is clear from Fig. 16.1 that

1 2

< γ < 1. In fact,

γ = 0.57721566490153286060651209008240243104215933593992 . . .

1.8 It follows from Problem 1.7 with N = x that for x ≥ 1,   1 1 = log( x + 1) + γ + O . n

x + 1 n≤x (In fact, the result of Problem 1.7 shows that 1 is an acceptable value of the implied 1 constant.) Since x+1 ≤ x1 , the O-term above is O(1/x). So it will suffice to show that   1 log( x + 1) = log x + O . x This follows from  0 ≤ log( x + 1) − log x =

x+1

x

1 1 dt ≤ t x



x+1 x

1 dt ≤

1 . x

2 1.9 Square the recurrence relation and rearrange to get that an+1 − an2 = an−2 + 2. Adding from n = 1 to N,

2 aN +1 = 2N + 1 +

N 

an−2 .

n=1 2 + We immediately deduce that aN +1 ≥ 2N + 1, for each N ∈ Z . This inequality also holds when N = 0 (since a1 = 1), and so an−2 ≤ (2n − 1)−1 for all n ∈ Z+ . Referring back to the last displayed equation now gives

2 aN +1 ≤ 2N + 1 +

N  n=1

Since

1 2n−1

≤

1 n

and



1 n≤N n

1 . 2n − 1

= O(log N) for large N ,

2 aN +1 ≤ 2N + O(log N). 2 ≥ 2N − 1, we deduce that a 2 = Putting this together with the lower bound aN N 2N + O(log N ), as N → ∞.

Reference

69

To finish things off, write    log N . 2N + O(log N) = 2N 1 + O N √ √ Since 1+u−1 → 12 as u → 0, we see that 1 + u = 1 + O(u) when |u| is u sufficiently small. (This could also be deduced from the Maclaurin series expansion of (1 + u)1/2 .) Thus, as N → ∞,

  1/2   log N log N 1+O , =1+O N N and so !

   √ √ log N = 2N + O(N −1/2 log N), 2N + O(log N) = 2N 1 + O N

as desired. Remark. Problem 1.9 is adapted from [1] (see Problem 60 there).

Reference 1. Donald J. Newman, A problem seminar. Problem Books in Mathematics. Springer-Verlag, New York-Berlin, 1982.

Solutions to Step #2

2.10 The function f (t) = t −s is strictly decreasing for t > 0, and thus 

n+1

f (n + 1)
m

m1−s x 1−s < x −s + , s−1 s−1

< m−s + as desired.

Now let f (t) = log t. This time, f is increasing on (0, ∞), and so n+1 f (t) dt ≤ f (n) ≤ n f (t) dt for all natural numbers n. So if N is any n−1 natural number, then 2.12 n



N

f (1) + f (2) + · · · + f (N) = f (2) + · · · + f (N) ≥

log t dt, 1

while f (1) + f (2) + · · · + f (N) = f (1) + f (2) + · · · + f (N − 1) + log N  N ≤ log t dt + log N. 1

The last two displayed estimates together imply that log N! = log 1 + · · · + log N  N = log t dt + O(log N). 1

We take N = x to find that 

x

log x! = 1

 log t dt −

x

log t dt + O(log x)

N

  = x log x − x + 1 −

x

 log t dt + O(log x) .

N

x Notice that N log t dt ≤ (x − N) log x ≤ log x. Since log (ex) is larger than both 1 and log x, the entire parenthesized term above is O(log (ex)). This completes the proof. Why did we write log (ex) and not log x? One answer: In our proof, we used that log (ex) ≥ 1 for all x ≥ 1, and that inequality is not true without the factor of e. But maybe this is just our proof, and by a more clever argument we could get the result without the e. NO! It is simply not true that for some constant C, | log x! − (x log x − x)| ≤ C log x

for all x ≥ 1.

Solutions to Step #2

73

You can see this by thinking about values of x slightly larger than 1. For these, the left hand-side will be close to 1, while the right will be close to 0. Remark. When n is a positive integer, one can obtain much sharper estimates for log n!. For example, an explicit form of “Stirling’s approximation” states that for all n ∈ Z+ , √ √ 1 1 2π n(n/e)n e 12n+1 < n! < 2π n(n/e)n e 12n . This formula determines log n! to within O(1/n2 ), whereas the result of Problem 2.12 only determines log n! to within O(log n). Nevertheless, there is a sense in which the estimate of Problem 2.12 is optimal. Namely, when viewing log x! as a function of a real number x, there are jumps of size log x at all positive integers x. So one cannot hope for a continuous approximation to log x! with an error of size smaller than O(log x).

2.13 If p is any of p1 , . . . , pk , then p divides exactly one of a or b, and so does not divide a +b. But a +b is an integer larger than 1, and so has some prime factor. Thus, we have located a prime not among p1 , . . . , pk . Since this holds for any conceivable finite list of primes p1 , . . . , pk , there must be infinitely many primes. n

2.14 Let Fn := 22 + 1 denote the nth Fermat number. We begin by noting that Fn+1 − 2 =

n

Fi

i=0

for all nonnegative integers n. Indeed, the identity is easily checked for n = 0, n+2 −1 = and if it holds for some nonnegative integer n, then Fn+2 − 2 = 22



n+1 n+1 − 1)(22 + 1) = ( ni=0 Fi )Fn+1 = n+1 F . So the result follows by (22 i i=0 induction. An immediate consequence is that gcd(Fi , Fj ) = 1 whenever i = j : For if (say) i > j , then Fi − 2 ≡ 0 (mod Fj ). Thus, gcd(Fi , Fj ) = gcd((Fi − 2) + 2, Fj ) = j gcd(2, Fj ) = 1. In the last step, we used that Fj = 22 + 1 is odd. Finally as each of the Fermat numbers is greater than one and no two distinct Fermat numbers share a common prime divisor, the infinitude of the primes follows from that of the Fermat numbers. 2.15 An integer in [1, N] which is not squarefree is divisible by p2 for some prime p. For each p, the number of integers in [1, N] divisible by p2 is at most pN2 . Thus,  the number of non-squarefree integers in [1, N] is at most p pN2 , where the sum on p extends over all primes. Consequently, the number of integers in [1, N ] that are squarefree is at least N−

N  N ≥ N − = cN p2 m2 p m≥2

74

Solutions to Step #2

where   1  1  1 1 =1− − = 0. c =1− >1− m(m − 1) m−1 m m2 m≥2

m≥2

m≥2

The infinitude of primes is again immediate: If there are only k different primes p1 , . . . , pk , then there are only 2k squarefree numbers, namely the products i∈I pi for subsets I ⊆ {1, 2, . . . , k}. But there are infinitely many squarefree numbers, as we see by sending N to infinity in the above estimate. 2.16 The natural numbers n ≤ x divisible only by primes in P are in one-toone with the tuples (e1 , . . . , ek ) of nonnegative integers satisfying

k correspondence ei i=1 pi ≤ x. Upon taking logarithms, this last inequality becomes k 

ei log pi ≤ log x.

i=1

Thus, our counting problem is equivalent to the following exercise in discrete geometry: Estimate the number of lattice points (e1 , . . . , ek ) belonging to the simplex S = {(x1 , . . . , xk ) ∈ Rk :

k 

xi log pi ≤ log x, and all xi ≥ 0}.

i=1

It is helpful to view Rk as partitioned into unit hypercubes having their vertices at lattice points. Explicitly, for each (e1 , . . . , ek ) ∈ Zk , we let C(e1 ,...,ek ) denote the unit hypercube whose lower ‘lower left’ corner is (e1 , . . . , ek ). That is, C(e1 ,...,ek ) = {(x1 , . . . , xk ) ∈ Rk : each ei ≤ xi < ei + 1}. Then every (x1 , . . . , xk ) ∈ S belongs to a cube C(e1 ,...,ek ) for some lattice point (e1 , . . . , ek ) ∈ S. In the other direction, if (e1 , . . . , ek ) is a lattice point belonging to S, then every (x1 , . . . , xk ) ∈ C(e1 ,...,ek ) satisfies k 

xi log pi ≤

i=1

k 

(ei + 1) log pi ≤ log (xp1 · · · pk ),

i=1

so that (x1 , . . . , xk ) belongs to the simplex S  = {(x1 , . . . , xk ) ∈ Rk :

k  i=1

xi log pi ≤ log(xp1 · · · pk ), and all xi ≥ 0}.

Solutions to Step #2

75

Hence, S⊆

"

C(e1 ,...,ek ) ⊆ S  .

(e1 ,...,ek )∈ZK ∩S

Comparing (hyper)volumes, we see that if N is our lattice point count, then vol(S) ≤ N ≤ vol(S  ). It is a standard exercise in multivariable calculus to compute the (hyper)volumes of multidimensional tetrahedra: vol(S) =

(log x)k ,

k k! i=1 log pi

while (log x + log(p1 · · · pk ))k

k! ki=1 log pi k  k−j log(p · · · p )j k  1 k (log x)k j (log x) = k + .

k k! i=1 log pi j =1 k! i=1 log pi

vol(S  ) =

For each j = 1, 2, . . . , k − 1, we have (log x)k−j ≤ log(ex)k−j ≤ log (ex)k−1 . Thus, each term in the sum on j is O(log (ex)k−1 ), where the implied constant may depend on k and the pi . Hence, that entire sum is OP (log(ex)k−1 ), and we have the estimate claimed in the problem statement. The infinitude of primes is again an easy consequence. Indeed, if p1 , · · · , pk are all the primes, then no number has a prime factor outside {p1 , . . . , pk }. Hence, the total count of natural numbers up to x is O((log x)k−1 ), as x → ∞. But that count is x, and x is not O((log x)k−1 ). Contradiction. 2.17 We fix n ∈ Z+ and proceed by induction on r. The for    r =  result is immediate 0. Suppose that for some r with 0 ≤ r < n we have rj =0 (−1)j nj = (−1)r n−1 r . Then        r+1 r  n j n j n r+1 = + (−1) (−1) (−1) j j r +1 j =0

j =0

    n−1 n + (−1)r+1 r r +1     n−1 n − = (−1)r r r +1   n−1 . = (−1)r+1 r +1 = (−1)r

76

Solutions to Step #2

Here the last step is a consequence of Pascal’s identity in the form n−1 r+1 .



n  r+1

=

n−1 r

+

Alternative Solution An application of generating functions yields a one-line  solution. We observe that on the one hand, the right hand side (−1)r n−1 is the r coefficient of x r in (1 − x)n−1 , whereas on the other, ⎞ ⎛  n      n (1 − x)n−1 = (1 − x)n (1 − x)−1 = xi ⎝ (−1)i xj ⎠ i j ≥0

i=0

and the coefficient of x r in the rightmost expression is equal to          n n i n r n = − + · · · + (−1) . (−1) i 0 1 r i+j =r, i,j ≥0

2.18 The inclusion-exclusion formula (or “Principle of Inclusion-Exclusion”) states that |A \ (A1 ∪ · · · ∪ Ak )| = |A| − −





|Ai1 | +

1≤i1 ≤k



|Ai1 ∩ Ai2 |

1≤i1 N ms

is the tail of a convergent series. We conclude that  ∞

  1 1 1 1 + s + 2s + . . . , = ns p p p n=1

as desired. The final equality in the problem statement follows from the geometric series identity 1 + p1s + p12s + · · · = 1 1 . 1− ps

3.23 The previous problem allows us to write log ζ (s) =

 p

log

1 1− p1s

. Recalling

that log

1 1 1 = − log(1 − x) = x + x 2 + x 3 + . . . 1−x 2 3

whenever |x| < 1, we find that log ζ (s) =

 1 , kpks p k≥1

which is the first equality claimed in the problem statement. To obtain the second, it is enough to show that  1 = O(1). kpks p k≥2

In fact (keeping in mind that s > 1),  1   1  1 1 < = 1. < = ks k kp p p(p − 1) m(m − 1) p p p k≥2

k≥2

m≥2

3.24 By Problem 2.10, if 1 < s < 2 then log

s 1 1 < log ζ (s) < log < log 2 + log , s−1 s−1 s−1

so that log ζ (s) = log

1 + O(1). s−1

Solutions to Step #3

81

By Problem 3.23, log ζ (s) =

 1 + O(1). ps p

Comparing the estimates,  1 1 + O(1). = log s p s − 1 p  1 As s ↓ 1, we see that log s−1 → ∞ and so p p1s → ∞. This certainly implies that  1  1 p p diverges: If that series converged to A (say), we would have that p ps ≤ A for every s > 1.

3.25 By Problem 3.22, the identity ζ (s)−1 = p (1 − p1s ) holds for every s > 1. So it will be enough to find a sequence {c(n)} satisfying   ∞

 1 c(n) 1− s = p ns p

(3.5)

n=1

for all s > 1. If we formally expand out the left-hand side (“formally” meaning that we ignore issues of convergence), we obtain a weighted sum of terms n1s , where the (only) positive integers n that appear are the products of distinct primes, and where 1 ns appears with a coefficient of +1 or −1 according to whether n is the product of an even or an odd number of primes. This suggests (but does not prove) that we can take  0 unless n is squarefree, c(n) = (3.6) k (−1) if n = p1 · · · pk with the pi distinct. Note that c(1) = 1 with this definition, as 1 has a prime factorization into k = 0 distinct primes! Moreover, c(n) is multiplicative with the usual definition from elementary number theory: c(ab) = c(a)c(b) whenever gcd(a, b) = 1. To rigorously prove that (3.5) holds with c(n) defined as above, we imitate the solution to Problem 3.22. The series on the right of (3.5) converges absolutely when s > 1, and so we may rearrange or regroup terms at will. Writing n = 2e1 n1 , where e1 ≥ 0 and n1 is odd, ∞  c(n) n=1

ns

=

 c(2e1 ) 2e1 s

e1 ≥0

 n1 ≥1 gcd(n1 ,2)=1

c(n1 ) n1 s

82

Solutions to Step #3





= 

n1 ≥1 gcd(n1 ,2)=1



=

n1 ≥1 gcd(n1 ,2)=1

c(n1 ) n1 s c(n1 ) n1 s





c(2) c(4) 1 + s + s + ... 2 4



 1 1− s . 2

Iterating this procedure as in Problem 3.22, we find that for every N ≥ 2, ∞  c(n) n=1

ns





=

m≥1

gcd(m, p≤N p)=1

c(m) ms

   1 1− s . p

(3.7)

p≤N

The sum on m tends to 1 as N → ∞, since            |c(m)| c(m)   , − 1  ≤ s   m ms m>N m≥1    gcd(m, p≤N p)=1  with the right-hand sum being the tail of a convergent series. Finally, taking N → ∞ in (3.7) yields (3.5). The reader will have noticed that c(n) is precisely the “Möbius function”, familiar from elementary number theory. In future problem sets, it will be denoted μ(n) as usual.  c(n) Alternative Solution Suppose that ∞ n=1 ns converges absolutely for all real s > 1. Then, for these same values of s, ζ (s)

∞  c(n) n=1

ns

∞ ∞  c(n)  1  c(n) = ms ns (mn)s m=1 n=1 m,n≥1 ⎞ ⎛ ∞  ∞    1 c(n) ⎝ = = c(n)⎠ s Ns N

=

N =1 mn=N

= g(1) +

N =1

n|N

g(2) g(3) + s + ..., 2s 3

where g(N ) :=

 n|N

c(n).

Solutions to Step #3

83

This leads us to ask: Under what conditions on the sequence g(1), g(2), g(3), . . . is g(1) +

g(2) g(3) + s + ··· = 1 2s 3

for all s > 1? Certainly this holds if g(1) = 1 and g(N ) = 0 for N > 1. (Later we will prove the less obvious fact that this is the only way that equality can hold for all s > 1.) Thus, we will be done if {c(n)} can be chosen to satisfy 

 c(n) =

n|N

1 if N = 1,

(3.8)

0 otherwise,

 c(n) and also to make the series ∞ n=1 ns converge absolutely whenever s > 1. Is this possible? Yes! It is well known from elementary number theory that the Möbius function, defined by (3.6), satisfies (3.8). Since |c(n)| ≤ 1 for all n, the absolute convergence condition also holds. So c(n) as defined by (3.6) solves our problem. Remark. The method used in the first solution, and in Problem 3.22, is easily adapted to prove  f (n) the following very useful general theorem concerning series of the form ∞ n=1 ns : Let f be ∞ f (n) a complex-valued multiplicative function with f (1) = 1. Suppose that n=1 ns converges absolutely, for a certain s ∈ C. For this same s,   ∞  f (n) f (p 2 ) f (p) = + + . . . . 1 + ns ps p 2s p n=1

3.26 We imitate the solution to Problem 2.18. Why does stopping after an inclusion always give an overcount? It suffices to show that when r ≥ 0 is even, each x ∈ A is counted at least as many times by |A| −

 1≤i1 ≤k



|Ai1 | +

|Ai1 ∩ Ai2 |

1≤i1 δN.

n≤N, n∈A

Letting η = 12 δ, we see that for such N, ⎛  a∈A ηN x 1/a → 0, as the tail of a convergent series. So the desired result follows from the last display, upon dividing by x and sending x → ∞.

Solutions to Step #3

3.28

85

The integral test (or Cauchy condensation) implies that



1 m m(log m)1.01  then m a1m

converges. So if we were to have am ≥ m(log m)1.01 for all large m, would converge by the comparison test. So there must be infinitely many m with am < m(log m)1.01 . Write A(t) for n≤t, n∈A 1. Whenever am < m(log m)1.01 , m = A(am ) ≤ A(m(log m)1.01 ), so that if m is a large number of this kind and x := m(log m)1.01 , A(x) ≥ m = x/(log m)1.01 > x/(log x)1.01 , as desired. Many other functions can take the place of x/(log x)1.01 . For instance, running the above argument with m(log m)1.01 replaced by m(log m)(log log m)1.01 or m(log m)(log log m)(log log log m)1.01 allows us to take x/(log x(log log x)1.01 ) or x/(log x(log log x)(log log log x)1.01 ) here.   3.29 The key is to recognize that d|n g(d)h(n/d) = de=n g(d)h(e). Thus, 

g(d)h(n/d) =

n≤x d|n



g(d)h(e) =

n≤x de=n



g(d)h(e).

de≤x

Here the final sum is over all ordered pairs (d, e) of positive integers with de ≤ x. We can interpret this sum on pairs as a double sum by choosing to first sum on d or on e. In the former case, the e corresponding to a given d are precisely those with e ≤ x/d, so that  de≤x

g(d)h(e) =

 d≤x

g(d)



h(e).

e≤x/d

This yields the first equality claimed in the problem. The second is obtained similarly by first summing on e. Remark. We can recast the above solution in geometric terms. A sum on de ≤ x can be viewed as a sum over first-quadrant lattice points in the (d, e)-plane that lie on or below the hyperbola de = x. Summing on d (respectively, e) first corresponds to grouping together points on vertical (respectively, horizontal) lines.

3.30 

Taking g and h to be identically 1, the divisor function τ (n) satisfies g(d)h(n/d) = τ (n) for all n ∈ Z+ . So by the previous problem, d|n

86

Solutions to Step #3





τ (n) =

n≤x



1=

a≤x

ab≤x

=

1

 b≤x/a

1=

 x  a≤x

a

 1 + O(1) = x + O(x) = x log x + O(x). a a a≤x

 x a≤x

Here the last equality is a consequence of Problem 1.8.

 Remark. A more precise estimate for n≤x τ (n) was obtained by Dirichlet in 1849. √ √ Dirichlet’s starting point is the observation that ab ≤ x implies that a ≤ x or b ≤ x. This allows him to write        1= 1+ 1− 1. √ a≤ x b≤x/a

ab≤x

√ b≤ x a≤x/b

√ √ a≤ x b≤ x

The first two double sums coincide, both being equal to  √ m≤ x

x/m =

  x  1 √ + O(1) = x + O( x) m √ √ m

m≤ x

m≤ x

√ √ √ = x(log x + γ + O(1/ x)) + O( x) =

√ 1 x log x + γ x + O( x). 2

(To go from the√ first line to √ the second, we used again √ the result of Problem 1.8.) The third double sum is of size x2 = ( x + O(1))2 = x + O( x). Collecting our estimates, 

√ τ (n) = x log x + (2γ − 1)x + O( x).

n≤x

√ Voronoi showed in 1903 that the error term O( x) in the last line can be replaced by 1/3 O(x log (ex)). A folklore conjecture claims that this error is in fact O (x 1/4+ ), for any fixed  > 0. It is known (Hardy) that the “+” is necessary, in the sense that the error is not O(x 1/4 ).

3.31 (a) This is easy: τ (n) = k + 1 > k > k log 2 = log n. (b) Fix a positive integer m > A, and fix m distinct primes p1 , . . . , pm . For each positive integer k, τ ((p1 · · · pm )k ) = (k + 1)m . On the other hand, for C := (log(p1 · · · pm ))A , (log((p1 · · · pm )k ))A = Ck A . Since m is larger than A, the quantity (k + 1)m will exceed Ck A for all large enough values of k. So any n of the form (p1 · · · pm )k , with k large enough, satisfies the conclusion of (b).

Solutions to Step #3

87

3.32 The map √ d → n/d sets up√a one-to-one correspondence between the divisors of n below n and those above n. It follows that τ (n) = 2

 d|n √ d≤ n

1−

 1 0

if n is a square, otherwise.

√ The desired inequality τ (n) ≤ 2 n is an immediate consequence. Remark. It is natural to ask about the optimality of the exponent 12 in the bound τ (n) ≤ 2n1/2 . Is τ (n) ≤ Cn1/3 (for some C and all n)? If so, what about τ (n) ≤ C  n1/100 ? In fact, for each fixed  > 0, there is a C such that τ (n) ≤ C n

(3.11)

for all n ∈ Z+ . This result was first proved by Runge in 1885, who used it to show that, in a precise sense, almost all quintic polynomials x 5 + ux + v (u, v ∈ Z) are not solvable by radicals. A stronger result than (3.11) will be proved later (Problem 11.114), so we only give the idea of the proof here, leaving the reader to work out the details. First, one shows that if p > 21/ , then τ (p k ) = k+1 < p k for every positive integer k. Next, one argues that for any fixed p, the function τ (p k )/p k assumes a maximum value as k ranges over Z+ . Call this value Cp, . As noted already, Cp, < 1 when p > 21/ . On the other hand, it is easy to see (consider k = 1) that Cp, ≥ 1 for p ≤ 21/ . Thus, for each positive integer n,

τ (p k )

τ (n) ≤ ≤ Cp, ≤ Cp, .  k n p k 1/ p n p≤21/

So we can take C =

p≤21/

p|n p≤21/

p≤2

Cp, .

3.33 This is equivalent to showing that for each N, there is a positive integer d that occurs at least N times as a difference of primes q − p.   With x a large real number to be specified shortly, we consider all π(x) pairs of 2 primes (p, q) with p < q ≤ x. Since each difference q − p < x, there are fewer than x possible values for the difference q − p. Hence some difference occurs for at least   1 π(x) x 2 such pairs. We will argue that this quantity can be made sufficiently large by a suitable choice of x. From Problem 3.24, the sum of the reciprocals of the primes diverges. By Problem 3.28, there are arbitrarily large of x with  values  > x/(log x)2.1 π(x) > x/(log x)1.01 . For sufficiently large x of this kind, x1 π(x) 2 (say). This lower bound tends to infinity with x, and the result follows. Remark. In 2013, Y. Zhang proved the breakthrough result that there are positive integers d that occur infinitely often as a difference q − p. The “twin prime conjecture”, still open, makes the claim that d = 2 has this property.

Solutions to Step #4

4.34 (a) The first inequality is obvious, since (x/p) ≥ 1 for every prime p over which the sum is taken. The second inequality then follows from Problem 3.24 with s = 1 +  (∈ (1, 2)). 1 (b) We take  = log x·log log x , which belongs to (0, 1) for large enough x. Since exp(u) = 1 + O(u) as u → 0, we see that  x  = exp

1 log log x



 =1+O

1 log log x

 .

Moreover, log(1/) = log(log x · log log x)   log log log x . = log log x 1 + log log x Therefore,     log log log x 1 1 x log = log log x 1 + 1+O  log log x log log x 

= log log x + log log log x + O(1). So from part (a), keeping in mind that x  = O(1), we infer that 1 ≤ log log x + log log log x + O(1). p p≤x

© Springer Nature Switzerland AG 2021 P. Pollack, A. Singha Roy, Steps into Analytic Number Theory, Problem Books in Mathematics, https://doi.org/10.1007/978-3-030-65077-3_19

89

90

Solutions to Step #4

This is slightly sharper than the upper bound claimed in the problem statement. Remark. How did we come up with this ? We viewed x as fixed and chose  to nearly minimize x  log 1 . Viewed as a function of , the quantity x  log 1 blows up as  ↓ 0 and as  → ∞; being a continuously varying function of , it must therefore assume a global minimum on (0, ∞). Taking derivatives, we see that the global minimum is attained at the unique solution  to  log  = −

1 . log x

Equivalently, putting t = 1 , the minimum occurs when t/ log t = log x. This last equation implicitly defines t as a function of x. While there is no simple closed form solution for t in terms of x, it is not hard to approximate t. Observe that as x → ∞, so does our solution t. Moreover, taking logarithms, log t − log log t = log log x. Thus,   log log t −1 log t = log log x 1 − . log t Multiplying with t/ log t = log x, we see that   log log t −1 t = log x · log log x 1 − . log t The parenthesized factor tends to 1 as x (and hence t) tends to infinity. Thus, log x · log log x is a 1 reasonable approximation to t for large x, making log x·log log x a reasonable choice for .

4.35 (a) Clearly, p1+ > p for all primes p, so that taking s = 1 +  in Problem 3.24 yields     1 1  1  1 1 1 ≥ − = − ≥ log + O(1). 1+ 1+ 1+ 1+ p  p p p n p≤x p≤x p p>x n>x  (b) We insert into (a) the upper bound of Problem 2.11 for n>x that   1 1 1+ ≥ log − x − + O(1) p   p≤x = log

1 x − − + O(1).  

1 . n1+

This gives

Solutions to Step #4

91 −

Now we look to maximize the function log 1 − x  for  ∈ (0, ∞). Taking derivatives, the global maximum occurs at an  with 1 log x x + 2 = .    Equivalently, writing  = t/ log x, (log x)2 (t −1 + t −2 ) = et (log x)t −1 , or (1 + t −1 ) log x = et . The left side here is larger than log x, and so for equality to hold, t must be large (certainly, t > log log x). Thus, 1 + t −1 is close to 1, and now referring back to the same equation, et is roughly equal to log x. This suggests that the maximum is attained when t ≈ log log x, so that  ≈ log log x/ log x. And indeed if we take  = logloglogx x , then log

1 = log log x − log log log x, 

while 1 − log x log x − log log x 1 − log log x x = · x log x = e = O(1). =  log log x log log x log log x Thus, as x → ∞, 1 ≥ log log x − log log log x + O(1), p p≤x which is again slightly stronger than the claimed estimate. Remark. Combining the results of Problems 4.34 and 4.35, we deduce that  1 = log log x + O(log log log x). p p≤x Later we will see how to establish still sharper estimates for

4.36 The identity φ(n) = n



d|n μ(d)/d

f (n) =

 d|n

=





1 p≤x p .

d|n μ(d)(n/d)

g(d)h(n/d)

is of the form

92

Solutions to Step #4

with f (m) = φ(m), g(m) = μ(m), and h(m) = m. So by Problem 3.29, 



φ(n) =

n≤x

μ(a)

a≤x



b=

 μ(a)  x  a≤x

b≤x/a

2

a

+

 x 2  a

.

This last sum 1 = μ(a) 2 a≤x



  2  x x + O(1) + + O(1) a a

 x   x2 +O a a2   1 x 2  μ(a) = +O x . 2 a≤x a 2 a a≤x 1 = μ(a) 2 a≤x

To finish, recall that



1 a≤x a



= O(log (ex)) for all x ≥ 1.

4.37 Let x ≥ 1. By the triangle inequality and Problem 2.11,      μ(a)   1 1  ≤ . = O   2 2 x a a a>x a>x Now recalling the result of Problem 3.25,  μ(a) a≤x

a2

=

∞  μ(a) a=1

a2

−

 μ(a) a>x

a2

=

1 +O ζ (2)

  1 . x

To finish, insert this estimate into the result of Problem 4.36. 4.38 Say that (a, b) ∈ Z2 is a “coprime lattice point” if gcd(a, b) = 1. We start by considering those coprime lattice points in (0, N] × (0, N] that lie on or below the diagonal y = x. The number of these is 

#{1 ≤ b ≤ a : gcd(b, a) = 1} =

a≤N



φ(a).

a≤N

The number of coprime lattice points on or above the diagonal is exactly the same, by the symmetry swapping a and b. Finally, there is precisely one coprime lattice point on the diagonal, namely (1, 1). Hence, the total count of coprime lattice points in (0, N] × (0, N] is 2

 a≤N

φ(a) − 1,

Solutions to Step #4

93

which, by the last problem, is 1 N 2 + O(N log (eN )). ζ (2) To find the limiting probability, we divide by N 2 (the total number of lattice points 1 in the square) and send N → ∞; this gives ζ (2) , as desired. Remark. This result has an interesting geometric interpretation. It is easy to see that a lattice point P is coprime precisely when there is no lattice point on the interior of the segment OP , where O = (0, 0) denotes the origin. For this reason, coprime lattice points are often called “visible points”, meaning “visible from the origin”. The result we just proved says (in rough terms) that a 1 randomly chosen lattice point in the plane is visible with probability ζ (2) . It is natural to ask for the corresponding probability in higher dimensions. The general theorem is that a randomly chosen lattice point in Rn is coprime with probability 1/ζ (n). How might one prove this? We have already handled the case n = 2, so assume that n ≥ 3. Just as when n = 2, (x1 , . . . , xn ) ∈ Zn is visible ⇐⇒ gcd(x1 , . . . , xn ) = 1. To detect the latter condition, we apply directly the following fundamental property of the Möbius  function: If m is a positive integer, then d|m μ(d) = 1 or 0, according to whether m = 1 or not. As a consequence, for any choice of integers x1 , . . . , xn , not all 0, 

μ(d) =

d|x1 ,...,xn





μ(d) =

d|gcd(x1 ,...,xn )

1 if (x1 , . . . , xn ) is visible, 0 otherwise.

Hence, the total count of visible lattice points in (0, N ]n is precisely 



μ(d) =

x1 ,...,xn ≤N d|x1 ,...,xn



μ(d) N/dn .

d≤N

1 We leave to the reader to work out that this last quantity has size ζ (n) N n + O(N n−1 ), as N → ∞. There are many possible variations on this problem that are worth considering. For instance: What is the probability that a randomly chosen (a1 , a2 , a3 ) ∈ (Z+ )3 is “pairwise coprime”, meaning gcd(ai , aj ) = 1 for all pairs i = j ? what is the analogous probability in (Z+ )n ? Answers can be found in [1].

4.39 If a ≡ 0 (mod m) then e of a geometric series,  k mod m

e

2π ika m

2π ia m

=

= 1. By the familiar formula for the partial sums

m−1  k=0

(e

2π ia m

) = k

(e

2π ia m

e

On the other hand, if a ≡ 0 (mod m), then the sum equal to 1, and the result is immediate.

)m − 1

2π ia m



−1

= 0.

k mod m e

2π ika m

has all its terms

 2π ika 4.40 The identity in the previous problem tells us that m1 k mod m e m is the indicator function for the property “m divides a” (that is, it takes the value 1 when m | a, and takes the value 0 otherwise). Hence, for a given positive integer m, and

94

Solutions to Step #4

given integers a and n,  1  2π ik(n−a2 ) 1 if a 2 ≡ n (mod m), e m = m 0 otherwise. k mod m Summing on a, we deduce that the number of square roots of n modulo m is given by 1  m



e

2π ik(n−a 2 ) m

=

a mod m k mod m

1  2π ikn e m m k mod m



e

−2π ika 2 m

,

a mod m

exactly as claimed.

 4.41 We begin with some general considerations. Suppose that ∞ n=1 an is any convergent series of complex numbers. Then limn→∞ an = 0 (the nth term test). Thus there is positive integer N such that |an | ≤ 1 whenever n > N. Then |an | ≤ C for all natural numbers n if we set C := max{1, |a1 |, . . . , |aN |}. Applying this general observation with an = a(n) ns , we get the first assertion of the problem. The next assertion then follows from the triangle inequality, the 1 comparison test, and the convergence of ∞ n=1 ns−s for all s > s0 + 1. It remains to prove the final claim, giving a limit expression for a(m) for every natural number m. The claim can be rephrased as the assertion that 0

0

lim ms

s→∞

 a(n) = 0. ns n>m

We know already that |a(n)| ≤ Cns0 for every natural number n. So by the triangle inequality, for every s > s0 + 1,     1  a(n)  s ≤ Cm 0 ≤ ms .  ns ns−s0 n>m n>m Since

1 t s−s0

is decreasing for t > 0,  n>m

1 ns−s0

 ≤

∞ m

dt t s−s0

=

1 ms0 −s+1 , s − s0 − 1

Reference

95

and Cms



1 C ms0 +1 . ≤ s−s0 n s − s − 1 0 n>m

The right-hand side clearly tends to 0 as s → ∞, and the claim follows. 4.42 Every perfect power N can be written uniquely in the form nm , where n, m ≥ 2 and n is not a perfect power. (Here m may be determined as the gcd of the exponents in the prime factorization of N.) Therefore, using  for a sum over non-perfect powers,  N perfect power

    1/nm 1 1 = = N −1 nm − 1 1 − 1/nm n≥2 m≥2

n≥2 m≥2

   1    1   1 . = = = km km k k n n n (n − 1) n≥2 m≥2 k≥1

n≥2 k≥1 m≥2

n≥2 k≥1

Every integer from 2 onward has a unique representation in the form nk , where n ≥ 2, n is not a perfect power, and k ≥ 1. Hence,   n≥2 k≥1

1 nk (nk

− 1)

=

 r≥2

1 = 1. r(r − 1)

Here our carefree approach to series manipulations is justified by our dealing only with series of positive terms.

Reference 1. László Tóth, The probability that k positive integers are pairwise relatively prime. Fibonacci Quart. 40 (2002), 13–18.

Solutions to Step #5

  5.43 Suppose first that pk = 1. Then each nonzero square modulo p occurs twice as a value of ka 2 mod p, as a runs mod p, while 0 occurs once. This description of the  2 values assumed by ka 2 is independent of k, and so the value of a mod p e2π ika /p k  is the same for all k with p = 1. In particular, since 1 is a square, 

2π ika 2 /p

e



=

a mod p

2π i·1·a 2 /p

e

a mod p

  k G =G= p

  in this case. Now suppose that pk = −1. Then every residue class appears twice when we take the concatenation of the lists {a 2 } and {ka 2 }, where again a runs mod p. Thus, 

e2π ia

2 /p

+

a mod p



e2π ika

2 /p

=2

a mod p



e2π ib/p = 0.

b mod p

(Here the final relation is from Problem 4.39.) Hence, 

2π ika 2 /p

e

=−

a mod p



2π ia 2 /p

e

a mod p

  k G = −G = p

in this case as well. 5.44 By Problem 4.40, the number of square roots of n modulo p is 1  2π ikn/p  −2π ika 2 /p e e . p k mod p

a mod p

© Springer Nature Switzerland AG 2021 P. Pollack, A. Singha Roy, Steps into Analytic Number Theory, Problem Books in Mathematics, https://doi.org/10.1007/978-3-030-65077-3_20

97

98

Solutions to Step #5

We isolate the contribution of the terms with k ≡ 0 and then insert the result of the last problem: 1  2π ikn/p  −2π ika 2 /p e e p k mod p

a mod p

⎛ =

⎞

  ⎟ 1⎜ 2 ⎜p + e2π ikn/p e−2π ika /p ⎟ ⎝ ⎠ p k mod p k≡0

=1+

a mod p

  G  −k 2π ikn/p e . p p k mod p k≡0

The value of the sum on k is unchanged if we remove the restriction   that k ≡ 0, and the first assertion of the problem follows. The expression for pn is a simple consequence, since the number of square roots of n modulo p is also given by the  formula 1 + pn .   5.45 Starting from the identity of Problem 5.44 for pn and taking the squared modulus of both sides, we find that  2 ¯   −k  2π ikn/p   −k   −2π ik  n/p  n G·G . e e = p p p p2  k mod p

k mod p

Summing both sides over n mod p yields p−1=

¯ G·G 2 p

 k,k  mod p



kk  p

 



e2π i(k−k )n/p

n mod p

¯   k2  G·G = p p k mod p

=

¯ G·G (p − 1). p

(Here we used Problem 4.39 to evaluate the sum on n in the first line.) Rearranging, ¯ = p. G·G √ √ Remark. Gauss showed in 1805 that G = p when p ≡ 1 (mod 4) and G = i p when p ≡ 3 (mod 4). There are by now several proofs of this result (see Chapter VI of [1] for four different arguments), but none that can really be called easy. We sketch one of the simpler arguments, due to Schur. The reader familiar with linear algebra is invited to fill in the details, consulting [1] for hints when necessary.

Solutions to Step #5

99

We let ω = e2π i/p , and we introduce the p × p matrix

A = (ωij )0≤i,j . n ζ (2) n≤x n ζ (2) 2

(In the last step we used that ζ (2) = 1 + logarithms,



1 n≥2 n2

log log x − log 2, p p≤x which is slightly sharper than the inequality that was claimed.  5.48 We can write σ (n) = d|n g(d)h(n/d), where g and h are the arithmetic functions defined by setting g(m) = m and h(m) = 1 for all m ∈ Z+ . By Problem 3.29, 

σ (n) =

n≤x

  b≤x a≤x/b

=

a=

1

x/b( x/b + 1) 2 b≤x

   1 1 1  1 . ((x/b)2 + O(x/b)) = x 2 + O x 2 2 b b2 b≤x

b≤x

b≤x

The O-term here is O(x log (ex)), while    1  1 1 . = ζ (2) − = ζ (2) + O 2 2 x b b b≤x

b>x

Substituting these estimates above gives the desired result. 5.49 Since each of 2ω(n) , τ (n), and 2(n) is multiplicative, we can assume that n is a prime power, say n = pk . In that case, the claim is that 2 ≤ k + 1 ≤ 2k . The first inequality is trivial, while the second is easily proved by induction on k. 5.50 Let gk (n) denote the number of representations of n as a product of k integers > 1, where order matters. Clearly, g(n) =

 k≥0

gk (n).

102

Solutions to Step #5

Now notice that for every integer k ≥ 0, ⎛

⎞k  1 ⎠ = (ζ (s) − 1)k = ⎝ ms m≥2

 m1 ,...,mk ≥2

∞  1 gk (N ) = , s (m1 · · · mk ) Ns N =1

so that summing on k, 

(ζ (s) − 1)k =

∞ ∞   1  g(N ) g (N ) = . k s N Ns

N =1

k≥0

N =1

k≥0

We have so far ignored issues of convergence. This is not as much of a problem as it may first appear. Since all of our series manipulations have been on series of nonnegative terms, all of the identities claimed above hold without any restriction on s, provided equality is interpreted in the following sense: If one side is finite, the other is the same finite real number, while if one side is infinite, then both are infinite. If we suppose now that s > ρ, then 1 < ζ (s) < 2, so that 0 < ζ (s) − 1 < 1. Thus, 

(ζ (s) − 1)k =

k≥0

1 1 = , 1 − (ζ (s) − 1) 2 − ζ (s)

and we obtain the identity claimed in the problem. 5.51 First notice that a(1) = b(1), since by Problem 4.41, a(1) = lim

s→∞

∞  a(n) n=1

ns

= lim

s→∞

Since a(1) = b(1), we can remove the terms series and deduce that ∞  a(n) n=2

ns

=

a(1) 1s

∞  b(n) n=1

and

∞  b(n) n=2

ns

ns b(1) 1s

= b(1).

from our starting Dirichlet

(5.15)

for all large enough real s. Multiplying both sides by 2s and sending s → ∞ gives— again using Problem 4.41—that a(2) = b(2). Now remove the terms a(2)/2s and b(2)/2s from (5.15), multiply by 3s and send s → ∞ to get a(3) = b(3). Continuing in this way, we obtain (by induction) that a(n) = b(n) for all positive integers n.

Solutions to Step #5

103

5.52 For every integer m ≥ 2, 

π/2



π/2

sin x dx = m

0



0 π/2

= 

(sinm−2 x)(1 − cos2 x) dx 

0 π/2

=

π/2

sinm−2 x dx − 

0 π/2

sinm−2 x dx −

0

cos x · (sinm−2 x cos x) dx  cos x · d

0

 1 sinm−1 x . m−1

Now integrating by parts, 

π/2

 cos x · d

0

1 sinm−1 x m−1



π/2  π/2  1 1 m−1  cos x sin = x + sinm x dx m−1 m−1 0 0  π/2 1 = sinm x dx. m−1 0

Substituting this back above and rearranging, m m−1



π/2



π/2

sin x dx = m

0

sinm−2 x dx,

0

and so 

π/2

sinm x dx =

0

m−1 m



π/2

sinm−2 x dx.

0

Both of the claimed identities now follow by induction, starting from 

π/2

sin0 x dx =

0

π 2

and 

π/2

sin x dx = 1.

0

π/2 5.53 For brevity, write Im for 0 sinm x dx. For x from the interval [0, π/2], 0 ≤ sin x ≤ 1, so that sinm+1 x ≤ sinm x. It follows immediately that Im+1 ≤ Im for all m. Consequently,

104

Solutions to Step #5

1=

I2n+1 I2n I2n−1 2n + 1 ≤ ≤ = I2n+1 I2n+1 I2n+1 2n

for every positive integer n. By the squeeze theorem, I2n /I2n+1 → 1. From Problem 5.52, I2n I2n+1

=

  n   n  2n 

2i − 1 2 π π j + 1 2i − 1 2 · (2n + 1) · = 2 2i 2 j 2i j =1

i=1

π = 2

n 

i=1

(2i)(2i + 1) (2i − 1)(2i)

 n  i=1

2i − 1 2i

2

π = 2

i=1

n 

i=1

 2i − 1 2i + 1 · . 2i 2i

Hence,    n 

2i 2i π I2n −1 · = · . 2i − 1 2i + 1 2 I2n+1 i=1

Send n to infinity to conclude that  ∞ 

2i 2i π · = , 2i − 1 2i + 1 2 i=1

as desired.

Reference 1. Edmund Landau, Elementary number theory. Chelsea Publishing Co., New York, N.Y., 1958.

Solutions to Step #6

6.54 Summing the geometric series, 

e2π ikn/p =

n≤N

=

e2π ik(N +1)/p − e2π ik/p e2π ik/p − 1 e2π ik/p · eπ ikN/p eπ ikN/p − e−π ikN/p · π ik/p . eπ ik/p e − e−π ik/p

The first right-hand fraction has absolute value 1. The second has numerator 2i sin(π kN/p) and denominator 2i sin(π k/p). Thus,       2π ikn/p  | sin π kN 1 p |  = ≤ , e   πk | sin p | | sin πpk | n≤N  as desired. 6.55 We begin by establishing the inequalities indicated in the hint. We could prove (i) by mucking about with derivatives. But here is a slicker proof. Let S(θ ) = sinπ πθ θ , with S(0) = 1. Then for all real θ , 

1

S(θ ) =

cos(π θ t) dt. 0

Since cos(π u) is a decreasing function of u on [0, 12 ], we infer that whenever 0 ≤ θ ≤ θ  ≤ 12 , 

1

S(θ ) = 0

 cos(π θ t) dt ≥

1

cos(π θ  t) dt = S(θ  ).

0

© Springer Nature Switzerland AG 2021 P. Pollack, A. Singha Roy, Steps into Analytic Number Theory, Problem Books in Mathematics, https://doi.org/10.1007/978-3-030-65077-3_21

105

106

Solutions to Step #6

That is, S is also decreasing on [0, 12 ]. In particular, S(θ ) ≥ S( 12 ) = π2 for all θ ∈ [0, 12 ]. This gives (i) in the range 0 ≤ θ ≤ 12 ; that (i) also holds when − 12 ≤ θ ≤ 0 follows immediately upon noting that S(θ ) = S(−θ ). To prove (ii), observe that 

1

−1

1 dt = 2n + t

1

 0



1

= 0

 1 1 + dt 2n + t 2n − t  1  1 1 4n 1 1 dt = . dt = dt > 2 t n n 4n2 − t 2 0 n− 0 4n

With these out of the way, we can proceed with the main event,    preliminaries estimating n≤N pn . From Problem 5.44,     n G  2π ikn/p −k = , e p p p k mod p

where G =

 a mod p

e2π ia

2 /p

 n p

n≤N

Recalling that |G| =

is the Gauss sum. Thus, =

  −k  2π ikn/p G  e . p p k mod p

n≤N

√ p and keeping in mind the last problem,

         √      n  1 2π ikn/p   ≤ p  e   ≤ √p   p p   n≤N k mod p n≤N k≡0

≤

√

p

 |k| p0 (), there is an integer smaller than p 2 + generating the cyclic group Up . For this and much more, see Chapter 8 of Shapiro’s textbook [6].

6.57 The expression for π(x, y) in terms of alternating sums follows from the Principle of Inclusion-Exclusion, by the same reasoning as in Problem 2.19. Now we drop the floor functions. Replacing x with x incurs  an error of at most 1 (in absolute value). By the triangle inequality, replacing p1 ≤y px1  with  x p1 ≤y p1 incurs an error of size at most π(y)—the number of terms in the sum. In general, when we replace  p1 ρ, ∞  ge (n) − go (n) n=1

ns

= 1 − (ζ (s) − 1) + (ζ (s) − 1)2 − . . . ∞

=

 μ(n) 1 1 = = . 1 + (ζ (s) − 1) ζ (s) ns n=1

(Here the final equality is from the solution to Problem 3.25.) Now we appeal to Problem 5.51: If two Dirichlet series represent the same function for all large enough s, then their coefficients are equal. Consequently, ge (n) − go (n) = μ(n) for all positive integers n.

114

Solutions to Step #6

References 1. Raymond Ayoub, Euler and the zeta function. Amer. Math. Monthly 81 (1974), 1067–1086. 2. Robin J. Chapman, Evaluating ζ (2). Published online at https://empslocal.ex.ac.uk/people/staff/ rjchapma/etc/zeta2.pdf. 3. Loo Keng Hua, Introduction to number theory. Translated from the Chinese by Peter Shiu. Springer-Verlag, Berlin-New York, 1982. 4. Kenneth Ireland and Michael A. Rosen, A classical introduction to modern number theory. Second edition. Graduate Texts in Mathematics, vol. 84. Springer-Verlag, New York, 1990. 5. Konrad Knopp, Theory and application of infinite series. Blackie and Son, London–Glasgow, 1954. 6. Harold N. Shapiro, Introduction to the theory of numbers. John Wiley & Sons, Inc., New York, 1983.

Solutions to Step #7

2 LCM(2n+1) 7.65 Let p be a prime dividing LCM(2n+1) LCM(n+1) . We must show that p  LCM(n+1) . Let pk be the highest power of p dividing LCM(2n + 1), so that k ≥ 1. Then pk ≤ 2n + 1 (in fact, pk is the largest power of p with pk ≤ 2n + 1). Thus, pk−1 ≤ 12 pk < n + 1, and so pk−1 | LCM(n + 1). As pk is the highest power of p dividing LCM(2n + 1) and pk−1 | LCM(n + 1), we conclude that p2  LCM(2n+1) LCM(n+1) . LCM(2n+1) LCM(n+1)

is squarefree, it is enough to show that every prime 2n+1 dividing this ratio also divides n+1 . For this we will use the following well-known theorem of Legendre: If p is prime and x ≥ 1, then the exponent on the highest power of p dividing x! is

7.66 Since the ratio

   x x x + + .... + p p2 p3 Let p be a prime dividing LCM(2n+1) LCM(n+1) . Then the highest power of p less than or equal to 2n + 1 cannot also be less than or equal to n + 1. Thus there is some power of p, say pk , in the interval(n +1, 2n + 1]. By Legendre’s formula, the exponent (2n+1)! on p in the factorization of 2n+1 n+1 = (n+1)!n! is   2n + 1 r≥1

pr



  n+1 n − − . r p pr

By the choice of k, the term of the sum corresponding to r = k is positive. So to finish the proof, it is enough to show that every term in the sum is nonnegative. This is easy: First observe that

© Springer Nature Switzerland AG 2021 P. Pollack, A. Singha Roy, Steps into Analytic Number Theory, Problem Books in Mathematics, https://doi.org/10.1007/978-3-030-65077-3_22

115

116

Solutions to Step #7



 n+1 n n+1 n 2n + 1 + ≤ + r = . pr pr pr p pr

      n 2n+1 + ≤ Since the left-hand side is an integer, the inequality n+1 r r r p p p follows. For completeness, we give a proof of Legendre’s formula. For each positive integer  m, the exponent on the highest of power of p dividing m can be written as

j ≥1: pj |m 1. Hence, the exponent on p appearing in the factorization of x! = n≤x n is 

1=

n≤x j ≥1 pj |n

 x . 1= pj n≤x

 j ≥1

pj |n

j ≥1

    7.67 The first inequality is clear, since 2n+1 = 2n+1 , and both of those terms n n+1 2n+1 2n+1  1 n appear in = 4 follows from the binomial k≥0 k . That 2n+12 k≥0 k 2n+1 theorem, noting that k≥0 k = (1 + 1) . 7.68 If the proposed inequality fails, let n be the smallest counterexample. If n is even, then n2 ≤ n − 1. Thus, n = 2 · n2 divides 2 · LCM(n − 1). Clearly, each of 1, 2, . . . , n − 1 also divides 2 · LCM(n − 1). Hence, LCM(n) ≤ 2 · LCM(n − 1) ≤ 2 · 4n−1 < 4n , contrary to hypothesis. So n must be odd. Write n = 2m + 1, where m ∈ Z. Clearly, n > 1, and so m ∈ Z+ . From Problems 7.66 and 7.67, LCM(2m + 1) ≤ 4m , LCM(m + 1) so that LCM(n) = LCM(2m + 1) ≤ 4m · LCM(m + 1) ≤ 4m · 4m+1 = 42m+1 = 4n . So n is not a counterexample to the inequality after all, a contradiction. 7.69 It is easy to prove (with calculus or without) that 0 ≤ t (1 − t) ≤ t ∈ [0, 1]. That 1 ≥ 4n

 0

1

t n (1 − t)n dt.

1 4

for all

Solutions to Step #7

117

follows immediately. If we expand out t n (1 − t)n , we obtain a sum of terms At m , where A ∈ Z and n ≤ m ≤ 2n. For each of these terms, 

1

LCM(2n + 1)

At m dt = A ·

0

LCM(2n + 1) ∈ Z, m+1

and so 

1

LCM(2n + 1)

t n (1 − t)n dt ∈ Z.

0

To replace Z with Z+ on the right-hand side, it remains only to note that the left-hand side is positive (which is clear, since t n (1 − t)n is continuous on [0, 1], nonnegativevalued, and not identically 0). Since 1 is the least positive integer, LCM(2n + 1)

1 ≥ LCM(2n + 1) 4n



1

t n (1 − t)n dt ≥ 1,

0

and so LCM(2n + 1) ≥ 4n . 7.70 Every block of m consecutive positive integers contains precisely one integer belonging to a given congruence class mod m. The set of positive integers ≤ x contains the first x/m such blocks and is contained within the first x/m such blocks. Both x/m and x/m are within 1 of x/m, and the result follows. 7.71 Let Ad denote the set of n ≤ x for which d | n(n + 2), so that Ad = #Ad . In this notation,   π2 (x, y) = A1 \ ∪p≤y Ap  . Each Ap is a subset of A1 , so by inclusion-exclusion, our desired count is |A1 | −

 p1 ≤y

|Ap1 | +

 p1 δm log m. Suppose that π(x) > (log 4+) logx x . We will see that this leads to a contradiction for large x, provided δ was chosen sufficiently close to 1 to start with. Indeed, for all sufficiently all large x, log π(x) > log x − log log x + log log 4 > δ log x, and so   x · δ log x log π(x)! > δπ(x) log π(x) > δ (log 4 + ) log x = δ 2 (log 4 + )x. This contradicts (8.21) if δ 2 >

log 4 log 4+ .

8.79 A prime p divides LCM(x) precisely when p ≤ x. For each such p, the exponent on p in the factorization of LCM(x) is the largest integer k with pk ≤ x. That k is nothing other than log x/ log p. Thus, LCM(x) =

p

log x/ log p

,

and

p≤x

log x . log LCM(x) = log p log p p≤x 



Clearly, 



log x log p log p p≤x

≤

 p≤x

log p ·

 log x = log x = π(x) log x. log p p≤x

8.80 Let n be the largest integer with 2n + 1 ≤ x. For x ≥ 3, we have n ≥ 1, so that by Problems 7.69 and 8.79, n log 4 ≤ log LCM(2n + 1) ≤ log LCM(x) ≤ π(x) log x. But n = 12 (x − 1) > 12 (x − 3), and so n > 12 (1 − )x for all large enough x. Thus, for all large x,

Solutions to Step #8

125

π(x) ≥ log 4 ·

x x n > log 2 · (1 − ) > (log 2 − ) . log x log x log x

8.81 Suppose K > 2. For every fixed  > 0 and all large enough x, π(Kx) (log 2 − ) · Kx/ log(Kx) > π(x) (log 4 + ) · x/ log x =K Since K > 2 and

log 2 log 4

=

1 2,

log 2 −  log x · . log 4 +  log Kx

(8.22)

we can fix  > 0 small enough to ensure that

2− K log log 4+

log x > 1. Since the ratio log Kx tends to 1, (8.22) is then larger than 1 for all large x. In other words, π(Kx) > π(x), and so the interval (x, Kx] contains at least one prime.

8.82 We have already said that the highest power of p dividing LCM(x) is pk , with k the largest integer satisfying pk ≤ x. Musing on this for a moment, we see that LCM(x) admits the factorization

LCM(x) =

p,

pk ≤x

where the product runs over all the numbers pk with p prime, k ≥ 1, and pk ≤ x. It follows that for x ≥ 1, ⎛ ⎞  

 

LCM(x/m) = log ⎝ p⎠ = log p log m≤x

m≤x pk ≤x/m

=

 pk ≤x

log p

 m≤x/pk

m≤x pk ≤x/m

1=

 pk ≤x

x . log p pk 

This last expression can be rewritten as    x x + ... , + log p p p2 p≤x 

which we recognize as log x! (see the end of the solution to Problem 7.66). Now exponentiate. 8.83 (a) We begin by showing that χ is multiplicative. In fact, χ has the stronger property that

126

Solutions to Step #8

χ (nm) = χ (n)χ (m)

for all pairs of integers n, m.

(8.23)

To see this, note that if 3 divides n or m, then both sides of (8.23) vanish. If both n, m ≡ 1 (mod 3), or both n, m ≡ −1 (mod 3), then both sides are 1. If n ≡ 1 (mod 3) while m ≡ −1  (mod 3), or vice-versa, both sides are −1. χ (n) Next, we notice that ∞ n=1 ns converges absolutely for real s > 1, since |χ (n)| 1 ns ≤ ns for all n.  χ (n) These observations allow us to factor ∞ n=1 ns , by the procedure used in the solution to Problem 3.22. (See the remark made at the end of the solution to Problem 3.25 for a general statement along these lines.) We obtain in this way that for all real s > 1, ∞  χ (n) n=1

ns



 χ (p) χ (p2 ) 1+ = + + ... . ps p2s p

(8.24)

Because of (8.23), the factor corresponding to p in (8.24) is a geometric 1 series with first term 1 and common ratio χp(p) s , and so has sum χ (p) . 1−

ps

(b) For each s > 1, the series for L(s) is alternating with terms decreasing in absolute value. Thus,     1 1 1 1 − + ... L(s) = 1 − − − 2s 4s 5s 7s ≤ 1 − 0 − 0 − 0 − · · · = 1, and       1 1 1 1 1 L(s) = 1 − s + + + ... − − 2 4s 5s 7s 8s   1 1 ≥ 1− + 0 + 0 + 0 + ··· = . 2 2 That log L(s) = O(1) is now immediate: log 12 ≤ log L(s) ≤ 0. (c) Taking logs, log L(s) = log

1−

p 1 Recalling that log 1−x =x+

 p

log

x2 2

1

+

x3 3

1 1−

χ (p) ps

χ (p) ps

=

 p

log

1 1−

χ (p) ps

+ . . . when |x| < 1, =

  χ (pk ) p k≥1

kpks

.

.

Solutions to Step #8

127

This double sum on p and k converges absolutely for each s > 1, since   |χ (pk )| p k≥1

kpks

≤

 1  1 = log = log ζ (s). ks kp 1 − p1s p p k≥1

Thus, we may rearrange the terms to sum on k first. Doing so, we find that   χ (pk ) p k≥1

kpks

=

 χ (p) p

ps

+

  χ (pk ) k≥2 p

kpks

,

where       χ (pk )    1  ≤ < 1.  ks  ks  k≥2 p kp  k≥2 p kp (See the solution to Problem 3.23 for the final inequality.) Hence, log L(s) =

 χ (p) ps

p

+ O(1)



=

p≡1 (mod 3)

1 − ps

 p≡−1 (mod 3)

1 + O(1). ps

From (b), log L(s) = O(1). Therefore,  p≡1 (mod 3)

1 − ps

 p≡−1 (mod 3)

1 = O(1). ps

(d) For 1 < s < 2, we have from Problem 3.24 that  p≡1 (mod 3)

1 + ps

 p≡−1 (mod 3)

   1 1 1 = − s s s p p 3 p = log

1 + O(1). s−1

Adding or subtracting the result of (c) from this, we obtain that for either choice of sign,  p≡±1 (mod 3)

1 1 1 = log + O(1). s p 2 s−1

128

Solutions to Step #8

 8.84 Letting s ↓ 1 in Problem 8.83(d), we see that p≡±1 (mod 3) p1s → ∞, for  either choice of sign. Hence, p≡±1 (mod 3) p1 diverges.  To get the stated estimates for p≤x, p≡±1 (mod 3) p1 , we adapt the solutions of Problems 4.34 and 4.35. First, let  =  p≤x p≡±1 (mod 3)

1 ≤ p =

 p≤x p≡±1 (mod 3)

p≤x p≡±1 (mod 3)

1 ≥ p

≥

  x = x p

Then for large x, 

1

p≡±1 (mod 3)

p1+

1 1  1 1 x log + O(x  ) = log log x + log log log x + O(1). 2  2 2

On the other hand, with  = 

1 p

1 log x·log log x .

log log x log x ,



1

p≤x p≡±1 (mod 3)

p1+



>

p≡±1 (mod 3)

1 p1+

−

 n>x

1 n1+

1 x − 1 1 1 log − + O(1) = log log x − log log log x + O(1). 2   2 2

8.85 Almost everything follows at once when we pick the correct χ . This time, we define χ : Z → {−1, 0, 1} by ⎧ ⎪ ⎪ ⎨ 1 χ (n) = −1 ⎪ ⎪ ⎩ 0

if n ≡ 1 (mod 4), if n ≡ −1 (mod 4), if 2 | n.

We let L(s) =

∞  χ (n) n=1

=1−

ns 1 1 1 + s − s + .... 3s 5 7

Then the assertion of Problem 8.83(a) holds as stated, with the same proof. Turning to (b), our previous proof now shows that 1 ≥ L(s) ≥ 23 for all s > 1; again, this implies log L(s) = O(1). Parts (c) and (d) hold, by essentially the same proofs as before, once “3” is changed to “4”. With these assertions in hand, the proofs in Problem 8.84 go through essentially unchanged. 8.86 The sum expression for S follows quickly from Problem 4.39. Interchanging the order of summation, we find that

Solutions to Step #8

129

S =

1  −2π ik/p e p k mod p



e2π ik(x1 +···+x )/p 2

2

x1 ,...,x mod p

      1 2 p + = e−2π ik/p e2π ikx /p p k mod p k≡0

x mod p

      k 1  −2π ik/p , p + G = e p p k mod p k≡0

where we used the result of Problem 5.43 for the final step. To continue, we take cases according to the parity of . Suppose first that  is odd. Then 

−2π ik/p

e

k mod p k≡0

    k k  G =G e−2π ik/p p p k mod p

=G



       −k  k 2π ik/p 2π ik/p  −1 e e =G . p p p

k mod p

k mod p

  The number of square roots of k modulo p is given by 1 + pk . Hence, viewing a 2 ≡ k,       k  k 2π ia 2 /p 2π ik/p G= e = = 1+ e e2π ik/p . p p a mod p

k mod p

k mod p

Putting this back into the last display, and recalling that G2 = that (for  odd) S = p

−1

+p

−1

 +1

G

−1 p

(8.25) −1 p p, we deduce



= p−1 + G−1  =p

−1

+

  −1 2 −1 p p

= p−1 + (−1) (We used here that

−1 p

= (−1)

p−1 2

p−1 −1 2 · 2

p

−1 2

.

.) If instead  is even, then

(8.26)

130

Solutions to Step #8



e−2π ik/p

k mod p k≡0

    /2   k −1 G = G p e−2π ik/p = −G = − , p p k mod p k≡0

and so  S = p−1 −

−1 p

/2

1

p 2 −1 .

Remarks. (a) The formula (8.26) for S ( odd) yields a slick proof of the famous Law of Quadratic Reciprocity. Assume  and p are odd primes,  = p. From (8.26) and Euler’s criterion for quadratic residues,   p−1 −1 p S ≡ 1 + (−1) 2 · 2 (mod ).  Here is another way of computing S (mod ). Observe that if (x1 , . . . , x ) is a tuple counted in the definition of S , so is its cyclic shift (x2 , . . . , x−1 , x1 ). Call two tuples equivalent if one can be obtained from the other by a finite number of cyclic shifts; it is easy to see that this is an equivalence relation, and that each equivalence class has size dividing , so is either 1 or . Therefore, S is congruent, mod , to the number of classes of size 1. That number is just #{x ∈ Zp : x 2 = 1} = #{square roots of 1/ mod p} = #{square roots of  mod p} = 1 +

   . p

Thus, modulo , 1+

    p−1 −1  p ≡ 1 + (−1) 2 · 2 . p 

Subtracting 1 and multiplying both sides by

p  , we obtain

   p−1 −1  p ≡ (−1) 2 · 2 .  p Since both sides here are ±1, this congruence mod  must be a genuine equality of integers, completing the proof of Quadratic Reciprocity. (b) Problem 8.86 demonstrates how the seemingly trivial act of rewriting a quantity as an exponential sum can yield nontrivial results. Here is another entertaining example. To set up, recall that every element of Zp is a sum of two squares. Here is an easy proof: We assume p is odd, since the result is obvious when p = 2. Now take any r ∈ Zp . The sets p A = {r − a 2 : a ∈ Zp } and B = {b2 : b ∈ Zp } both have size p+1 2 > 2 , so A and B cannot be disjoint. Hence, there are a, b ∈ Zp with r − a 2 = b2 , and r = a 2 + b2 . By contrast, it is not always the case that every element of Zp is a sum of two cubes. For instance, when p = 7, the cubes mod p are −1, 0, and 1, so that 3 and 4 are not representable. We now show that 7 is the exception rather than the rule: For all sufficiently large primes p,

Solutions to Step #8

131

every element of Zp is a sum of two cubes. Our argument works as soon as p ≥ 233; with more effort, one can show that p > 7 suffices. If p = 3, every element of Zp is a cube, the cube of itself. When p ≡ 2 (mod 3), it is still the case that every a ∈ Zp is a cube: (a (2p−1)/3 )3 = a 2p−1 = a p a p−1 = a p = a. So the problem is only interesting when p ≡ 1 (mod 3), which we now assume. Let a ∈ Zp with a = 0. (Clearly, 0 is a sum of two cubes.) We look at the solutions to x 3 + y 3 − az3 = 0, with x, y, z ∈ Zp . Note that if there is a solution with z = 0, then (x/z)3 + (y/z)3 = a. The number of solutions to x 3 +y 3 −az3 = 0 having z = 0 is easily seen to be 1+3(p−1): There is the solution x = y = 0, and for each nonzero x ∈ Zp , there are 3 values of y with x 3 + y 3 − a · 03 = 0 (namely, y = −ωx, where ω ∈ Zp satisfies ω3 = 1). Thus, we are guaranteed a is a sum of two cubes as long as the count N of solutions to x 3 + y 3 − az3 = 0 exceeds 1 + 3(p − 1). We write N as an exponential sum: N=

1 p





3 +y 3 −az3 )/p

e2π ik(x

.

x,y,z mod p k mod p

Now setting 

˜ k := G

e2π ika

3 /p

a mod p

(these are “cubic Gauss sums”), we see that N=

1  ˜2 ˜ 1  ˜2 ˜ Gk · G−ak = p 2 + Gk · G−ak . p p k mod p

˜ k| ≤ We show below that |G

√

k mod p k≡0

6p for all k ≡ 0 (mod p). It follows immediately that

! 1 (p − 1)( 6p)3 p √ 2 > p − ( 6)3 p 3/2 > p 2 − 15p 3/2 .

N ≥ p2 −

The last expression exceeds 1 + 3(p − 1) for all p ≥ 233. So for p ≥ 233, every a ∈ Zp is a sum of two cubes. ˜ k |, we observe that To prove the claimed inequality for |G 

⎛



˜ k| = ⎝ |G 2

k mod p k≡0

⎞ ˜ k| |G

2⎠

− p2 ,

k mod p

and that 

˜ k |2 = |G

k mod p



˜k ·G ˜k G

k mod p

=





k mod p x,y mod p

e2π ik(x

3 −y 3 )/p

132

Solutions to Step #8 =





e2π ik(x

3 −y 3 )/p

x,y mod p k mod p

= p · #{(x, y) ∈ (Zp )2 : x 3 = y 3 } = p(1 + 3(p − 1)). Thus, 

˜ k |2 = p(1 + 3(p − 1)) − p 2 = 2p(p − 1). |G

k mod p k≡0

˜ k depends only on the coset of k with respect to the subgroup of cubes in Now the value of G Up . So using a, b, c for representatives of the three cosets, ˜ a |2 + |G ˜ b |2 + |G ˜ c |2 ) · p − 1 = 2p(p − 1), (|G 3 and so ˜ a |2 + |G ˜ b |2 + |G ˜ c |2 = 6p. |G √ ˜ b |, |G ˜ c | ≤ 6p, finishing the proof of the claim. ˜ a |, |G It is now immediate that |G It is not difficult to modify the above arguments above to prove that for each fixed d, and all primes p > p0 (d), every element of Zp is a sum of two dth powers. For an extensive discussion of how Gauss sums (and closely allied concepts) can be used to count solutions to equations over finite fields, see Chapter II of [2] or Chapter 8 of [1].

References 1. Kenneth Ireland and Michael A. Rosen, A classical introduction to modern number theory. Second edition. Graduate Texts in Mathematics, vol. 84. Springer-Verlag, New York, 1990. 2. Wolfgang M. Schmidt, Equations over finite fields. An elementary approach. Lecture Notes in Mathematics, vol. 536. Springer-Verlag, Berlin-New York, 1976.

Solutions to Step #9

9.87 Let n ∈ Z+ , and factor n = p pep . A given d ∈ Z+ has d 2 dividing n if and only the

exponent of p in d is at most ep /2, for each p, i.e., if and only if d divides n := p|n p ep /2 . Therefore, 

μ(d) =



 μ(d) =

d|n

d 2 |n

1

if n = 1,

0

otherwise.

 Now n = 1 if and only if n is squarefree. Hence, d 2 |n μ(d) is the indicator function of the squarefree numbers—but that description also fits |μ(n)|. 9.88 From the expression for |μ(n)| in the last problem, 

|μ(n)| =

n≤x



μ(d) =

n≤x d 2 |n

 √ d≤ x

μ(d)



1=

n≤x d 2 |n

 √ d≤ x

μ(d)

x  . d2

Continuing,  √ d≤ x

μ(d)

x  x   = μ(d) + O(1) d2 d2 √ d≤ x

=x

∞  μ(d) d=1

d2

−x

 μ(d) √ + O( x). 2 d √

d> x

    √   By Exercise 2.11, x d>√x μ(d)  ≤ x d>√x d12 = O( x). And from the d2  μ(d) −1 solution to Exercise 3.25, ∞ d=1 d 2 = ζ (2) . Thus,

© Springer Nature Switzerland AG 2021 P. Pollack, A. Singha Roy, Steps into Analytic Number Theory, Problem Books in Mathematics, https://doi.org/10.1007/978-3-030-65077-3_24

133

134

Solutions to Step #9



|μ(n)| =

n≤x

√ 1 x + O( x). ζ (2)

The probability claim, interpreted as a statement about the limiting probability of a randomly chosen number from [1, N] being squarefree (as N → ∞), follows immediately. All of this is easily generalized to kth-power-free numbers, for any fixed k ≥ 2.  In this case, the characteristic function can be written as d k |n μ(d). Following the above arguments shows that the count of kth-power-free numbers in [1, x] is x + O(x 1/k ), ζ (k) so that the probability of being kth-power-free is 1/ζ (k). 9.89 Since ω(d  ) ≤ ω(d), the first inequality is clear. That  2ω(d)  2ω(d) ≤ eω(d)−k d d

d|P ω(d)>k

d|P

is also easy: Each d contributing to the left sum contributes at least as much to the sum on the right, since eω(d)−k > e0 for these d. And those d contributing to the right-hand sum but not the left make a nonnegative contribution, so are harmless for the desired inequality. Continuing on, the factorization  2ω(d) d|P

d

eω(d)−k = e−k



 2e 1+ p p≤y

follows immediately upon expanding out the right-hand side. Thus, it remains only to prove that 

 2e 1+ ≤ (log y)6 p p≤y for large enough x. Since y is a function of x that tends to infinity as x → ∞, it is enough to prove the inequality for large enough y. For this we use Problem 4.34 and the inequality 1 + t ≤ et : For large y,    

 2e 1 1+ ≤ exp 2e p p p≤y p≤y ≤ exp(2e(log log y + 2 log log log y)) = (log y)2e (log log y)4e < (log y)6 .

Solutions to Step #9

135

In the final line, it is important that 2e = 5.43 · · · < 6. 9.90 If d divides P , then d is a squarefree product of primes not exceeding y. Writing d = p1 · · · pr with r = ω(d), it is then immediate that d ≤ y r . Thus, d ≤ y k when ω(d) ≤ k, and E2 =







2ω(d ) ≤

d|P ω(d)≤k

2ω(d) ≤

d|P ω(d)≤k



2ω(d) ≤ y k max 2ω(d) . d≤y k

d≤y k

(At the last step, we bounded the sum by the number of summands multiplied by the largest term.) Since 2ω(d) ≤ τ (d) ≤ d (the final bound being trivial), maxd≤y k 2ω(d) ≤ y k , finishing the proof. 9.91 Piecing together what we know so far, π2 (x, y) ≤

  1 2 1− x + xE1 + E2 2 p 2 x/(log x)1.01 . But this contradicts the upper bound on π2 (x) shown in Problem 9.91. Remark. The bound on π2 (x) in Problem 9.91 was published by Brun in 1919. A year later, he succeeded in proving the somewhat sharper estimate π2 (x) = O(x/(log x)2 ). It is widely believed π2 (x) that limx→∞ x/(log exists and is positive (a special case of very-plausible seeming conjectures x)2 of Hardy–Littlewood), so that Brun’s sharper upper bound is probably best possible. However, it remains a notorious open problem (the “twin prime conjecture”) to even show that π2 (x) → ∞ as x → ∞.

9.93 Let a/b and a  /b be reduced fractions with 0 < a, a  , b, b
12 p for large p, whereas the number of nonzero squares in Zp is p−1 2 < 2 p. Remark. By keeping careful track of the error terms, one can show by this method that the least √ positive nonsquare modulo p is smaller than p for every prime p > 23.

9.95 We recognize that  n∈Z+ n=m

When N ≥ m,

1 n2 −m2

=

1 2m



1 1 = lim N →∞ 2m n2 − m2

1 n−m

−

  1≤n≤N n=m

1 n+m



and rewrite

  1 1 − . n−m n+m 1≤n≤N n=m

Solutions to Step #9

 1≤n≤N n=m

137

   1 1 1 1 − =− + + ··· + 1 n−m n+m m−1 m−2 1≤n≤N n=m

    1 1 1 1 1 1 − + + ··· + − . + 1 + + ··· + 2 N −m m+1 m+2 m+N 2m

Now suppose that N ≥ 2m + 1. Then in the second displayed line, the fractions 1 1 1 m+1 , m+2 , . . . , N −m appear with a + sign in the first group of terms but with a − sign in the second. Canceling, the above right-hand side becomes  −

   1 1 1 1 + + ··· + 1 + 1 + + ··· + m−1 m−2 2 m   1 1 1 1 . − + + ··· + − N −m+1 N −m+2 N + m 2m

In the first line of this new display, everything cancels except m1 . The subtraction of 1 1 1 1 1 in the second line gives us an additional 2m , while N −m+1 , N −m+2 , . . . , N +m − 2m make a contribution bounded in absolute value by 2m/(N − m + 1). Therefore, 1 2m

  1≤n≤N n=m

     1 1 1 m 1 1 − = + +O n−m n+m 2m m 2m N −m+1 1≤n≤N n=m

= Letting N → ∞, we conclude that

3 +O 4m2



1 n∈Z+ n2 −m2 n=m

=



 1 . N −m+1

3 . 4m2

9.96 The series converges if and only if λ ≥ 1. Take first the case when λ = 1. We reorder the terms according to the value of P (n) (reordering is harmless as we deal with a series of positive terms):  n>1

1 1 = nP (n) p p = =

 n: P (n)=p

 1 p2 p

1 n

 m=1 or P (m)≤p

1 m

  1  1 1 1 + + + . . . ,  2 p2 p ≤p

where in this final product  runs over the primes not exceeding p. For large enough p,

138

Solutions to Step #9

    

 1 1 2 1 1 + + 2 + ... = 1+ 1+ ≤   −1 

≤p

≤p

≤p

  2 ≤ exp(2(log log p + 2 log log log p)) < (log p)3 . ≤ exp  ≤p

3 3   The sum p (logp2p) certainly converges, since the larger sum n≥2 (logn2n) converges. This establishes convergence of the initial series for λ = 1, and so also for λ ≥ 1. If λ ≤ 0, then the terms nλ P1(n) do not tend to zero (consider n of the form 2k ), so the series diverges. Finally, we suppose that 0 < λ < 1. Proceeding as above,

 n>1

  1  1 1 1 1 + λ + 2λ + . . . . = nλ P (n)  p1+λ  p ≤p

1 The series expansion log 1−t = t + 12 t 2 + . . . tells us that

 

 1 1 1 1 + λ + 2λ + . . . = log log   1− ≤p

≤p

1 p 2 log p

We know that π(p) > all large primes p,

1 λ

≥

 1 1 ≥ λ π(p). λ  p

≤p

for all large p (see Problem 8.80). Consequently, for

1 1 π(p) > p1−λ / log p > 2 log p. pλ 2 For these p, the product of the series

 ≤p 1 +



 p

1 p1+λ

1 λ

+

1 2λ

 + . . . > p2 . It follows that the terms



 1 1 1 + λ + 2λ + . . .  

≤p

tend to infinity (not zero!), and so we again have the desired divergence.

Solutions to Step #10

10.97 (a) By Legendre’s formula, log x! = log



p

k≥1

x/p k 

p≤x

⎛ ⎞  x ⎝ ⎠ log p = k p p≤x 

k≥1

  x x . log p + log p = p pk p≤x p≤x k≥2

Since 0≤



log p

p≤x

  x  1 ≤x log p k p pk p≤x k≥2

k≥2

=x

 log n  log p ≤x , 2 p −p n2 − n p≤x n≥2

and



log n n≥2 n2 −n

< ∞, it follows that

    log p  x x − − + O(x). log p log x! = x p p p p≤x p≤x Finally, 0≤

 p≤x

 log p

   x x − log p ≤ log LCM(x) ≤ x log 4, ≤ p p p≤x

© Springer Nature Switzerland AG 2021 P. Pollack, A. Singha Roy, Steps into Analytic Number Theory, Problem Books in Mathematics, https://doi.org/10.1007/978-3-030-65077-3_25

139

140

Solutions to Step #10

 so that p≤x log p( px − px ) = O(x). (b) Recall from Problem 2.12 that log x! = x log x + O(x). (In fact, the estimate  there was somewhat sharper.) Comparing with (a), x p≤x logp p = x log x + O(x). Divide by x. 10.98 We apply the summation by parts formula established in Problem 6.60. To set things up, we let an = logn n when n is prime, and an = 0 otherwise. We let f (t) = log1 t for t ≥ 2. We extend f to all of [1, ∞) in such a way that f  (t) is continuous for all t ≥ 1. (The choice of extension is unimportant, as will become  clear shortly.) Finally, we let S(x) := p≤x logp p . Since S(t) vanishes when t < 2, the summation by parts formula yields  x 1  = an f (n) = S(x)f (x) − S(t)f  (t) dt p 1 p≤x n≤x  x  x S(t) S(x)  + S(t)f (t) dt = dt. = S(x)f (x) − 2 log x 2 2 t (log t) Write S(t) = log t + E(t), so that E(t) = O(1) for all t ≥ 2. Then, for x ≥ 2, E(x) S(x) =1+ =1+O log x log x



1 log x

 .

Also, 

x

2

 1 E(t) dt + t log t t (log t)2 2  x E(t) = log log x − log log 2 + dt. t (log t)2 2

S(t) dt = t (log t)2



x



Since E(t) = O(1), the improper integral  2

x

E(t) dt = t (log t)2

 

∞ 2 ∞

= 

2 ∞

= 2

Piecing the puzzle back together,

∞ 2

E(t) t (log t)2

E(t) dt − t (log t)2



dt converges absolutely, and ∞

E(t) dt t (log t)2 x   ∞ E(t) dt dt + O t (log t)2 t (log t)2 x   E(t) 1 . dt + O log x t (log t)2

Solutions to Step #10

141

   1 1 = 1+O + p log x p≤x   log log x − log log 2 + 2

∞

E(t) dt + O t (log t)2



1 log x



= log log x + C + O



1 log x

 .

Remark. The constant C appearing here is usually referred to as the “Meissel–Mertens constant”. It can be shown that     1 1 C=γ + log 1 − + p p p where γ is the Euler–Mascheroni constant from Exercise 1.7. A consequence of this representation of C (or really, an equivalent form of that same result) is “Mertens’ product theorem”, which states that 

 1 lim eγ log x = 1. 1− x→∞ p p≤x Numerically, C = 0.26149721 . . . .

10.99

√ (a) A number n ≤ x can have at most one prime factor > x, for if it had two, say p and q, then √ we would have n ≥ pq > x. So to count the √ n ≤ x with a prime factor > x, it suffices to add up, for each prime p ∈ ( x, x], the number of multiples of p ≤ x. For large x, this is  √

x 1 and 1 ≤ u ≤ 2. This is √good as far as it goes, but we would also like to also understand (x, y) when y is smaller than x, perhaps significantly smaller. To start the discussion, suppose that u ∈ [2, 3], and notice that (x, x 1/u ) = (x, x 1/2 ) −



(x/p, p).

x 1/u log n.

 It is easily proved by induction that i (1 − ai ) ≥ 1 − i ai , for any finite sequence of ai in [0, 1]. Thus, 

  1 1 1 1− ≥1− ≥1− · ω>log n (n), p p log n

p|n p>log n

p|n p>log n

where ω>log n (n) denotes the number of prime factors of n exceeding log n. Noting

that n ≥ p|n, p>log n p ≥ (log n)ω>log n (n) , we deduce that ω>log n (n) ≤ logloglogn n .

Solutions to Step #10

145

Putting this back above, we see that for large n, 

 1 1 1 1− ≥1− > , p log log n 2

p|n p>log n

and 

 1 −1 ≤ 2. 1− p

(10.31)

p|n p>log n

Turning now to the primes p ≤ log n, clearly  



 1 −1 1 −1 1− 1− ≤ p p

p|n p≤log n

p≤log n

 1+ = p≤log n

1 p−1

 ≤ exp

  p≤log n

 1 . p−1

(10.32)

Moreover,  p≤log n

 1  1 1 ≤ + ≤ log log log n + O(1), p−1 p p(p − 1) p p≤log n

so that exp

  p≤log n

1 p−1

 = O(log log n).

(10.33)

The desired upper bound follows from combining (10.30)–(10.33). 10.103 (a) Since o(2 mod ) is a proper divisor of n, there is at least one prime p dividing n to a higher power than that to which it divides o(2 mod ). Any such prime p has the desired property. (b) Let p be a prime satisfying (a), meaning that p | n and o(2 mod ) divides n/p. n Then  divides 2n/p − 1, while also  | 22n/p−1 from the definition of Bn . So −1 working modulo , 0≡

2n − 1 2n/p − 1

146

Solutions to Step #10

= (2n/p )p−1 + (2n/p )p−2 + · · · + 2n/p + 1 ≡ 1 + · · · + 1 ≡ p. % &' ( p times

This forces  = p. So the prime p in (a) is unique, and p = . Note that as a byproduct we get that  | n. (c) Let q be a prime dividing n with q = . Then q does not satisfy the conclusion of (a). Hence, o(2 mod ) does not divide n/q. But o(2 mod ) does divide n; the only way this is possible is if q | o(2 mod ). (d) From (c), if q is a prime dividing n with q = , then q | o(2 mod ). But o(2 mod ) |  − 1, so that q < . Since  is a prime dividing n and all other prime divisors of n are less than , we conclude that  is the largest prime factor of n. n (e) It is enough to show that 2  22n/−1 . Write 2n/ − 1 = r q, where r ≥ 1 and −1   q. Note, as will be useful momentarily, that  is odd and so  ≥ 3. Then 2 = (1 +  q) = 1 +  n

r



r+1

q+

  j ≥2

j

j r q j .

All of the terms in the sum on j are divisible by 2r+1 : When 2 ≤ j < , one sees this by noting that 2r divides j r , while  | j . When j = , the corresponding term is divisible by 3r (since  ≥ 3), and 3r ≥ 2r + 1. Thus, ≡0 (mod )

 2n − 1 = r+1

q %&'(

≡0 (mod )

' (% &   j r−r−1 q j , + j j ≥2

and the parenthesized factor is not a multiple of . Hence, r+1 is the largest n power of  dividing 2n − 1, and 2  22n/−1 . −1 10.104 We show that if n is repellent and Bn > 1, then Bn is the largest prime factor of n. Let  be any prime dividing Bn . By assumption, o(2 mod ) < n and so, from the last problem,  is the largest prime factor of n. Since this holds for every prime dividing Bn , it must be that Bn is a power of . But we know that 2  Bn , and so Bn = . In the table of values of Bn for 2 ≤ n ≤ 18, we see that the only time Bn is 1 or the largest prime factor of n is n = 6. And 6 is indeed repellent: If  is a prime with o(2 mod ) = 6, then  | B6 = 3, so we need  = 3—but o(2 mod 3) = 2. 10.105 Recall that in Problem 7.74, you showed that

Solutions to Step #10 k−1 

147

ζ (2i)ζ (2k − 2i)

i=1

= (k − 1)ζ (2k) +

  m,n≥1 m=n

1

1 1 1 + 2k−2 2 m2k−2 n2 − m2 n m − n2

 .

The double sum is symmetric in m and n, which suggests that   m,n≥1 m=n

1

1 1 1 + 2k−2 2 2k−2 2 2 m n −m n m − n2

 =2

∞   m=1 n∈Z+ n=m

1 m2k−2

n2

1 . − m2

(10.34) We will give a rigorous justification of (10.34) at the end of this solution. Assuming this for now, we observe that by Problem 9.95, 2

∞   m=1 n∈Z+ n=m

∞

 1 1 3 3 = 2 · = ζ (2k). 2 m2k−2 n2 − m2 m2k−2 4m2 1

m=1

Substituting this above, k−1  i=1

  1 ζ (2k), ζ (2i)ζ (2k − 2i) = k + 2

which is the desired identity. Plugging k = 2 into our identity shows that ζ (2)2 = π2

π4

5 2 ζ (4).

Recalling that

ζ (2) = 6 , we infer that ζ (4) = 90 . The proof that ζ (2k) ∈ π 2k Q is a straightforward argument by complete induction. The claim holds when k = 1. If it holds for all positive integers < k, then ζ (2k − 2i)ζ (2i) ∈ (π 2k−2i Q)(π 2i Q) ⊆ π 2k Q for all i = 1, 2, . . . , k − 1. Hence, (k + 12 )ζ (2k) ∈ π 2k Q, and so ζ (2k) ∈ π 2k Q. What about (10.34)? A naive argument for (10.34) would proceed in two steps. First, splitting the sum into obvious “halves” and exploiting the symmetry,   m,n≥1 m=n

1

1 1 1 + 2k−2 2 m2k−2 n2 − m2 n m − n2

 =2

 m,n≥1 m=n

1 1 . m2k−2 n2 − m2

(10.35) Now rewriting the right-hand double sum as an iterated sum, with the sum on m outside, gives (10.34). 1 1 The problem with this “proof” is that the terms m2k−2 are not all positive. n2 −m2 This renders both of the above steps somewhat suspect. All such concerns vanish if

148

Solutions to Step #10

we show that the right-hand sum in (10.35) converges absolutely; that is,  m,n≥1 m=n

1 m2k−2

|n2

1 < ∞. − m2 |

(10.36)

 1 Thankfully, (10.36) is not so difficult. Since m≥1 m2k−2 < ∞, it suffices to prove  1 that for every positive integer m, the sum n∈Z+ , n=m |n2 −m 2 | = O(1), with an implied constant independent of m. This sum on n has m − 1 terms corresponding 1 1 1 to n < m, and for each of these, |n2 −m 2 | ≤ m2 −(m−1)2 = 2m−1 , so these terms

m−1 contribute at most 2m−1 < 1. Now suppose n > m, say n = m + r. Then n2 − m2 =  2mr + r 2 ≥ r 2 , and so the terms with n > m contribute at most r r12 < 2. Hence,  1 n∈Z+ , n=m |n2 −m2 | < 3, for every m.

Remarks. (a) It was known already to Euler that the values ζ (2k) appear in the coefficients of the power series expansion of cotangent. Specifically, for x near 0, ∞

 1 1 ζ (2k)x 2k . − π x cot(π x) = 2 2

(10.37)

k=1

Euler’s proof of (10.37) starts by establishing the beautiful (and independently interesting) identity sin(π x) = π x



 x2 1− 2 , n

(10.38)

n≥1

for all real numbers x. This approach to (10.37) can be found in many textbooks. Usually the product formula (10.38) is proved using ideas from complex analysis, but this can be avoided, as shown (for example) by Koblitz in [5]. Alternatively, the identity (10.37) can be deduced from Problem 10.105 coupled with our previous determination of ζ (2). For x sufficiently close to 0, 1 1 1 1 − π x cot(π x) = − π x 2 2 2 2 =





c2k x 2k ,

π 2x2 π 4x4 π 6x6 2! + 4! − 6! + . . . 3 3 5 5 7 7 − π 3!x + π 5!x − π 7!x + . . .

1− πx

for certain real numbers c2k .



(10.39)

k≥1 2

By a direct calculation, c2 = π6 , which we recognize as ζ (2). Now differentiate both sides of (10.39) and multiply by x. Since cot = −1 − cot2 , one can use (10.39) to determine a power series expansion of the resulting left-hand side. The resulting right-hand side is, of course,  simply k≥1 2k · c2k x 2k . Comparing coefficients of x 2k , one finds after a little computation  that (k + 12 )c2k = k−1 i=1 c2i c2k−2i for k ≥ 2. Since c2k and ζ (2k) satisfy the same recurrence, and agree for k = 1, we conclude that c2k = ζ (2k) for all k. (b) A celebrated 1882 theorem of Lindemann asserts that π is “transcendental”, that is, not a root of a nonzero polynomial with rational coefficients. (We recommend Hadlock’s book [3] for a

Solutions to Step #10

149

readable account of Lindemann’s theorem.) This result together with Problem 10.105 implies that ζ (2k) is also transcendental, for every positive integer k. Much less is known about the values ζ (2k + 1). It was only in 1978 that Apéry succeeded in proving the irrationality of ζ (3) (irrationality being a much weaker property than transcendence), and it is still an open problem to decide the irrationality of ζ (5). There are some glimmers of hope, though: Ball and Rivoal have shown that ζ (2k + 1) is irrational for infinitely many k, and Zudilin has shown that at least one of ζ (5), ζ (7), ζ (9), ζ (11) is irrational. (c) Problems 7.74, 9.95, and 10.105 were based on Murty’s article [7].

10.106 (a) Suppose n is odd and perfect, and factor n into primes, say n = 2n = σ (n) =



p|n p

ep .

Then

σ (pep ).

p|n

Since 2n is twice an odd number, σ (pep ) is odd for every prime power pep appearing in the prime factorization of n, with precisely one exception. But for each odd prime p and positive integer e, σ (pe ) = 1 + p + · · · + pe ≡ e + 1 (mod 2). It follows that ep is even for every prime p dividing n, with exactly one exception.

1 The claim in (a) follows with m = p|n, 2|ep p 2 ep . (b) Suppose that n √ is an odd perfect number with n ≤ x, and write n = pk m2 as in (a). Then m ≤ x, and so it suffices to prove that m determines n. To this end, observe that 2pk m2 = 2n = σ (n) = σ (pk )σ (m2 ), so that σ (pk ) 2m2 = . pk σ (m2 ) The right-hand fraction is reduced, as its numerator 1 + p + · · · + pk is coprime to p. Since each rational number has a unique lowest-terms representation, pk , and hence also n = pk m2 , is determined by m: Namely, pk is the denominator when 2m2 /σ (m2 ) is reduced. Remark. In 1957, 2 years after Hornfeck proved the x 1/2 upper bound of Problem 10.106(b), Hornfeck and Wirsing [4] proved that the count of odd perfect numbers up to x is at most x  , for any fixed  > 0 and all large x. It is widely believed that there are no odd perfect numbers, so that the “correct” upper bound in this problem is not x 1/2 or x  , but 0 ! But that goal appears rather distant. It would seem to require new ideas to prove, for example, that the count of odd perfect 10 numbers up to x is O((log x)10 ), as x → ∞.

150

Solutions to Step #10

References 1. George E. Andrews and Bruce C. Berndt, Ramanujan’s lost notebook. Part IV. Springer, New York, 2013. 2. Andrew Granville, Smooth numbers: computational number theory and beyond. Algorithmic number theory: lattices, number fields, curves and cryptography, 267–323, Math. Sci. Res. Inst. Publ., vol. 44, Cambridge Univ. Press, Cambridge, 2008. 3. Charles Robert Hadlock, Field theory and its classical problems. Carus Mathematical Monographs, vol. 19. Mathematical Association of America, Washington, D.C., 1978. 4. Bernhard Hornfeck and Eduard Wirsing, Über die Häufigkeit vollkommener Zahlen. Math. Ann. 133 (1957), 431–438. 5. Neal Koblitz, p-adic numbers, p-adic analysis, and zeta-functions. Second edition. Graduate Texts in Mathematics, vol. 58. Springer-Verlag, New York, 1984. 6. Pieter Moree, Integers without large prime factors: from Ramanujan to de Bruijn. Integers 14A (2014), Paper No. A5, 13 pp. 2 7. M. Ram Murty, A simple proof that ζ (2) = π6 . Math. Student 88 (2019), 113–115.

Solutions to Step #11

1

1

1

11.107 Note that p 3 > (p 2 + ) 2 . Now the same reasoning as in the solution to 1 Problem 10.99 shows that the limiting proportion of n ≤ p 2 + possessing a prime 1 factor > p 3 is 1

1

1

log log p 2 + − log log p 3 = log < log

log p 2 + log p

1 3

= log

1/2 +  1/3

1/2 + 1/50 = 0.444 · · · < 0.45. 1/3

11.108 Let  ∈ (0, 1/50). By Exercise 6.56, once p is large enough at least 49% of 1 the integers in [1, p 2 + ] are nonsquares modulo p. Fewer than 45% of the integers 1 from the same interval have a prime factor > p1/3 . So there is integer n ∈ [1, p 2 + ] that is both a nonsquare modulo p and factors completely into primes not exceeding 1 p 3 . Not every prime factor of n can be a square, otherwise n itself would be a square. 1 So there is a prime < p 3 reducing to a nonsquare modulo p. 1

11.109 Fix  ∈ (0, 1/2). We look at the integers in [1, p 2 + ] which have a prime 1 √

factor > p 2

e

+

. The limiting proportion of these integers (as p → ∞) is η := log

1 2 + 1 √ + 2 e


0. Let N be the number of n ≤ x with |ω(n) −  log log x| >  log log x, so that N/ x represents the relative frequency of such n ≤ x. For every n ∈ Z+ , we have (ω(n) − log log x)2 ≥ 0. Moreover, if |ω(n) − log log x| >  log log x, then (ω(n) − log log x)2 >  2 (log log x)2 . Therefore,  2 (log log x)2 N ≤

 (ω(n) − log log x)2 = O(x log log x). n≤x

Hence, N = O( −2 x/ log log x), and the ratio N/ x → 0 as x → ∞. Remark. Those who have seen some probability theory will recognize this proof as an application of “Chebyshev’s inequality”.

11.112 Fix  > 0. Let N, N  , and N  denote (respectively) the counts of n ≤ x with |(n) − log log x| >  log log x, 1  log log x, and 2 1 |(n) − ω(n)| >  log log x. 2

|ω(n) − log log x| >

Solutions to Step #11

153

Then N ≤ N  + N  . From Problem 11.111, we have that N  / x → 0 (as x → ∞). Moreover, from Problem 10.101,  1 N  ·  log log x ≤ |(n) − ω(n)| 2 n≤x  = ((n) − ω(n)) = O(x). n≤x

Thus, N  = O( −1 x/ log log x) and N  / x → 0. We conclude that N/ x → 0 as well. 11.113 Fix  > 0. By Problems 11.111 and 11.112, we have that (1 − ) log log x ≤ ω(n), (n) ≤ (1 + ) log log x for all n ≤ x, apart from an exceptional set of n with limiting frequency 0. It now suffices to observe that for all n ≤ x not in this exceptional set, τ (n) ≥ 2ω(n) ≥ 2(1−) log log x = (log x)(log 2)(1−) > (log x)log 2− , and τ (n) ≤ 2(n) ≤ 2(1+) log log x = (log x)(log 2)(1+) < (log x)log 2+ . Here we used the bounds for τ (n) proved in Problem 5.49. Since log 2 < 1, a typical n ≤ x has many fewer divisors than the average of ≈ log x. Remark. How can it be that the average order of τ (n) is so much larger than its typical order? The explanation is that the average is dominated by a relatively sparse set of n on which τ (n) is much larger than usual. While a typical n ≤ x has ω(n) ≈ log log x, it can be shown that the average of τ (n) on n ≤ x is essentially determined by those n with ω(n) ≈ 2 log log x: For each  > 0, there is a δ > 0 such that, for all x > x0 (), 

τ (n) < x(log x)1−δ .

n≤x |ω(n)−2 log log x|> log log x

See [1] for a general theorem containing this and many similar results.

154

Solutions to Step #11

11.114 (a) Write A =

p≤log n/(log log n)3

pep . Since A divides n, it is certainly the case

that each pep ≤ n, and so each ep ≤

log n log 2 .

τ (A) =

Thus, for large enough n,

(ep + 1)

p≤log n/(log log n)3

log n/(log log n)3 log n +1 ≤ log 2    log n log n + 1 = exp · log log 2 (log log n)3   log n < exp · 2 log log n (log log n)3   log n . = exp 2 · (log log n)2 

(b) Since every prime dividing B exceeds  n≥

log n , (log log n)3

log n (log log n)3

(B) ,

so that for large n, (B) ≤

log n n log (loglog log n)3

log n log log n − 3 log log log n ⎞ ⎛ 1 log n ⎝ ⎠ = log log n 1 − 3 log log log n

=

log log n

  log log log n log n 1+6 < log log n log log n =

log n · log log log n log n +6 . log log n (log log n)2

The bound on τ (B) now follows from Problem 5.49.

Solutions to Step #11

155

(c) From (a) and (b), we have for large n that τ (n) = τ (AB) = τ (A)τ (B) log n

≤ 2 log log n

n +O( log n·log log log 2 ) (log log n)

,

and the exponent here is eventually smaller than (1 + ) logloglogn n . (d) We start by noting that τ (LCM(x)) ≥ 2ω(LCM(x)) = 2π(x) . So it suffices to show that for each fixed  > 0, and all large x, π(x) > (1 − )

log LCM(x) . log log LCM(x)

We will prove this in the equivalent form (1 − ) log LCM(x) < π(x) · log log LCM(x). Taking 2n + 1 as the largest odd integer not exceeding x, we see from Problem 7.69 that for large x, LCM(x) ≥ LCM(2n + 1) ≥ 4n > 4x/3 . Thus, log log LCM(x) ≥ log x + log

log 4 > (1 − ) log x 3

and π(x) log log LCM(x) > (1 − )π(x) log x ≥ (1 − ) log LCM(x), using in the last step the result of Exercise 8.79. Remark. A sharper result on the maximal size of τ (n) was obtained by Ramanujan in [3]. Fix any nonnegative integer k. Then, for all n ≥ 3,   k log τ (n) j! log n log n  ; + O ≤ k log 2 log log n (log log n)j (log log n)k+2

(11.40)

j =0

moreover, equality holds here along a sequence of n tending to infinity. Note that taking k = 0 in (11.40) already gives a more precise upper bound for τ (n) than the one proved in part (c).

11.115 It suffices to show that n (2) | 22n/p−1 for each prime divisor p of n. −1 We know that

d (2) = 2n − 1, while d (2) = 2n/p − 1. n

d|n

d|n/p

156

Solutions to Step #11

Hence,

2n − 1 = d (2). n/p 2 −1 d|n dn/p

All of the terms in the right-hand product are integers, and one of them is n (2). The result follows. Remark. With more effort, one can show that Bn = n (2). Compare with the remark on p. 6 of [5].

11.116 We write n (2) =

(2 − 1) d

μ(n/d)

=

d|n

Notice that



d μ(n/d)

(2 )

d|n



(2d )μ(n/d) = 2

d|n dμ(n/d)



 1 μ(n/d) 1− d . 2 d|n



=2

d|n (n/d)μ(d)

= 2φ(n) ,

d|n

and so we have the equality claimed in the problem. To show the inequality, it suffices to prove that 

 1 μ(n/d) 1 1− d > . 2 3 d|n

We take cases. First, if μ(n) = 0 or −1, then  n   n  

  1 1 μ( d ) 1 μ( d ) 1 1 1− d 1− d 1− d ≥1− ≥ ≥ = , 2 2 2 2d 2 d|n

d|n d>1

d>1

d>1

so we may assume that μ(n) = 1. If n = 1, the result is easily verified, so we also assume that n > 1. Let p be the smallest prime dividing n. If d > 1 and d divides n, then d ≥ p. Moreover, when d = p, we have μ(n/d) = −1, so that the d = p term makes a contribution of size > 1 to our product. We conclude that

 d|n

1 1− d 2

μ( n ) d

    1 1 1− d > 1− 2 2 d>p

⎞      1 1 1 1 1 1⎝ 1 ⎠ 1− p ≥ 1− > . ≥ = 1− 2 2d 2 2 2 4 3 ⎛

d>p

Remark. In fact, n (2) ≥ 12 2φ(n) for all n (see [2] for much more general results).

References

157

11.117 If n > 1 and there is no prime  with o(2 mod ) = n, then Bn = 1 or p, where p is the largest prime factor of n (Problem 10.104). Thus, it will suffice to prove that Bn > p for every integer n > 6. Since p − 1 | φ(n), we get from Problems 11.115 and 11.116 that Bn ≥ n (2) >

1 φ(n) 1 p−1 2 ≥ 2 , 3 3

which is larger than p once p ≥ 5. So we can assume p = 2 or p = 3. If p = 2, then n = 2k with k ≥ 3 (we are assuming n ≥ 6), and k

Bn ≥ 2k (2) =

22 − 1 k−1 22

−1

= 22

k−1

+ 1 > 2.

So suppose now that p = 3. If n = 3k , then k ≥ 2, and k

23 − 1

Bn ≥ 3k (2) =

k−1 23

−1

= 22·3

k−1

+ 23

k−1

+ 1 > 3.

It remains to consider those cases when p = 3 and 6 | n. Write n = 6m, where no prime greater than 3 divides m. Since n > 6, we have m ≥ 2. Then Bn ≥ n (2) >

1 φ(n) 1 1 1 = 2n/3 = 22m ≥ 24 > 3. 2 3 3 3 3

Therefore, Bn > p for all n > 6. Remark. One can push these same methods a bit further to establish the following generalization of Bang’s theorem, published by Zsigmondy in 1892: Let a, b be positive coprime integers with a > b. For each n ≥ 1, there is a prime number p which divides a n − bn but which does not divide a k − bk for any positive integer k < n, except in the following cases: • n = 1 and a − b = 1, • n = 2 and a + b is a power of 2, • n = 6, a = 2 and b = 1. A clean exposition can be found in [4].

References 1. Paul Pollack, A generalization of the Hardy–Ramanujan inequality and applications. J. Number Theory 210 (2020), 171–182. 2. Carl Pomerance and Simon Rubinstein-Salzedo, Cyclotomic Coincidences. Experimental Mathematics, https://doi.org/10.1080/10586458.2019.1660741. 3. Srinivasa Ramanujan, Highly composite numbers, Proc. London Math. Soc. 14 (1915), 347–400. 4. Moshe Roitman, On Zsigmondy primes. Proc. Amer. Math. Soc. 125 (1997), 1913–1919. 5. Nairi Sedrakian and John Steinig, A particular case of Dirichlet’s theorem on arithmetic progressions. Enseign. Math. (2) 44 (1998), 3–7.

Solutions to Step #12

12.118 Observe that |(mn) − 2 log log N| = |((m) − log log N) + ((n) − log log N)| ≤ |(m) − log log N| + |(n) − log log N |. Hence, if |(mn) − 2 log log N | >  log log N, either m or n belongs to E := {k ≤ N : |(k) − log log N| >

1  log log N }. 2

Recall from Problem 11.112 that E contains asymptotically 0% of the integers in [1, N], as N → ∞. So given δ > 0, we may choose N0 (δ) such that #E < δN whenever N > N0 (δ). Then for N > N0 (δ), the number of pairs (m, n) ∈ [1, N ]2 with m ∈ E or n ∈ E is smaller than δN · N + N · δN = 2δ · N 2 , and so a random pair from [1, N]2 has this property with probability < 2δ. Since δ > 0 is arbitrary, the result follows. 12.119 Notice that log log (N 2 ) − log log N = log 2. Hence, if N is large and |(n) − log log N| >  log log N, then |(n) − log log (N 2 )| ≥ |(n) − log log N | − log 2 >  log log N − log 2 >

1  log log (N 2 ). 2

By Problem 11.112, the proportion of n ∈ [1, N 2 ] with this property tends to 0, as N → ∞. ) 12.120 We fix  > 0 and argue that M(N < 2 for all large enough N . N2 Suppose that a appears in the N × N multiplication table.

© Springer Nature Switzerland AG 2021 P. Pollack, A. Singha Roy, Steps into Analytic Number Theory, Problem Books in Mathematics, https://doi.org/10.1007/978-3-030-65077-3_27

159

160

Solutions to Step #12

If |(a) − log log N | > 12 log log N, then a belongs to a certain set of size < N 2 by Problem 12.119 (once N is large enough). Suppose instead that |(a) − log log N| ≤ 12 log log N. By assumption, we can write a = mn with (m, n) ∈ [1, N]2 . Then |(mn) − 2 log log N| = |(a) − 2 log log N| ≥

1 log log N. 2

By Problem 12.118, there are fewer than N 2 possibilities for the pair (m, n) (once N is large enough), and so also fewer than N 2 possibilities for the product mn. 12.121 (a) By Problem 8.82, (2n)! =

LCM(2n/m)

m≤2n

and n! =

LCM(n/m) =

m≤n

LCM(2n/m).

m≤2n m even

Hence,   2n (2n)! = =

n (n!)2



m≤2n LCM(2n/m)



m≤2n LCM(2n/m) m even



m≤n LCM(n/m)



m≤2n LCM(2n/m)

= m odd

m≤n LCM(n/m)

;

  this is the claimed representation of 2n from n . The inequality follows 2n−1 = the unimodality of the binomial coefficients: Indeed, since n−1 2n−1 max0≤r≤2n−1 r , 

    2n−1  2 · 22n−1 2n 2n − 1 2  2n − 1 4n = =2 ≥ = . r n n−1 2n 2n 2n r=0

(b) For all pairs of positive integers m, n, we have that 2n n ) ≤ LCM( m ). Thus, from part (a), LCM( 2m+1

2n 2m+1

≤

2n 2m

=

n m,

and so

Solutions to Step #12

161

  2n

LCM( 2m+1 ) 2n LCM(2n) 4n ≤ = LCM(2n/3) · n 2n n LCM(n) LCM( m ) m≥2

LCM(2n) LCM(2n/3). ≤ LCM(n) (c) By Problem 7.68, LCM(2n/3) = LCM( 2n/3) ≤ 4 2n/3 ≤ 42n/3 . That LCM(2n) 4n/3 LCM(n) > 2n is now immediate from (b). (d) We argue as in Problem 7.65. Let p be a prime dividing LCM(2n), and let pk be the highest power of p dividing LCM(2n). Then pk ≤ 2n (indeed, pk is the highest power of p not exceeding 2n). Thus, pk−1 = pk /p ≤ pk /2 ≤ 2n/2 = LCM(2n) n. Hence, pk−1 | LCM(n), and so p2  LCM(2n) is squarefree. LCM(n) . Thus, LCM(n) √ To finish off, it is enough to show that no prime p ∈ ( 2n, n] divides LCM(2n) 1 LCM(n) . For those primes p, the power p is the highest power of p not exceeding n and also the highest power of p not exceeding 2n. Hence, p appears in the factorizations of both LCM(2n) and LCM(n) to the first power, and so p  LCM(2n) LCM(n) . (e) Suppose by way of contradiction that there is no prime in the interval (n, 2n] for some n ∈ Z+ . Then by parts (c) and (d), √ √

√ 4n/3 LCM(2n) ≤ ≤ p ≤ LCM( 2n) ≤ 4 2n ≤ 4 2n . 2n LCM(n) √ p≤ 2n

n/3

√

However, the inequality 42n ≤ 4 2n fails once n ≥ 34 (see below). Thus, as long as n ≥ 34, there is a prime in (n, 2n]. Finally, for each n < 34, one of the primes 2, 3, 5, 7, 13, 23, 43 belongs to (n, 2n]. x/3 How can we see that the inequality fails for n ≥ 34? Let g(x) = 4 √2x . It 2x·4 is enough to prove that g(x) > 1 for all real x ≥ 34. For x ≥ 10, (log g(x)) =

1 log 4 log 4 1 log 4 log 4 − −√ ≥ − −√ > 0, 3 x 3 10 2x 2 · 10

and so g is increasing on [10, ∞). It remains only to note that g(34) = 1.06199 . . . > 1. 12.122 When n = 6, there is a prime  for which o(2 mod ) = n (Exercise 11.117). Then n |  − 1, so that  ≡ 1 (mod n), while  | 2n − 1, so that  < 2n . When n = 6, it is enough to observe that  = 7 is both 1 mod 6 and smaller than 26 . Remark. The following deep result on the least prime in a general coprime progression is due to Linnik: For certain positive constants c1 and c2 , and every pair of coprime integers a and m, with m > 0, there is a prime p ≡ a (mod m) with p ≤ c1 mc2 . That is, the least prime in a coprime progression is polynomially bounded by the size of the modulus. Xylouris has shown that c2 can be taken as 5, for a suitable choice of c1 .

162

Solutions to Step #12

12.123 We deduce this from a lemma of independent interest: For each fixed  > 0, and all large enough x, the interval (x, (1 + )x] contains a prime. (For  = 1, it follows from Exercise 12.121 that all x ≥ 1 are “large enough”, but in this problem we will need the lemma for arbitrarily small values of .) To prove the lemma, it suffices to show that π((1 + )x) = 1 + . x→∞ π(x) lim

Rewrite π((1 + )x) (1 + )x/ log((1 + )x) x/ log x π((1 + )x) = · · . π(x) (1 + )x/ log((1 + )x) x/ log x π(x) By the Prime Number Theorem, the first and third right-hand factors tend to 1, as x → ∞. The middle factor is (1 + )

log x 1 = (1 + ) , log log((1 + )x) 1 + (1+) log x

which tends to 1 + . n n−i d be the (fixed) integer whose digits, from left-toNow let A := i i=1 10 right, are d1 , . . . , dn . For each positive integer k, the number of primes whose decimal expansion begins with A and which have exactly n + k digits is π((A + 1) · 10k ) − π(A · 10k ). This is positive for all large enough k, by the lemma with  = A1 . 12.124 (a) We apply Abel summation. Let an = 1 when n is prime and an = 0 otherwise, and take f (x) = x. Then with S(x) having its usual meaning, S(x) = π(x) and f  (x) = 1. Hence,  p≤x

p=





x

an f (n) = S(x)f (x) −

S(t)f  (t) dt

1

n≤x

 = x · π(x) −

x

π(t) dt. 2

(We start the final integral at x = 2, since π(x) vanishes when x < 2.) By the π(x) Prime Number Theorem, limx→∞ xx·π(x) 2 / log x = limx→∞ x/ log x = 1. Thus, what needs to be shown is that x π(t) dt 1 = . lim 22 x→∞ x / log x 2

Solutions to Step #12

163

Define E(t), for real numbers t ≥ 2, by the equation π(t) = L’Hôpital’s rule,

t log t

+ E(t). By

x lim

x→∞

t/ log t dt x/ log x 1 = lim = . 2 1 x→∞ 2 x / log x (2x/ log x)(1 − 2 / log x)

2

x

Thus, all we have left to do is to show that limx→∞ the stronger assertion that

2 E(t) dt x/(log x)2

(12.41)

= 0. We will prove

x

|E(t)| dt = 0. x 2 / log x

2

lim

x→∞

(12.42)

The Prime Number Theorem says that t/E(t) log t → 0, as t → ∞. So given t  > 0, we have |E(t)| <  log t for all sufficiently large t. It follows that is a positive constant C (depending on ) with |E(t)| ≤ C +  logt t for all t ≥ 2. Keeping (12.41) in mind, we deduce that for all large enough x, x

x

2

2

|E(t)| dt ≤ x 2 / log x

x (C + t/ log t) dt Cx 2 t/ log t dt ≤ +  ≤ . x 2 / log x x 2 / log x x 2 / log x

Since this holds for each  > 0, (12.42) follows. 2 A(n) = A0 (n) + A1 (n), where A0 (n) = (b) Let  y = x/(log x) and decompose  p|n, p≤y p, and A1 (n) = p|n, p>y p. Then 

A0 (n) =

n≤x

 n≤x p|n p≤y

p=



p



1≤

n≤x p|n

p≤y



p·

p≤y

x ≤ xy = x 2 /(log x)2 , p

so that  lim

x→∞

n≤x A0 (n) x 2 / log x

= 0. 

A (n)

1 So the problem reduces to computing limx→∞ xn≤x 2 / log x . We rewrite      A1 (n) = p= p=

n≤x

Clearly,

n≤x p|n p>y

m,p mp≤x p>y

m≤x/y y 1. Then for each positive integer n, |ns | = |nσ | · |niτ | = nσ · | cos(τ log n) + i sin(τ log n)| = nσ . 1 Hence, | χn(n) s | ≤ nσ for each n (with equality if n and m are coprime). The  ∞  χ (n)  1 convergence of n=1  ns  now follows by comparison with ∞ n=1 nσ . To get the product decomposition, we argue as in the solution to Problem 3.22 (or apply directly the remark at the end of the solution to Problem 3.25).

13.130 If we replace each term of the double sum with its absolute value, we arrive at   |χ (pk )| kpks

p k≥1

.

This is dominated, term by term, by the double sum  1 , kpks p k≥1

which was shown to converge in Problem 3.23 (in fact, to converge to log ζ (s)). So we have absolute convergence. 1 = x + 12 x 2 + 13 x 3 + . . . for all real x with |x| < 1. When Recall that log 1−x s > 1, the absolute value | χp(p) s | ≤

1 ps

  χ (pk ) p k≥1

Exponentiating yields

kpks

< 1 for all primes p. Hence, =

 p

 log

1 1−

χ (p) ps

 .

Solutions to Special Step A

169

⎛ ⎞   χ (pk )

1 ⎠= exp ⎝ = L(s, χ ). ks χ (p) kp s p p 1− k≥1

(A.45)

p

Since L(s, χ ) can be written as the exponential of a real number, L(s, χ ) > 0. The expression claimed for log L(s, χ ) is now immediate from (A.45). 13.131 (a) This is a direct application of Abel summation, with each an = χ (n) and f (t) = t −s .  (b) Examining the table, we find that g∈U8 χ (g) vanishes for every nontrivial character χ of U8 . As a consequence, for every nontrivial Dirichlet character χ mod 8, the sum of χ (n) over any 8 consecutive integers vanishes. Therefore, it is enough to verify that |Aχ (t)| ≤ 2 for t ≤ 7. This is easy to check directly from the values in the table. (c) The bound on |Aχ (t)| proved in (b) implies that the improper integral ∞ Aχ (t) dt converges absolutely, and that (for any s > 0) Aχ (x)x −s → 0 1 t s+1 as x → ∞. Now sending x to infinity in the result of (a), we get that the series ∞ Aχ (t) dt. defining L(s, χ ) converges to s 1 t s+1 (d) We may deduce from (a)–(c) that for all x ≥ 1 and all s ≥ ,     χ (n)      = s L(s, χ ) −  ns  

∞

x

n≤x



∞

≤ 2s x

  Aχ (t) −s  dt − x A (x) χ  t s+1 dt t s+1

+ 2x −s = 4x −s ≤ 4x − .

Given any δ > 0, we have 4x − < δ whenever x > (4/δ)1/ . That bound on x is independent of s, proving uniform convergence for s ≥ . Since each summand χ (n)n−s is a continuous function of s, continuity of the sum function L(s, χ ) on s ≥  now follows from standard theorems. As  can be any positive number, L(s, χ ) is continuous for all s > 0. (e) By part (d), L(s, χ ) is continuous at s = 1. Since L(s, χ ) > 0 for s > 0, continuity guarantees that L(1, χ ) ≥ 0. We are supposing that L(1, χ ) = 0, and so L(1, χ ) > 0. Now consider the composite function log L(s, χ ). Since L(s, χ ) is continuous and positive-valued for s ≥ 1, this composite function is continuous for s ≥ 1. Being continuous, log L(s, χ ) is certainly bounded on the closed interval [1, 2]. That is, log L(s, χ ) = O(1) for 1 ≤ s ≤ 2. 13.132 (a) For s > 1, we have from Problem 13.130 and our solution to Problem 3.23 that

170

Solutions to Special Step A

       χ (p) χ (p)   1    ≤ − log L(s, χ ) =  < 1.  ps kpks  kpks p p,k k≥2

p,k k≥2

(b) By definition of χ0 , L(s, χ0 ) =

∞  χ0 (n) n=1

ns

=

 1 ns

n odd

   1  1 1 = − = 1 − s ζ (s). ns (2m)s 2 n≥1

m≥1

By part (a) and Problem 3.24, we obtain, for all s ∈ (1, 2),  χ0 (p) p

ps

= log L(s, χ0 ) + O(1)   1 1 + O(1). = log ζ (s) + log 1 − s + O(1) = log 2 s−1

    Here we used that log 1 − 21s ∈ log 12 , log 34 when 1 < s < 2. (c) This is an obvious corollary of part (a) and Problem 13.131(e). 13.133 For every odd integer a, and every s > 1, the computation preceding Problem 13.132 tells us that  p≡a (mod 8)

 χ (p) 1 1  χ0 (p) 1  = + χ (a)−1 . s s p 4 p p 4 ps p χ =χ0

Substituting in the estimates from the last problem,  p≡a (mod 8)

1 1 1 + O(1) = log s p 4 s−1

for as desired. Now letting s ↓ 1 yields the divergence of the series  1 < s < 2, 1 and, consequently, the infinitude of primes p ≡ a (mod 8). p≡a (mod 8) p 13.134 For χ1 , the grouping of terms         1 1 1 1 1 1 1 + − + − + − L(1, χ1 ) = 1 − 3 5 7 9 11 13 15 makes it obvious that L(1, χ1 ) > 0. To handle χ3 , we rewrite

Solutions to Special Step A

171

        1 1 1 1 1 1 1 − + + + − + + .... L(1, χ3 ) = 1 + 3 5 7 9 11 13 15 This is an alternating series with terms strictly decreasing in absolute value; hence, L(1, χ3 ) > (1 + 13 ) − ( 15 + 17 ) > 0. Finally,  L(1, χ2 ) = 1 −

1 1 + 3 5



 +

1 1 + 7 9



 −

1 1 + 11 13

 + ...,

so that  1 − L(1, χ2 ) =

1 1 + 3 5



 −

1 1 + 7 9

 + ....

The right-hand side is an alternating series with terms strictly decreasing in absolute value, and so is < 13 + 15 < 1. Thus, L(1, χ2 ) > 0. 13.135 The groups U8 and U12 are isomorphic, both being abstractly the same as Z2 ⊕ Z2 . An explicit isomorphism maps the two independent generators 3 mod 8 and 5 mod 8 of U8 to the two independent generators 5 mod 12 and 7 mod 12 of U12 . From this, we infer that U12 has the “same” character table as U8 , if we relabel the row headers with 1 mod 12, 5 mod 12, 7 mod 12, and 11 mod 12: χ0 1 1 1 1

1 mod 12 5 mod 12 7 mod 12 11 mod 12

χ1 1 −1 1 −1

χ2 1 −1 −1 1

χ3 1 1 −1 −1

All of our subsequent arguments go through with only minor changes. Perhaps the only difference worth noting is that the formula L(s, χ0 ) = (1 − 2−s )ζ (s), from the start of Problem 13.132(b), is now replaced by the identity L(s, χ0 ) = (1 − 2−s )(1 − 3−s )ζ (s). To prove this, notice that χ0 (2) = χ0 (3) = 0, while χ0 (p) = 1 for all primes p > 3; therefore, L(s, χ0 ) =

p

1 1−

χ0 (p) ps

=

p=2,3

= (1 − 2−s )(1 − 3−s )

1 1−

p

1 ps

1 1−

1 ps

= (1 − 2−s )(1 − 3−s )ζ (s).

172

Solutions to Special Step A

Note that this change to the first half of Problem 13.132(b) does not affect the  1 deduction that p χ0p(p) = log s−1 + O(1). s

References 1. Stephen Abbott, Understanding analysis. Second edition. Undergraduate Texts in Mathematics. Springer, New York, 2015. 2. Joseph Bak and Donald J. Newman, Complex analysis. Third edition. Undergraduate Texts in Mathematics. Springer, New York, 2010.

Solutions to Special Step B

14.136 By Fermat’s little theorem, g −1 = 1 mod  for all g ∈ U . Thus, for all g ∈ U , χ (g)−1 = χ (g −1 ) = χ (1 mod ) = 1, meaning that χ (g) is an ( − 1)th root of unity. 14.137 Let ω be a complex ( − 1)th root of unity. If there is any character χ of U with χ (g) = ω, then χ (g k ) = ωk

for all k ∈ Z.

(B.46)

Since g generates U , the description (B.46) completely determines χ . Thus, there can be at most one character χ of U with χ (g) = ω. It remains to check that there is any such χ . For this, we simply define χ by (B.46). One might worry about whether χ is well-defined, but this is easily checked: If g k1 = g k2 , then k1 ≡ k2 (mod  − 1), and so ωk1 = ωk2 . That χ is a homomorphism is immediate, and clearly χ (g) = ω. Since there are  − 1 distinct ( − 1)th roots of unity, there are  − 1 characters of U . 14.138 Let a ∈ U with a = 1 mod . We first argue that we can find a character ψ of U with ψ(a) = 1. To see this, fix a generator g of U , and (using the previous problem) choose a character ψ of U with ψ(g) = e2π i/(−1) . Since a is not the identity of U , a = g m for some m not divisible by  − 1, and m

ψ(a) = ψ(g m ) = ψ(g)m = e2π i −1 = 1.

© Springer Nature Switzerland AG 2021 P. Pollack, A. Singha Roy, Steps into Analytic Number Theory, Problem Books in Mathematics, https://doi.org/10.1007/978-3-030-65077-3_29

173

174

Solutions to Special Step B

Now we show that the row sum vanishes in the row headed by a. Call that row sum Sa . Then Sa =



χ (a)

χ

where χ runs over the characters of U . Fix a character ψ of U with ψ(a) = 1. Then  ψ(a)Sa = ψ(a)χ (a). χ

For each character χ of U , the function ψχ on U defined by (ψχ )(a) = ψ(a)χ (a) is itself a character of U . Moreover, distinct χ give rise to distinct functions ψχ . It follows that ψχ runs over all the characters of U as χ does, and so  χ

ψ(a)χ (a) =

  (ψχ )(a) = χ (a) = Sa . χ

χ

Comparing the last two displays, we find that Sa is invariant under multiplication by ψ(a). Since ψ(a) = 1, this forces Sa = 0. Thus, every row sum vanishes, except in the row headed by 1 mod . Of course, in that row every entry is 1, and so that row sum is  − 1. 1 From the above, we deduce that I1 mod  = −1 χ χ . The claimed identity for Ia mod  follows exactly as in Problem 13.128, since (for integers a and b coprime to ) a mod  = b mod  ⇐⇒ (a mod )−1 (b mod ) = 1 mod . 14.139 Every compact subset of (s) > 1 is contained within some half-plane (s) ≥ 1 + , where  > 0. Thus, it is enough to show uniform convergence on each of these half-planes. The terms in our double sum have the form pk p−ks , where |pk | ≤ 1/k ≤ 1. When (s) ≥ 1 + ,    p k  1    pks  ≤ (pk )1+ .  1 The sum p,k (pk )1+ converges—and as a positive series, does so under an arbitrary ordering of its terms. So by the Weierstrass M-test, our series converges uniformly for (s) ≥ 1 + , under any ordering of its terms. 14.140 (a) By  definition of χ0 , the sum of the entries in the column headed by χ0 is g∈U χ0 (g) = |U | =  − 1. Now let χ be a nontrivial character of U .

Solutions to Special Step B

175

Choose b ∈ U with χ (b) = 1. Writing Sχ for the sum of the entries in the column headed by χ , χ (b)Sχ = χ (b)



χ (a) =

a∈U





χ (ba) =

a∈U

χ (a) = Sχ .

a∈U

Here we used that ba runs over the distinct elements of U as a does. Since χ (b) = 1, it must be that Sχ = 0. (b) It follows from (a) that the sum of χ (n) vanishes when  n runs over any block of  consecutive integers. So it is enough to prove that | n≤t χ (n)| <  when t < . But this is immediate from the triangle inequality. (c) It is enough to show uniform convergence on each set of the form {s ∈ C : (s) ≥ , |s| < 1/},

(B.47)

where  > 0. We will prove this by borrowing some ideas from our solution to Problem 13.131. Throughout the next paragraph, we assume s belongs to the set (B.47). Abel summation shows that for every real x ≥ 1, 

χ (n)n−s = x −s Aχ (x) + s

 1

n≤x

x

Aχ (t) dt, t s+1

(B.48)

∞ Aχ (t)  dt converges where Aχ (t) := n≤t χ (n). The improper integral 1 t s+1 −s−1 −1− −s absolutely, since |Aχ (t)t | ≤ t . Moreover, |Aχ (x)x | ≤ x − , and −s so Aχ (x)x → 0 as x → ∞. Thus, sending x → ∞ in (B.48),  L(s, χ ) = s 1

∞

Aχ (t) dt. t s+1

Moreover, writing σ for the real part of s, we have (again for x ≥ 1)       χ (n)    ∞ Aχ (t)   −s  dt − x Aχ (s) L(s, χ ) −  = s s s+1   n t x n≤x  ∞ dt ≤ |s| + x −σ σ +1 t x   |s| + 1 x −σ = σ   1 ≤  2 + 1 x − . 

176

Solutions to Special Step B

This final expression can be made arbitrarily small by choosing x large enough in terms of  (independently of the precise value of s). So the series defining L(s, χ ) converges uniformly on the region (B.47). Finally, since each summand χ (n)n−s is an entire function of s, standard results imply that L(s, χ ) is holomorphic for (s) > 0. 14.141 As mentioned in the problem statement, there is a logarithm F (s) of L(s, χ ) defined and analytic on open ball U centered at s = 1. On the other hand, by Exercise 14.139, the function Log L(s, χ ) itself is analytic for (s) > 1. Let V = U ∩ {s ∈ C : (s) > 1}. Then V is a complex domain (a nonempty connected open set). Since F (s) and Log L(s, χ ) are both logarithms of L(s, χ ) on V , their difference F (s) − Log L(s, χ ) takes all its values on V from the set {2π ik : k ∈ Z}. Since F (s) − Log L(s, χ ) is analytic on V (and in particular, continuous), there is a single integer k with F (s) − Log L(s, χ ) = 2π ik for all s ∈ V . Replacing F (s) by F (s) − 2π ik, we can assume that F (s) = Log L(s, χ ) on V . We now extend Log L(s, χ ) to U ∪ {s ∈ C : (s) > 1} by defining Log L(s, χ ) to equal F (s) when s ∈ U \ {s ∈ C : (s) > 1}. This provides an analytic continuation of Log L(s, χ ) to U . The (continued version of the) function Log L(s, χ ), being analytic in a neighborhood of 1, is bounded on each sufficiently small closed ball centered at 1. So if we fix δ positive but sufficiently small, then Log L(s, χ ) = O(1) for all s ∈ [1, 1 + δ]. On the other hand, by Problem 14.139, Log L(s, χ ) is analytic for (s) > 1, and so in particular is continuous for real s > 1. Therefore, Log L(s, χ ) is bounded on the closed interval [1 + δ, 2]. Putting these facts together, Log L(s, χ ) = O(1) for 1 ≤ s ≤ 2. Note in these arguments, the implied constants may depend on χ . 14.142 This is true even without the assumption that  is prime: For 1 < s < 2, Problem 3.24 gives us that  χ0 (p) p

ps

=

 p: gcd(p,)=1

 1  1 1 1 + O (1). = − = log s s s p p p s − 1 p

In the final step, we used that 0 ≤

p|



1 p| ps

≤



1 p| p ,

so that



1 p| ps

= O (1).

14.143 (a) The proof is the same as that of Problem 13.132(a). (b) This is an immediate consequence of part (a) and Problem 14.141. 14.144 Let a be an integer not divisible by . By Problems 14.142, 14.143, and the computation preceding Problem 14.142, when 1 < s < 2,  p≡a (mod )

 χ (p) 1 1  χ0 (p) 1  = + χ (a)−1 s s p −1 p p −1 ps p χ =χ0

=

1 1 log + O (1). −1 s−1

Letting s ↓ 1, we deduce that there are infinitely many primes p ≡ a (mod ).

Solutions to Special Step B

177

14.145 (a) Let χ be either χ1 or χ2 . Since L(s, χ ) is analytic at s = 1 and L(1, χ ) = 0, lim s↓1

L(s, χ ) L(s, χ ) − L(1, χ ) = lim = L (1, χ ). s↓1 s−1 s−1

L(s, χ1 ) L(s, χ2 ) and remain bounded as s ↓ 1. s−1 s−1 For s > 1,

Thus,

L(s, χ0 ) =

p

1 1−

χ0 (p) ps

= (1 − −s )

p: gcd(p,)=1

p

= 1 1−

1 ps

1 1−

1 ps

= (1 − −s )ζ (s).

Since ζ (s)(s − 1) → 1 as s ↓ 1 (see Problem 2.10), we infer that (s − 1)L(s, χ0 ) → 1 − −1 as s ↓ 1. Hence, (s − 1)L(s, χ0 ) remains bounded as s ↓ 1. To finish off, write L(s, χ1 ) L(s, χ2 ) 1 · · L(s, χ ) = ((s − 1)L(s, χ0 )) · s−1 χ s−1 s−1

L(s, χ ).

χ =χ0 ,χ1 ,χ2

The right-hand side is a finite product of terms remaining bounded as s ↓ 1, and so itself remains bounded as s ↓ 1. (b) Observe that when s > 1,  χ

Log L(s, χ ) =

  χ (pk )  1  = χ (pk ) ks ks kp kp χ χ p prime k≥1

= ( − 1)

p prime k≥1

 p,k pk ≡1 (mod )

1 ≥ 0. kpks

(Since there are only finitely many characters χ mod , we encounter no difficulties when interchanging the order of summation.) Exponentiating,

1

L(s, χ ) ≥ 1. Thus, L(s, χ ) → ∞ as s ↓ 1, contrary to what χ χ s−1 we found in (a). This shows that there can be at most one nontrivial Dirichlet character χ modulo  satisfying L(1, χ ) = 0.

178

Solutions to Special Step B

14.146 (a) Suppose for the sake of contradiction that L(1, χ ) = 0 where χ (Z) ⊆ R. Then the function χ(n), defined by χ (n) = χ (n) for all n, is a Dirichlet character modulo  distinct from χ . Furthermore, 0 = L(1, χ ) =

∞  χ (n) n=1

n

=

∞  χ(n) n=1

n

= L(1, χ ).

This contradicts the conclusion of Problem 14.145(b). (b) Suppose χ is a nontrivial real-valued Dirichlet character modulo . Let g be an integer whose reduction mod  generates U . Then χ (g) is a real number of modulus 1, so that χ (g) = ±1. If χ (g) = 1, then χ (a) = 1 whenever a coprime to , forcing χ = χ0 , contrary to hypothesis. So χ (g) = −1. But then on integers a coprime to , χ (a) = 1 ⇐⇒ a is congruent, mod , to an even power of g ⇐⇒ a is a square modulo .   This shows that χ (a) = a when gcd(a, ) = 1. Equality also holds when  | a, a   since then χ (a) and  both vanish. Hence, χ (·) = · . 14.147 We preface the solution with a comment on notation. Throughout this solution, “log” denotes the “principal branch” of the complex logarithm, defined for nonzero complex z by log z = log |z| + i arg z, where arg z is the argument of z belonging to (−π, π ]. It is well-known that − log(1 − z) = z +

z3 z2 + + ... 2 3

for all z with |z| ≤ 1, z = 1.1 Coming to the proof proper, recall that the Gauss sum G admits the expression

|z| < 1, this follows in a routine way from the theory of Taylor expansions of holomorphic functions. The cases when |z| = 1, z = 1 are more delicate. Here one must apply Abel’s lemma (see, e.g., Theorem 21.0 on p. 368 of [1]). Specifically, suppose that |z| = 1, z = 1. The partial sums of zn are easily seen to be bounded (e.g., by 2/|1 − z|); this implies,  via summation by parts, n n that n≥1 zn converges. Abel’s lemma now shows that the power series n≥1 zn t n represents a  n continuous function of t on [0, 1], so that n≥1 zn = limt↑1 (− log(1 − tz)) = − log(1 − z).

1 For

Solutions to Special Step B

179

 b e2π ib/ . G=  b mod 

(See Eq. (8.25) from the solution of Problem 8.86.) Thus,  G·L=

 ∞     b  n 1 2π ib/ e   n

b mod 

=

∞  n=1

n=1

   b n 2π ib/ 1  e . n   b mod 

n

When  | n, the Legendre symbol  = 0. So there is no harm in removing the terms     where  | n. Now writing n¯ for the inverse of n mod , and noting that n¯ = n , we see that      ∞  b bn¯ 2π ib/ n 2π ib/  1  1  e e = . n   n  b mod 

n=1

b mod 

n≥1 n

We make the change of variables a ≡ bn, ¯ so that b ≡ an (mod ). This shows that  1   bn¯   1  a  e2π ib/ = e2π ian/ . n  n  n≥1 n

b mod 

n≥1 n

We now restore the terms where  | n, noting that n. We find that

a mod 



  2π ian/ e = 0 for these

a a mod  

  ∞   1  a  a 2π ian/ 1  2π ian/ e e = n  n  n≥1 n

a mod 

n=1

a mod 

  ∞ ∞   a   a 2π ian/ 1  1 2π ian/ e e = = . n   n n=1

a mod  a≡0

a mod  a≡0

(B.49)

n=1

The inner sum in the last expression is precisely − log(1 − e2π ia/ ). Collecting our results,  a  log(1 − e2π ia/ ). G·L=− (B.50)  0 π(x) for all x shown, Littlewood showed that there are arbitrarily large values of x where the reverse inequality holds

x 103 104 105 106 107 108 109 1010 1011 1012 1013

π(x) 168 1229 9592 78,498 664,579 5,761,455 50,847,534 455,052,511 4,118,054,813 37,607,912,018 346,065,536,839

Li(x) 177 1245 9629 78,627 664,917 5,762,208 50,849,234 455,055,614 4,118,066,400 37,607,950,280 346,065,645,809

Li(x) − π(x) 9 16 37 129 338 753 1700 3103 11,587 38,262 108,970

N ×N multiplication table. You showed in Problem 12.120 that M(N)/N 2 → 0 as N → ∞. It was proved by Kevin Ford (2008) that M(N) lies between two positive constant multiples of M≈ (N ) :=

N2 1 + log log 2 (≈0.086), , where δ = 1 − δ 3/2 log 2 (log N) (log log N)

once N is large enough. Question: Does M(N)/M≈ (N ) tend to a limit as N → ∞ ? Extensive computations on this problem have been carried out recently by Brent, Pomerance, Purdum, and Webster, but it remains unclear what the numbers are trying to tell us. The distribution of prime numbers. We alluded in Step #12 to the Prime Number Theorem, which we formulated as a result comparing π(x) to x/ log x. But it was recognized already by Gauss that π(x) ought to be compared not to x/ log x x dt x/ log x but to the “logarithmic integral” Li(x) := 2 log t . Since Li(x) → 1 as x → ∞, the Prime Number Theorem—being an assertion about a limit of ratios— can be stated either way. However, as far as absolute numerical difference with π(x) is concerned, the logarithmic integral is a much closer approximation than x/ log x (Table 17.1). How close? Conjecture |π(x) − Li(x)|
m1+ suffices, but known results do not guarantee equidistribution unless x > exp(m ).

Combinatorial methods. In Step #9, you finished the proof of Brun’s theorem:  1 < ∞, where p runs over the primes for which p + 2 is also prime. p p Brun’s argument (which we followed) was the first important example (1919) of what is called a “sieve method” in number theory. Sieve methods have seen extensive development, and they continue to be the source of the sharpest results to date on the twin prime conjecture. We mention two: (a) (Chen, 1966) There are infinitely many primes p for which p + 2 is either prime or a product of two primes (counting multiplicity). (b) (D.H.J. Polymath, 2014, building on ideas of Zhang, Maynard, Tao) There are infinitely pairs of primes that differ by no more than 246.

Reference 1. Harold Davenport, Multiplicative Number Theory (3rd ed.), 2000: Springer, New York.

Suggestions for Further Reading

1. Tom M. Apostol, Introduction to Analytic Number Theory, 2010: Springer, New York. 2. Alina Cojocaru and M. Ram Murty, An Introduction to Sieve Methods and their Applications, 2005: Cambridge University Press, Cambridge. 3. Harold Davenport, Multiplicative Number Theory (3rd ed.), 2000: Springer, 4. Adolf J. Hildebrand, Introduction to Analytic Number Theory: Lecture Notes, 2013. Online resource: https://faculty.math.illinois.edu/~hildebr/ant/index.html 5. Edmund Hlawka, Johannes Schoißengeier, and Rudolf Taschner, Geometric and Analytic Number Theory, 1991: Springer-Verlag, Berlin. 6. Loo-Keng Hua, Introduction to Number Theory, 1987: Springer-Verlag, Berlin. 7. Edmund Landau, Elementary Number Theory, 1999: American Mathematical Society, Providence, RI. 8. Florian Luca and Jean-Marie De Koninck, Analytic Number Theory: Exploring the Anatomy of Integers, 2012: American Mathematical Society, Providence, RI. 9. Hugh L. Montgomery and Robert C. Vaughan, Multiplicative Number Theory I. Classical Theory, 2012: Cambridge University Press, Cambridge. 10. M. Ram Murty, Problems in Analytic Number Theory (2nd ed.), 2007: Springer, New York. 11. Melvyn B. Nathanson, Additive Number Theory: The Classical Bases, 1996: Springer, New York. 12. Paul Pollack, Not Always Buried Deep: A Second Course in Elementary Number Theory, 2009: American Mathematical Society, Providence, RI. 13. Harold N. Shapiro, Introduction to the Theory of Numbers, 2008: Dover, New York. 14. Gerald Tenenbaum, Introduction to Analytic and Probabilistic Number Theory (3rd ed.), 2015: American Mathematical Society, Providence, RI.

© Springer Nature Switzerland AG 2021 P. Pollack, A. Singha Roy, Steps into Analytic Number Theory, Problem Books in Mathematics, https://doi.org/10.1007/978-3-030-65077-3

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