Diophantine equations and power integral bases theory [2 ed.] 9783030238643, 9783030238650

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István Gaál

Diophantine Equations and Power Integral Bases Theory and Algorithms Second Edition

Diophantine Equations and Power Integral Bases

István Gaál

Diophantine Equations and Power Integral Bases Theory and Algorithms Second Edition

István Gaál Institute of Mathematics University of Debrecen Debrecen, Debrecen, Hungary

ISBN 978-3-030-23864-3 ISBN 978-3-030-23865-0 (eBook) https://doi.org/10.1007/978-3-030-23865-0 Mathematics Subject Classification: 11D57, 11D59, 11D61, 11R04, 11Y50 1st edition: © Birkhäuser Boston 2002 © Springer Nature Switzerland AG 2019 This work is subject to copyright. All rights are reserved 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, express 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 book is published under the imprint Birkhäuser, www.birkhauser-science.com by the registered company Springer Nature Switzerland AG. The registered company address is: Gewerbestrasse 11, 6330 Cham, Switzerland

To Gabi, Zsuzsi, and Szilvi

Foreword

In the mid-1980s, the author started to investigate power integral bases in algebraic number fields and related Diophantine equations. To construct feasible methods requires both to develop the underlying theory of Diophantine equations and also to create algorithms. We described relations of index form equations and Thue equations and developed reduction and enumeration algorithms, involving Diophantine approximation tools and LLL algorithm. Starting with cubic number fields, within more than 10 years, we had feasible methods up to number fields of degree 5. The first edition of this book in 2002 (Diophantine Equations and Power Integral Bases, New Computational Methods, Birkhäuser Boston, 2002) was actually a thesis of the author. The book turned out to be useful in teaching university courses and in the research work of interested colleagues and PhD students. And the development of the methods did not stop. In the last 15 years, we developed several useful methods on higher degree number fields and relative extensions, considering specific number fields and also infinite parametric families of number fields. In this new edition of the book, we extended the material of the first edition with a great variety of new results, both on the theory of Diophantine equations and on the algorithms for their resolution. Our purpose was to arrange all these results into a unique context, to give an up-to-date overview of this area, and to inspire further research. Debrecen, Hungary January 2019

István Gaál

vii

Preface

One of the classical problems in algebraic number theory, going back among others to Dedekind [De78], Hensel [He08], and Hasse [Ha63], is to decide if an algebraic number field K of degree n has a power integral basis, that is, an integral basis of type (1, α, . . . , α n−1 ). The existence of such an integral basis means that ZK , the ring of integers of K, is monogenic and can be generated by the single element α over Z, that is ZK = Z[α]. This problem usually leads to Diophantine equations, the so-called index form equations. One of the purposes of this book is to describe the structure of index form equations and to point out their relation to other classical types of Diophantine equations like Thue equations. As we shall see, various types of Thue equations play an essential role in the resolution of index form equations. Therefore, we provide a detailed discussion of these equations and some related equations like norm form equations, as well. The other aim of the book is to get acquainted with various types of number fields. The methods for the resolution of index form equations heavily depend on the structure of the number field they are related to. We also discuss infinite parametric families of number fields, like the families of simplest cubic, simplest quartic, and simplest sextic fields, pure fields of various degrees, and some other families. The resolution of the index form equations in these parametric families leads to parametric index form equations, Thue equations, etc. Hence, we are very much interested in the solutions of these parametric families of equations. It yields to describe the solutions of infinitely many Diophantine equations; hence, this area promises very exciting results. The third important purpose of this book is to describe efficient algorithms for determining generators of power integral bases. Therefore, we provide algorithms for the resolution of various types of index form equations, related Thue equations, and norm form equations. To perform these algorithms, one needs computer algebra systems like Maple [CG88] and algebraic number theory packages like Kash [DF97], Magma [BCP97], or Pari [Pa]. In case we want to determine all solutions of the index form equation, i.e., all generators of power integral bases, then a standard way is to apply Baker’s method ix

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combined with reduction methods and enumeration methods, often together with certain sieve methods. To give a rough idea, Baker’s method gives an extremely high upper bound (e.g., exp(1020 )) for the solutions, which implies the finiteness of the number of solutions but no feasible way to determine the solutions themselves. Therefore, numerical reduction methods are applied to show that the solutions are in fact under a bound of relatively low magnitude (say 1000). If the number of variables is larger than 5 or 6, then it still yields a high number of possible cases whence we need some enumeration method utilizing special properties of the equation. Note that the reduction and enumeration methods are numerical, which gives an exciting experimental aspect to the resolution of the equations. In addition, we describe several other methods to calculate the solutions of index form equations. In some cases, it is simple to exclude the existence of generators of power integral bases or the solutions of certain Diophantine equations, by using special properties of the equation, by utilizing the structure of the actual number field, or by applying congruence considerations. An important feature of the book is to provide the reader with detailed numerical examples which give a good insight into the application of the methods. It is important to emphasize that the methods and algorithms we develop in the book are applicable for a wide class of related types of Diophantine equations, as well. The knowledge of power integral bases in a number field has important applications. The most straightforward benefit of a power integral basis is to have an easy way of performing arithmetic calculations in ZK , especially an easy way of multiplication. But it has many other, more or less, independent applications: just to mention only one quite characteristic application, let us recall the result of Kovács and Peth˝o [KP91], who proved that in ZK , there exist a generalized number system if and only if ZK is monogenic; moreover, the generators of power integral bases are needed to construct these number systems. This was one of the reasons why it became interesting not only to decide the monogenity of ZK but also to determine all possible generators of power integral bases. The algorithm for determining power integral bases in cubic fields was one of the first important practical applications of the methods for solving Thue equations. The simple paper [GSch89] on this cubic case is one of the most frequently cited papers of the author, which shows what kind of novelty such computations had at the end of the 1980s. It was a great challenge to try to extend the algorithms for computing all generators of power integral bases to higher degree number fields, which was finally successful at least up to degree 6 in general and for many special higher degree fields up to about degree 15, where we reached the capability of the present methods and the limits of capacity of the present computing machinery. Imagine that for a number field of degree n, the index form equation has n − 1 variables and degree n(n − 1)/2. This means that the index form equation is mostly (already in quartic fields) a very complicated equation that does not even fit onto one printed page. It is now about 30 years ago that the author of this book began to study constructive methods for determining power integral bases in algebraic number

Preface

xi

fields. The material of this book was formed by several lectures held by the author at various conferences (cf. [Ga91, Ga96b, Ga98b, Ga99, Ga00a]). Also, some parts of it appeared at special university courses and PhD courses held regularly by the author. The book has a transparent structure. In Chap. 1, we fix our notation, describe the basic concepts, and summarize some important results on power integral bases. We also treat the relative case, power integral bases in relative extensions, that will be useful in the sequel. In Chap. 2, we collect the main tools for solving our equations. We describe results on Baker’s method, reduction methods, and enumeration algorithms that will be used throughout. In Chaps. 3–5, we apply these tools to various types of Thue equations, starting with classical Thue equations continuing with inhomogeneous Thue equations until relative Thue equations, respectively. In Chap. 6, we describe how to apply these tools to norm form equations. Chapter 7 gives a general overview of the structure of index forms. We detail how to deal with fields having subfields or how much easier the problem is if the field is the composite of its subfields. In Chaps. 8–11, we describe those algorithms that can be used for solving index form equations in cubic, quartic, quintic, and sextic number fields, respectively. In addition to several efficient methods that work in special types of fields, we have general algorithms up to degree 6. We also include several infinite parametric families of fields and solve the corresponding index form equations in a parametric form. Chapter 12 is devoted to pure fields, with a lot of interesting properties. In Chaps. 13 and 14, we consider the problem of relative power integral bases in cubic and quartic relative extensions of fields. In Chap. 15, we consider power integral bases in some special types of higher degree number fields. Finally, in Chap. 16, we provide the results of our computations: tables of solutions of Diophantine equations and tables of generators of power integral bases in cubic, quartic, sextic fields. These tables might have many applications. At the end of the chapters, the reader will find notes on further results related to the topic of the chapter. The material of the book can be easily followed by graduates and even undergraduates. We only use standard algebra and number theory contained in usual university courses. Basic properties of continued fractions are in the book of Niven and Zuckerman [NZ80]. The LLL basis reduction algorithm is detailed in Pohst [Po93]. Further concepts are given in Chaps. 1 and 2. Starting with the classical book of Baker [Ba90], there are some other volumes related more or less to some topics discussed here, for example, Shorey and Tijdeman [200], de Weger [We89], Smart [Sm98], and Evertse and Gy˝ory [EGy17a]. In the present book, however, we intended to provide a detailed algorithmic approach focusing on power integral bases and the resolution of index form equations, based

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on the methods for solving several types of Thue equations, norm form equations, etc., therefore, giving a wide extension of existing literature. Debrecen, Hungary January 2019

István Gaál

Acknowledgments

Professor George Anastassiou encouraged the publication of the first edition of this book. The author is thankful to his senior colleagues and coauthors, Professor Kálmán Gy˝ory, Attila Peth˝o and Gábor Nyul. He is indebted to Professor Michael Pohst for a long-standing cooperation and more than 25 joint papers until today. He is also thankful to Professor Borka Jadrijevi´c, László Remete, and Timea Szabó for their several joint researches on the topic of the book. The first edition of this book could not have been written without the support of Ann Kostant, editor of Birkhäuser Boston, who helped the author to construct the material to become a book. He thanks Christopher Tominich and Chris Eder, editors of Birkhäuser Boston, for their support to create the second edition. The author is also thankful the copy editor of the second edition for a very careful reading of the text and for several corrections.

xiii

Contents

1

Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.1 Some Basics of Number Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.2 The Index of Algebraic Integers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.3 Relative Extensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.4 Factors of the Index in Composite Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.5 Results on the Field Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

1 1 3 6 9 11

2

Auxiliary Results and Tools. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.1 Baker’s Method and Effective Finiteness Theorems. . . . . . . . . . . . . . . . 2.2 Reduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.2.1 Davenport Lemma . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.2.2 The General Case . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.3 Enumeration Methods . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.4 Software and Hardware . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.5 Related Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

13 13 15 15 17 18 23 24

3

Thue Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.1 Elementary Estimates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.2 Thue’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.3 Fast Algorithm for Finding “Small” Solutions. . . . . . . . . . . . . . . . . . . . . . 3.4 Binomial Thue Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.4.1 Results on the Solutions of Binomial Thue Equations . . . . 3.4.2 Calculation of “Small” Solutions of Binomial Thue Equations. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.4.3 Computational Aspects . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.5 Effective Methods for Thue Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.6 The Method of Bilu and Hanrot . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.7 Experiences by Solving a Thue Equation with “Large” Entries . . . 3.8 Further Results on Thue Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

25 25 26 26 27 27 29 32 32 34 35 37

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4

Inhomogeneous Thue Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.1 Elementary Estimates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.2 Baker’s Method. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.3 Reduction, Test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.4 An Analogue of the Bilu-Hanrot Method . . . . . . . . . . . . . . . . . . . . . . . . . . .

39 39 41 42 43

5

Relative Thue Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.1 A Fast Algorithm for Finding “Small” Solutions of Relative Thue Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.1.1 The Reduction Procedure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.1.2 Enumerating Tiny Values of the Variables . . . . . . . . . . . . . . . . 5.1.3 The Complete Algorithm . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.1.4 Computational Aspects . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.1.5 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.2 Quartic Binomial Thue Equations over Imaginary Quadratic Fields. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.2.1 Specialties of the Actual Calculations . . . . . . . . . . . . . . . . . . . . . 5.3 Effective Methods for Relative Thue Equations . . . . . . . . . . . . . . . . . . . . 5.3.1 Baker’s Method and Reduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.3.2 Enumeration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.3.3 An Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.4 Totally Real Thue Equations over Imaginary Quadratic Fields . . . . 5.4.1 How to Apply Theorem 5.9 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.4.2 An Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.5 Simplest Quartic and Simplest Sextic Thue Equations over Imaginary Quadratic Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.5.1 Proof of Theorem 5.11 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.5.2 Proof of Theorem 5.12 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.5.3 Computational Aspects . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.6 Further Results on Relative Thue Equations . . . . . . . . . . . . . . . . . . . . . . . .

45 45 48 50 52 52 53 56 56 57 57 60 61 63 69 70 71 73 79 79 79

6

The Resolution of Norm Form Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.1 Preliminaries. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.2 Solving the Unit Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.3 Calculating the Solutions of the Norm Form Equation . . . . . . . . . . . . . 6.4 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.5 Further Results on Norm Form Equations . . . . . . . . . . . . . . . . . . . . . . . . . .

81 81 83 85 86 92

7

Index Form Equations in General . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.1 The Structure of the Index Form. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.2 Using Resolvents . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.3 Factorizing the Index Form When Proper Subfields Exist . . . . . . . . . 7.4 Linear Combinations of the Factors of the Index Form . . . . . . . . . . . .

93 93 95 96 96

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7.5

Composite Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 96 7.5.1 Coprime Discriminants . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 97 7.5.2 Non-coprime Discriminants 1. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 99 7.5.3 Non-coprime Discriminants 2. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101 Notes on Montes Algorithm . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103

7.6 8

Cubic Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.1 Arbitrary Cubic Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.2 Simplest Cubic Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.3 Pure Cubic Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.3.1 Integral Basis: Index Form . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.3.2 Frequency of Monogenic Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.4 Further Results on the Monogenity of Cubic Fields . . . . . . . . . . . . . . . .

105 105 107 108 108 109 111

9

Quartic Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.1 Algorithm for Arbitrary Quartic Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.1.1 The Resolvent Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.1.2 The Quartic Thue Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.1.3 Proof of Theorem 9.6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.1.4 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.2 Simplest Quartic Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.2.1 Power Integral Bases and Minimal Indices. . . . . . . . . . . . . . . . 9.2.2 Proof of Theorem 9.16 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.3 An Interesting Application to Mixed Dihedral Quartic Fields . . . . . 9.4 Totally Complex Quartic Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.4.1 Parametric Families of Totally Complex Quartic Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.4.2 A Family of Totally Complex Quartic Fields with Two Parameters . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.5 Bicyclic Biquadratic Number Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.5.1 Integral Basis and Index Form . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.5.2 The Totally Real Case . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.5.3 The Totally Complex Case . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.5.4 The Field Index of Bicyclic Biquadratic Number Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.6 Pure Quartic Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.7 Further Results on the Monogenity of Quartic Fields . . . . . . . . . . . . . .

113 113 114 115 118 123 124 124 127 132 132

Quintic Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.1 Algorithm for Arbitrary Quintic Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.1.1 Preliminaries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.1.2 Baker’s Method and Reduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.1.3 Enumeration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.1.4 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

151 151 152 153 155 156

10

134 136 139 139 141 143 144 147 150

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10.2

Lehmer’s Quintics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.2.1 Integer Basis and Unit Group . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.2.2 The Index Form . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.2.3 The Index Form Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.2.4 The Exceptional Case . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

161 162 164 166 168

Sextic Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.1 Arbitrary Sextic Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.1.1 The Unit Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.1.2 Enumeration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.1.3 An Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.2 Sextic Fields with a Quadratic Subfield . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.2.1 Real Quadratic Subfield . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.2.2 Totally Real Sextic Fields with a Quadratic and a Cubic Subfield . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.2.3 Imaginary Quadratic Subfield . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.2.4 Sextic Fields with an Imaginary Quadratic and a Real Cubic Subfield . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.2.5 Parametric Families of Sextic Fields with Imaginary Quadratic and Real Cubic Subfields . . . . . . . . . . . 11.3 Sextic Fields with a Cubic Subfield . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.4 Sextic Fields as Composite Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.4.1 A Cyclic Sextic Field . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.4.2 A Noncyclic Sextic Field . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.5 Integral Bases and Monogenity of the Simplest Sextic Fields . . . . . 11.5.1 Monogenity of an Order of ZKt . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.5.2 Integral Basis of ZKt . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.5.3 Monogenity of Simplest Sextic Fields . . . . . . . . . . . . . . . . . . . . . 11.5.4 Computational Remarks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.6 Further Results on the Monogenity of Sextic Fields. . . . . . . . . . . . . . . .

169 169 169 171 174 176 178

183 187 189 189 189 190 190 191 191 194 195

12

Pure Fields. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12.1 Integral Basis of Pure Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12.2 Monogenity of Pure Fields. . . . . . . . . . .√ ................................ 12.2.1 Pure Cubic Fields: K = Q( 3 √ m) . . . . . . . . . . . . . . . . . . . . . . . . . . 12.2.2 Pure Quartic Fields: K = Q(√4 m) . . . . . . . . . . . . . . . . . . . . . . . . . 12.2.3 Pure Sextic Fields: K = Q(√6 m) . . . . . . . . . . . . . . . . . . . . . . . . . . 12.2.4 Pure Octic Fields: K = Q( 8 m) . . . . . . . . . . . . . . . . . . . . . . . . . . . 12.3 Notes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

197 197 198 199 200 201 203 204

13

Cubic Relative Extensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13.1 The Cubic Relative Thue Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13.1.1 Example 1: Cubic Extension of a Quintic Field. . . . . . . . . . . 13.1.2 Example 2: Cubic Extension of a Sextic Field . . . . . . . . . . . . 13.1.3 Computational Experiences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

207 207 209 211 214

11

179 180 180

Contents

13.2

Nonic Fields with Cubic Subfields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13.2.1 The Relative Thue Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13.2.2 The Unit Equation over the Normal Closure . . . . . . . . . . . . . . 13.2.3 The Common Variables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13.2.4 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Composites of Cubic Fields with Real Quadratic Fields . . . . . . . . . . . 13.3.1 Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13.3.2 Computational Aspects . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

215 216 217 219 223 223 225 227

Quartic Relative Extensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14.1 Relative Quartic Extensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14.1.1 Preliminaries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14.1.2 The Cubic Relative Thue Equation . . . . . . . . . . . . . . . . . . . . . . . . 14.1.3 Representing the Variables as Binary Quadratic Forms. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14.1.4 The Quartic Relative Thue Equations . . . . . . . . . . . . . . . . . . . . . 14.1.5 An Example: Computing Relative Power Integral Bases in a Quartic Extension of a Cubic Subfield. . . . . . . . . 14.1.6 Computational Remarks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14.2 Octic Fields with a Quadratic Subfield . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14.2.1 Preliminaries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14.2.2 The Unit Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14.2.3 The Inhomogeneous Thue Equation . . . . . . . . . . . . . . . . . . . . . . . 14.2.4 Sieving . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14.2.5 An Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14.2.6 Computational Remarks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14.3 Composites of Quartic Fields with Real Quadratic Fields . . . . . . . . . 14.3.1 Composites of Totally Complex Quartic Fields with Real Quadratic Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14.3.2 An Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14.3.3 Computational Aspects . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14.4 Pure Quartic Extensions of Imaginary Quadratic Fields . . . . . . . . . . . 14.4.1 From Index Equations to Binomial Thue Equations . . . . . . 14.4.2 Generators of Relative Power Integral Bases. . . . . . . . . . . . . . 14.4.3 Absolute Power Integral Bases of Pure Quartic Extensions of Imaginary Quadratic Fields . . . . . . . . . . . . . . . . 14.5 Infinite Families of Octic Fields with Imaginary Quadratic Subfields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14.5.1 Example 1: A Galois Family . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14.5.2 Example 2: Simplest D4 Octics . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14.5.3 Example 3: An Infinite Family with Quadratic Algebraic Parameter . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

229 229 229 230

13.3

14

xix

232 234 234 237 237 237 238 240 241 242 245 245 247 249 252 252 253 254 255 256 256 260 262

xx

15

16

Contents

Some Higher Degree Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15.1 Power Integral Bases in Imaginary Quadratic Extensions of Totally Real Cyclic Fields of Prime Degree . . . . . . . . . . . . . . . . . . . . . . . . 15.2 Power Integral Bases in Imaginary Quadratic Extensions of Lehmer’s Quintics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15.3 One More Composite Field . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15.4 Further Results on the Monogenity of Higher Degree Number Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15.5 Notes on the Monogenity of Cyclotomic Fields . . . . . . . . . . . . . . . . . . . . Tables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16.1 Binomial Thue Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16.2 Binomial Thue Equations over Imaginary Quadratic Fields . . . . . . . 16.3 Cubic Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16.3.1 Totally Real Cubic Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16.3.2 Complex Cubic Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16.4 Pure Cubic Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16.5 Quartic Fields. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16.5.1 The Distribution of the Minimal Indices . . . . . . . . . . . . . . . . . . 16.5.2 The Average Behavior of the Minimal Indices . . . . . . . . . . . . 16.5.3 Totally Real Cyclic Quartic Fields . . . . . . . . . . . . . . . . . . . . . . . . . 16.5.4 Monogenic Mixed Dihedral Extensions of Real Quadratic Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16.5.5 Totally Real Bicyclic Biquadratic Number Fields . . . . . . . . 16.5.6 Totally Complex Bicyclic Biquadratic Number Fields . . . 16.5.7 Some More Quartic Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16.6 Sextic Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16.6.1 Totally Real Cyclic Sextic Fields . . . . . . . . . . . . . . . . . . . . . . . . . . 16.6.2 Sextic Fields with Imaginary Quadratic Subfields . . . . . . . . 16.7 Integral Basis of the Simplest Sextic Fields. . . . . . . . . . . . . . . . . . . . . . . . .

265 265 268 269 269 270 273 273 280 282 283 285 286 288 288 289 290 291 292 297 299 304 304 306 309

References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 313 Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 323

Notation

Q Z α, β, γ , . . . αj , α (j )

ω K, L, M, . . . ZK O DK DK/Q (α1 , . . . , αn ) NK/Q (α) DK/M NK/M (α) I (α) μ(K) m(K) IK/M (α) J (α) JL,M (α)

Field of rational numbers Ring of rational integers Algebraic numbers Conjugates of an algebraic number α: according to the actual context and notation, in some cases the lower index, in other cases the upper index, is more convenient The number theoretic conjugate of a quadratic algebraic number ω or the complex conjugate of a complex number ω Number fields The ring of integers of an algebraic number field K An order of the ring of integers of a number field The discriminant of the number field K The discriminant of the algebraic numbers α1 , . . . , αn ∈ K Norm of α ∈ K, that is, the product of the (relative) conjugates of α The relative discriminant of the number field K over the subfield M Relative norm of α ∈ K over M, that is, the product of the (relative) conjugates of α over M The index of a primitive element α of a number field The minimal index of the number field K The field index of the number field K The relative index of a primitive element α of a number field K over the subfield M A factor of I (α) of α ∈ K in case K has a subfield A factor of the index I (α) of α ∈ K in case K is the composite of its subfields L, M

xxi

xxii

Notation

|λ|

The size of λ, that is, the maximum absolute value of its conjugates The absolute logarithmic height of an algebraic number α A lattice The real part of a complex number s The imaginary part of a complex number s

h(α) L Re(s) Im(s)

Chapter 1

Introduction

Abstract It is a classical problem in algebraic number theory to decide if the ring of integers ZK of a number field K is monogenic, that is if it admits power integral bases of type (1, α, . . . , α n−1 ). In the 1960s Hasse (Zahlentheorie, AkademieVerlag, Berlin, 1963, §25.6., p. 438) asked to give an arithmetic characterization of those number fields which have power integral bases. The first example of a non-monogenic field was given by Dedekind (Abh. König. Ges. der Wissen. zu Göttingen 23:1–23, 1878). In this section we recall some basic notions of number fields, and then we give the most important concepts in connection with monogenic fields K having an integral basis (1, α, . . . , α n−1 ) that is a power integral basis. We describe this phenomenon both in the absolute case and in the relative case. We also discuss the specialties of the case when K is the composite of two subfields. Our main purpose is to determine generators of power integral bases. As we shall see, this algorithmic problem is satisfactorily solved for lower degree number fields (especially for cubic and quartic fields) and there are efficient methods for certain classes of higher degree fields. Our algorithms enable us in many cases to describe all power integral bases also in infinite parametric families of certain number fields.

1.1 Some Basics of Number Fields An algebraic number α of degree n is a root of a nonzero irreducible polynomial (defining polynomial) with coefficients in Z of degree n. The conjugates α (i) (i = 1, . . . , n) of α are the roots of its defining polynomial. An algebraic number field K is a finite extension of Q is C. It can be generated by a single element. If K is of degree n = [K : Q], then there exists an algebraic number α of degree n, such that K = Q(α). Any element β ∈ K can be represented as β = r0 +r1 α+. . .+rn−1 α n−1 with rational coefficients ri (i = 0, . . . , n−1). The (relative) conjugates of β are β (i) = r0 +r1 α (i) +. . .+rn−1 (α (i) )n−1 (i = 1, . . . , n). The ring of integers ZK of K is the set of algebraic integer elements of K (having defining polynomial with integers coefficients in Z and leading coefficient 1).

© Springer Nature Switzerland AG 2019 I. Gaál, Diophantine Equations and Power Integral Bases, https://doi.org/10.1007/978-3-030-23865-0_1

1

2

1 Introduction

An integral basis (1, ω2 , . . . , ωn ) of K is a set of Q-linearly independent elements of ZK having the property that any β ∈ K can be represented as β = r1 + r2 ω2 + . . . + rn ωn with ri ∈ Q (i = 1, . . . , n) and any β ∈ ZK can be represented as β = z1 + z2 ω2 + . . . + zn ωn with zi ∈ Z (i = 1, . . . , n). Remark that we also have β (i) = z1 + z2 ω2(i) + . . . + zn ωn(i) (i = 1, . . . , n). If α1 , . . . , αn are Q-linearly independent elements of K, then their discriminant is  (1) (1) α α ... 2  1  .. DK/Q (α1 , . . . , αn ) =  ... .   α (n) α (n) . . . 1 2

(1) 2

αn .. .

   .  (n)  α n

The discriminant of α ∈ K is defined as DK/Q (α) = DK/Q (1, α, . . . , α n−1 ) =



(α (i) − α (j ) )2 .

1≤i Sk . In each step we take S = Si , s = Si+1 and enumerate the lattice points in the corresponding ellipsoids. The initial constant is given by the reduced bound (2.10), the last constant Sk should be made as small as possible, so that the exponents with   1   ≤ β (I )  ≤ Sk for all I ∈ I ∗ Sk

(2.16)

can be enumerated easily. Observe that the set (2.16) is also contained in an ellipsoid, namely, by (2.8) we have in Rt ||g + a1 e1 + . . . + an en ||22 = ||b||22 ≤ t · (log Sk )2 .

(2.17)

For an optimal choice of the constants Si see Wildanger [Wild97]. According to our experience in the first step S1 can be much√smaller than S0 , for example S1 = 1020 . Then it is economical to have Si+1 = Si until Si decreases to about 103 . Then we choose Si+1 = Si /2.

2.4 Software and Hardware One of our main purposes is to perform computations in number fields using the algorithms described here. For all these algorithms the basic data of the number fields involved (such like integral bases, fundamental units) are considered as input data. These can be taken either from tables, or can be calculated by using algebraic number theory packages. We mostly used Kash [DF97] and Magma [BCP97] but one can also use Pari [Pa]. The methods can be applied to solving similar types of decomposable form equations, as well. Most of the algorithms were implemented in Maple [CG88] which is very convenient in developing the algorithms, trying different ideas which often helps to improve the procedures. The algorithms were executed mostly on an average laptop in Linux or Windows system. The clock speed of these laptops was growing approximately with a factor 10 during the time 1989–2019 when these calculations were performed. Some of our extensive computations were made on the supercomputer (high performance computer) network situated in Debrecen-Budapest-Pécs-Szeged in Hungary. Using the HPC the calculations can be distributed on a high number of

24

2 Auxiliary Results and Tools

parallel nodes. In such cases the CPU times are calculated as the sum of the CPU times of the nodes involved. However, the speed of a single node of the HPC is not much higher than the speed of the processors of a laptop. Introducing our algorithms we always indicate the year of calculations, this way trying to provide comparable CPU times.

2.5 Related Results A detailed explanation of unit equations can be found in Evertse and Gy˝ory [EGy15], see also Smart [Sm98]. In our calculations it was very easy to apply the estimates of Baker and Wüstholz [BW93] for linear forms in the logarithms of algebraic numbers. For sharper estimates see for example Matveev [Mat00]. Our experiences show that the result of the reduction methods is only weakly influenced by the linear form estimates. The finiteness of the number of solutions of Eq. (1.7) follows from the result of Birch and Merriman [BM71]. Using Baker’s method Gy˝ory [Gy76] gave the first explicit upper bounds for the absolute values of the solutions of Eq. (1.7), see Theorem 2.2. Solving our equations we directly use the Baker-type linear form estimates instead of applying the upper bounds of Theorem 2.2. This leads to sharper estimates. Combining Baker’s method with reduction methods (see Sect. 2.2) and enumeration algorithms (see Sect. 2.3) we shall be able to determine the solutions of the index form equation (1.7) in lower degree fields and in some classes of higher degree fields. These methods can be applied to several other types of Diophantine equations, as well.

Chapter 3

Thue Equations

Abstract The resolution of index form equations in cubic and quartic number fields is based on solving Thue equations. We give here an overview of the methods for solving these equations. We also consider binomial Thue equations that we shall apply in the sequel in pure quartic fields (Sect. 9.6).

3.1 Elementary Estimates Let F ∈ Z[X, Y ] be an irreducible homogeneous form of degree n ≥ 3 and let m be a nonzero integer. Consider the Thue equation F (x, y) = m in x, y ∈ Z.

(3.1)

We can assume that the leading coefficient of F is 1, otherwise if the leading coefficient is a, we multiply the equation by a n−1 , replace (x, y) by (ax, ay), and get the required situation. Let α (1) , . . . , α (n) be the roots of F (x, 1) = 0, (these are conjugates of the algebraic integer α = α (1) ) and let (x, y) be an arbitrary but fixed solution of (3.1). Then Eq. (3.1) can be written in the form n 

(x − α (j ) y) = m.

(3.2)

j =1

Let β (j ) = x − α (j ) y (1 ≤ √ j ≤ n) and denote by i the index with |β (i) | = minj |β (j ) |. Obviously |β (i) | ≤ n |m|, hence |β (j ) | ≥ |β (j ) − β (i) | − |β (i) | ≥ |α (j ) − α (i) ||y| − ≥

1 (j ) |α − α (i) ||y|, 2

© Springer Nature Switzerland AG 2019 I. Gaál, Diophantine Equations and Power Integral Bases, https://doi.org/10.1007/978-3-030-23865-0_3

n |m| (3.3)

25

26

3 Thue Equations

√ for all j = i if |y| > 2 n |m|/|α (j ) −α (i) | (small values of y can be directly checked). Using (3.2) inequality (3.3) implies    (i) x  α −  < c1 (3.4)  y  |y|n with c1 = 

√ n j =i

|m| . − α (i) |

|α (j )

3.2 Thue’s Theorem Equation (3.1) is called Thue’s equation in honor of Thue who gave [Thu09] an ineffective improvement of Liouville’s theorem in 1909. Theorem 3.1 If α is an algebraic integer of degree n, then for every ε > 0 there exist a positive constant c = c(α, ε), such that for every rational number x/y     c α − x  >  y  |y|n/2+1+ε holds. Thue’s proof does not make possible to calculate the constant c. The above inequality together with (3.4) implies the existence of an upper bound for |y|, which cannot be calculated explicitly. Such results are called “ineffective” in contrast with those “effective” results that provide explicit upper bounds. The above theorem implies Thue’s famous theorem [Thu09]: Theorem 3.2 Equation (3.1) has only finitely many solutions. The method of Thue does not enable one to find the solutions.

3.3 Fast Algorithm for Finding “Small” Solutions Consider now the solutions of the Thue inequality |F (x, y)| ≤ m in x, y ∈ Z, (x, y) = 1, |y| < C,

(3.5)

where C is a given large constant, say 10500 . For our further purposes we mention here an efficient algorithm of Peth˝o [Pe87] producing the solutions of (3.5). In practice the method gives all solutions but does not give a proof of the nonexistence of solutions with |y| ≥ C. Our experiences show that Thue equations usually have only small solutions. Therefore this method gives all solutions with a high

3.4 Binomial Thue Equations

27

probability, surely those solutions that are applicable in practice. Note that some properties of continues fractions used in the following can be found for example in Peth˝o [Pe99], Sect. 5.3. Observe that the inequalities (3.3) and (3.4) remain valid also in the present situation. If |y| is not very small, then (3.4) implies    (i) x  α −  < c1 < 1 .  y  |y|n 2|y|2 If α (i) is real (otherwise the solutions are easy to determine by considering the imaginary part of (x − α (i) y)), Legendre’s theorem implies that x/y is a convergent pk /qk in the continued fraction algorithm of α (i) . Using another well-known inequality, if ak is the corresponding partial quotient, then    (i) pk  1 ,  < α − qk  (ak+1 + 2)qk2 that is

   (i) pk  1 1  < c1 ,  ≤ < α − qk  |qk |n (A + 2)qk2 (ak+1 + 2)qk2

with A = max1≤k≤k0 ak , where k0 is the first index for which qk0 > C. Hence the above inequality implies |y| = qk < (c1 (A + 2))1/(n−2) .

(3.6)

Summarizing: |y| is either small, or x/y is a convergent in the continued fraction expansion to a real conjugate of α with (3.6). Thus in order to solve (3.5) we have to test small values of y, calculate the maximum of the indicated partial quotients to the real conjugates of α, and test the convergents satisfying (3.6). Inequality (3.6) implies also a kind of reduction of the bound C: having the bound in (3.6) we can repeat the calculation using the bound in (3.6) instead of C. Remember to perform the algorithm for all real roots α (i) .

3.4 Binomial Thue Equations 3.4.1 Results on the Solutions of Binomial Thue Equations Let m be a given positive integer and let n ≥ 3 be a given exponent. Consider the binomial Thue equations of type x n − my n = ±1 in x, y ∈ Z.

(3.7)

28

3 Thue Equations

Here we recall some important results on the solutions of Eq. (3.7) that we shall use in the sequel. If we know a positive solution of the binomial Thue equation, then the following result of Bennett [Be01] is useful to exclude any other positive solutions. Lemma 3.3 If , m are integers with m = 0 and n ≥ 3, then the equation x n − my n = ±1 (in x, y ∈ Z) has at most one solution in positive integers (x, y). The following theorem of Gy˝ory and Pintér [GyP12], describes all solutions of binomial Thue equation if its coefficients are composed of at most two primes less than or equal to 29. Lemma 3.4 If  < m are positive integers composed of at most two primes p, q with 2 ≤ p, q ≤ 29 and n ≥ 3, then the only solutions of the equation x n − my n = ±1 (in x, y ∈ Z) are those with  ∈ {1, 2, 3, 4, 7, 8, 16}, x = y = 1, and n = 3, (, x) = (1, 2), (1, 3), (1, 4), (1, 8), (1, 9), (1, 18), (1, 19), (1, 23), (2, 2), (3, 2), (5, 3), (5, 11), (11, 6), n = 4, (, x) = (1, 2), (1, 3), (1, 5), (3, 2), n = 5, (, x) = (1, 2), (1, 3), (9, 2), n = 6, (, x) = 1, 2).

Below the result of Bazsó et al. [BBGyP10] describes the solutions of binomial Thue equations for small values of m with  = 1. For simplicity we quote the case n = 4 only, what we shall use in Sect. 9.6. That is we consider equation x 4 − my 4 = 1 in x, y ∈ Z.

(3.8)

3.4 Binomial Thue Equations

29

Lemma 3.5 If 1 < m ≤ 400, then all positive integer solutions (x, y), with |xy| > 1 of Eq. (3.8) are (m, x, y) = (5, 3, 2), (15, 2, 1), (17, 2, 1), (39, 5, 2), (80, 3, 1), (82, 3, 1), (150, 7, 2), (255, 4, 1), (257, 4, 1).

Lemma 3.6 If 400 < m ≤ 2000 is odd, then all positive integer solutions (x, y), with |xy| > 1 of Eq. (3.8) are (m, x, y) = (915, 11, 2), (1295, 6, 1), (1297, 6, 1), (1785, 13, 2).

Remark 3.7 The solutions with m = 82, 915, 1295 of Lemmas 3.5 and 3.6 are indicated in the Corrigendum [BBGyP15] to [BBGyP10].

3.4.2 Calculation of “Small” Solutions of Binomial Thue Equations The algorithm of Sect. 3.3 is especially efficient if we consider binomial Thue equations. As an application of the algorithm of Sect. 3.3, our purpose here is to determine the “small” solutions max(|x|, |y|) < 10500 of Eq. (3.7) for n = 3, 4, 5, 7, 11, 13, 17, 19, 23, 29 with 1 < m < 107 . We omit those values of m for which the left-hand side is reducible. Obviously it is enough to consider square free exponents. Theorem 3.8 All solutions of Eq. (3.7) with max(|x|, |y|) < 10500 , for n = 3, 4, 5, 7, 11, 13, 17, 19, 23, 29, 1 < m < 107 , assumed the left-hand side of the equation is irreducible, are those listed in the tables of Sect. 16.1. Here we detail our calculation. To perform this calculation for such a huge number of equations we used especially sharp estimates that we detail here, following the arguments of Gaál and Remete [GRe15].

30

3 Thue Equations

Let ζ = exp(2π i/n). Let x, y be and arbitrary solution of (3.7). Set √ βj = x − ζ j −1 n m y for j = 1, . . . , n, then Eq. (3.7) can be written as β1 . . . βn = ±1.

(3.9)

We may assume y ≥ 0 since on the right side we have ±1 in our equation. Also, for y = 0 we only have the trivial solution x = ±1. Therefore in the following let y ≥ 1. For n = 4 the β2 , . . . , βn are complex, therefore √ √ |βj | = |x − ζ j −1 n m y| ≥ |Im(ζ j −1 )| n m y for j = 1, . . . , n. This yields that |β1 | = |x −

√ n

m y| ≤

1 , c1 y n−1

that is    √  n m− x ≤ 1 ,  y  c1 y n

(3.10)

 √ with c1 = ( n m)n−1 c2 , where c2 = nj=2 |Im(ζ j −1 )|. Obviously (x, y) = 1. If y > (2c1 )1/(n−2) ,

(3.11)

then (3.10) implies    √  n m− x < 1  y  2 y2 √ and applying Legendre’s theorem we confer that x/y is a convergent pj /qj to n m, that is x = pj , y = qj . For the corresponding partial quotients aj it is well known that    √ 1  n m − x. <  2 y (aj +1 + 2)qj

3.4 Binomial Thue Equations

31

Let A = max1≤i≤s ai where qs is the first denominator of a partial quotient exceeding C. Combining the above estimate with (3.10) we get    √ 1  n m− x ≤ 1 , <  2 y  c1 qjn (A + 2)qj whence we obtain y = qj < c3 = (c1 (A + 2))n−2 . We calculate all denominators of partial quotients up to C, we take their maximum, calculate the above bound, and check all possible x = pj , y = qj for qj < c3 running again the continued fraction algorithm. Now we return to the validity of (3.11). Simple calculation shows that (3.11) is satisfied if  y>

2 c2



1 n−2

n−1

m− n(n−2) .

(3.12)

If m is large enough, then condition (3.12) is satisfied for y ≥ 1. Some small values of y must be tested for the following values of m: n 3 5 7 11 13 17 19 23 29

2≤m≤4 2 ≤ m ≤ 10 2 ≤ m ≤ 29 2 ≤ m ≤ 314 2 ≤ m ≤ 1078 2 ≤ m ≤ 13489 2 ≤ m ≤ 48699 2 ≤ m ≤ 652798 2 ≤ m ≤ 7960210

For all these values of m we have to test all y with  y≤

2 c2



1 n−2

n−1

m− n(n−2) .

Easy calculation shows that this merely yields testing y = 1 for the values of m contained in the table. This can be done very fast.

32

3 Thue Equations

Remark 3.9 (The Case n = 4) The case n = 4 was considered in Gaál and Remete [GRe14]. Remark that in that case we may assume x > 0, y > 0. We have √ √ √ |x − 4 my| ≤ 1, |x ± i 4 my| ≥ 4 my, therefore |x + whence

√ 4

√ √ my| ≥ 2 4 my − 1 ≥ 4 my,

  √  1 1  4 m − x  ≤ √1 < 2,   4 4 3 y 2y ( m) y

where the last inequality is valid for all y ≥ 1.

3.4.3 Computational Aspects We were executing the algorithm of Sect. 3.3 with C = 10500 for n = 3, 4, 5, 7, 11, 13, 17, 19, 23, 29 using the above sharp estimates that made the procedure for binomial Thue equations much more efficient. This efficient algorithm allowed us to perform the calculations for all those 2 ≤ m ≤ 107 for which the left side of (3.7) is irreducible. The procedure was implemented in Maple [CG88], and we used 1200 digits accuracy. For each exponent this calculation involved almost 107 binomial Thue equations. The routines were running in 2013 on the supercomputer (high performance computer) network situated in Debrecen-Budapest-Pécs-Szeged in Hungary under Linux. For each exponent the total running time was about 120–200 h calculated for a single node which yields a few hours using parallel computing on a couple of nodes.

3.5 Effective Methods for Thue Equations Return now to Eq. (3.1). The first effective bounds for the solutions of Thue equations were given by Baker (see [Ba90]), and the best known bounds are due to Bugeaud and Gy˝ory [BGy96b]. Let η1 , . . . , ηr be a set of fundamental units of K = Q(α) and determine a full set of nonassociated elements μ of norm ±m in ZK . These data can be computed by Kash [DF97] or Magma [BCP97]. The following must be performed for all μ in this finite set (usually there are only some candidates). We have β = ξ · μ · η1a1 · · · ηrar

(3.13)

3.5 Effective Methods for Thue Equations

33

with a root of unity ξ and a1 , . . . , ar ∈ Z with A = max(|a1 |, . . . , |ar |). Taking conjugates, absolute values, and logarithms in (3.13), we get a system of linear equations in a1 , . . . , ar ,    β (j )         (j )   (j )  a1 log η1  + . . . + ar log ηr  = log  (j )  , μ 

(3.14)

where the conjugate j runs through all nonequivalent embeddings of K into C excluding the i-th conjugate. By using Cramer’s rule to solve the above system of linear equations we obtain      β (j )     A ≤ c2 max log  (j )  ≤ c3 log |y|,   μ 

(3.15)

(j )

where c2 is the row norm of the inverse matrix of (log |ηk |) that is the maximum sum of the absolute values of the elements in a row, and c3 = 2c2 , assuming |y| is large enough. Take indices j, k such that i, j, k are distinct, then we have the identity (α (i) − α (j ) )β (k) + (α (j ) − α (k) )β (i) + (α (k) − α (i) )β (j ) = 0. This is a crucial step in the application of Baker’s method (see [Ba90]). The linear combination of three conjugates of β is zero. The conjugates of β can be represented by (3.13) as a power product of certain algebraic numbers with the unknown exponents a1 , . . . , ar . Similar identities are called Siegel’s identity. We obtain (α (i) − α (j ) )β (k) (α (k) − α (j ) )β (i) − 1 = . (α (i) − α (k) )β (j ) (α (i) − α (k) )β (j )

(3.16)

Using the representation (3.13) and the estimates (3.3), (3.4), and (3.15) the above equation implies          (α (i) − α (j ) )μ(k)   η(k)   η(k)      1   r  Λ = log  (i)  + a1 log  (j )  + . . . + ar log  (j )    (α − α (k) )μ(j )  η   η r  1       (α (i) − α (j ) )β (k)   (α (i) − α (j ) )β (k)        ≤ 2  (i) ≤ 2 − 1 − 1      (α − α (k) )β (j )   (α (i) − α (k) )β (j )      (α (k) − α (j ) )β (i)    = 2  (i)  ≤ c4 |y|−n  (α − α (k) )β (j )  = c4 exp(−n log |y|) ≤ c4 exp(−c5 A).

(3.17)

34

3 Thue Equations

Here we used the elementary inequality | log x| < 2|x − 1|, holding for complex numbers with |x − 1| < 0.795, which is ensured if A is not very small,     α (j ) − α (k) n   c4 = 4c1  (i)  , c5 = . (k) (i) (j )  (α − α )(α − α )  c3 If the terms in the above linear form in the logarithm of certain algebraic numbers are linearly independent over Q, then using Theorem 2.1 of Baker and Wüstholz we can give a lower bound of the form exp(−C log A) < Λ for Λ with a huge positive constant C. Comparing it with (3.17) we get an upper bound A0 for A. This bound is very large, about 1018 even for the simplest cubic equations, 1035 for quartic equations etc. By (3.17), namely from          (α (i) − α (j ) )μ(k)   η(k)   η(k)      1   r  log  (i)  + a1 log  (j )  + . . . + ar log  (j )  ≤ c4 exp(−c5 A)   (α − α (k) )μ(j )  η   η r  1

we can derive an inequality of type (2.4). Using Lemma 2.4 we can reduce the large upper bound for A. The reduced bound AR for A is usually under 100 for small units ranks (for example, 2,3) and is under 1000 for larger unit ranks. Having the reduced bound for A we can either derive an upper bound for |y| by (3.13) and use the method of Sect. 3.3 to find the “small” solutions, or by (3.16) we get the unit equation   a1   (i)  (k) ar α − α (j ) ξ (k) μ(k) η1(k) ηr   ... (j ) α (i) − α (k) ξ (j ) μ(j ) η(j ) ηr 1  (j ) ar  a1   (j ) α − α (k) ξ (i) μ(i) η1(i) ηr  +  (i) ... = 1, (j ) (j ) (k) (j ) (j ) α −α ξ μ η1 ηr to which the enumeration methods of Sect. 2.3 can be applied. In the later case (3.13) enables one to calculate x, y from the exponents a1 , . . . , ar .

3.6 The Method of Bilu and Hanrot The most efficient method for solving  Thue equations is due to Bilu and Hanrot   (j )  [BH96]. If (ukj ) is the inverse matrix of log ηk  used above, then (3.14) implies

3.7 Experiences by Solving a Thue Equation with “Large” Entries

ak =

r  j =1

35

    r  β (j )    x − α (j ) y      ukj log  (j )  = ukj log   μ   μ(j )  j =1

  ⎛ ⎞  α (j ) − x    r r r     α (i) − α (j )    y     =⎝ ukj ⎠ log |y| + ukj log  (i) ukj log  + ,  α − α (j )   μ(j )  j =1 j =1   j =1 (where in the summation j runs through all nonequivalent embeddings of K into C excluding the i-th one) which in view of      α (i) − x  α (j ) − x          (i) x  2 y  y     log  (i) α −  , − 1 < (i)  = log  (i)  α − α (j )  α − α (j )   |α − α (j ) |  y     (3.4) and (3.15) imply |δk log |y| − ak + νk | < c6 exp(−c5 A) with

    r  α (i) − α (j )     δk = |ukj |, νk =  ukj log   , (j )   μ j =1 j =1 r 

 r     ukj   c6 = 2c1 ·  α (i) − α (j )  . j =1

Taking any two of these inequalities (say for k, l) and eliminating the term with log |y| we obtain an inequality of the form |ak ϑ + al − δ| < c7 exp(−c5 A) ≤ c7 exp(−c5 max(|ak |, |al |)) with ϑ =−

δk νl − δl νk c6 (|δk | + |δl |) δl . , δ= , c7 = δk δk |δk |

To this inequality the Davenport lemma (see Lemma 2.3) can be used. Thus both the reduction and enumeration processes become very easy. This enables one to solve Thue equations of very high degree, as well.

3.7 Experiences by Solving a Thue Equation with “Large” Entries In connection with Mordell equations Gaál et al. [GPP18] considered the cubic Thue equation x 3 − 206527662xy 2 + 845993618459y 3 = ±38557350 in x, y ∈ Z.

(3.18)

36

3 Thue Equations

The equation seems to be usual, maybe with somewhat larger coefficients and right-hand side. In the following we give an impression what kind of difficulties may appear, and how they can be settled using the well-known machinery. The fundamental units and the algebraic integer elements of given norm are usually input data for our computations. However, calculating them may cause the biggest problem. Having them explicitly, the reduction method works efficiently, even with extreme input data. Denote by K the number field generated over Q by ϑ with defining polynomial f (x) = x 3 − 206527662x + 845993618459. This is a totally real cubic field with integral basis   ϑ 2 + 118ϑ + 37 1, ϑ, . 2535 The fundamental units η1 , η2 were calculated by Magma [BCP97], they had coefficients of about 1600 digits in the above integral basis (we cannot explicitly give them here for brevity). Their conjugates were of magnitude up to 102200 . These data shifted the calculation into a new dimension. However, we followed the usual methods. The algebraic integer elements of K of given norm 38557350 = ±2 · 33 · 52 · 134 could not be computed with Magma. They were found by direct calculations on ideals. We found that there are altogether 32 principal ideals of that norm. Their conjugates were of magnitude up to 102000 . We then used standard arguments for solving Eq. (3.18). The challenge was to overcome the difficulties caused by the magnitudes of the units and the elements of given norm. Equation (3.18) can be written as NK/Q (x1 − ϑy) = ±38557350, whence for any solution x1 , y of (3.18) x1 − ϑy = γ · η1a1 · η2a2 ,

(3.19)

where γ ∈ ZK is one of the algebraic integer elements of norm ±38557350 and a1 , a2 ∈ Z. Set A = max(|a1 |, |a2 |). The usual arguments led us to linear forms in the logarithms of algebraic numbers of type          (α (i) − α (j ) )γ (k)   η(k)   η(k)      1   2  Λ = log  (i)  + a1  (j )  + a2  (j )  ,   (α − α (k) )γ (j )  η   η  1

2

3.8 Further Results on Thue Equations

37

where 1 ≤ i, j, k ≤ 3 are distinct indices. For nonzero y we had the upper estimate Λ < 0.0004533 · exp(−7705.901 · A + 46078.962)

(3.20)

and using the well-known Theorem 2.1 of Baker and Wüstholz we obtained the lower estimate Λ > exp(5.62671 · 1026 · log A). This yields A < 1025 . So far everything seems to be usual. But we had to use an extremely high precision. The reduction lemma of Baker and Davenport [BD69], cf. Sect. 2.2.1 can be used to diminish this bound using the continued fraction algorithm. In one reduction step we got the reduced bound A ≤ 5. The possible exponents a1 , a2 below this bound were tested if there are corresponding x1 , y satisfying (3.19). There were no solutions, as we expected. To perform this calculation in 2018 we used Maple [CG88] with 15000 digits precision for the reduction and 18000 digits precision for testing the possible small exponents. Surprisingly, with these extreme data and very high precision the standard algorithms worked well, the reduction was efficient and we got a final result. We had to perform the reduction procedure for all the 32 possible γ and for all possible i (for a given i the possible j and k play a symmetric role). The test of small exponents was performed for all possible γ and for −5 ≤ a1 , a2 ≤ 5. The total CPU time on an average laptop took about 10 min.

3.8 Further Results on Thue Equations Thue equations are discussed also by Tzanakis and de Weger [TW89], [TW92], de Weger [We89], Peth˝o [Pe99], Smart [Sm98], and Evertse and Gy˝ory [EGy15]. The program packages Kash [DF97] and Magma [BCP97] contain reliable routines for solving Thue equations. A delicate problem is to characterize all solutions of an infinite parametric family of Thue equations. The first result of this type was obtained by Thomas [Tho90], and then further families of Thue equations were considered among others by Mignotte and Tzanakis [MT91], Lettl and Peth˝o [LP95], Chen and Voutier [CV97], Lettl et al. [LPV98], Heuberger et al. [HPT98], and Hoshi [Ho12].

Chapter 4

Inhomogeneous Thue Equations

Abstract Let α be an algebraic integer of degree n ≥ 3, K = Q(α), and let 0 = m ∈ Z. In some applications for index form equations in sextic and octic fields (cf. Sects. 11.2.1, 11.2.2, and 14.2.3) we shall need to solve equations of type NK/Q (x + αy + λ) = m in x, y ∈ Z, λ ∈ ZK , where we assume that |λ| < X1−ζ , where X = max(|x|, |y|) and 0 < ζ < 1 is a given constant (|λ| denotes the size of λ, that is the maximum absolute value of its conjugates). Sprindžuk (J Number Theory 6:481–486, 1974) considered equations of this type. This equation might also be referred to as inhomogeneous Thue equation. Using Baker’s method he gave effective upper bounds for the solutions of the above equation. The variables x, y are called dominating variables, while λ is called non-dominating variable. In Gaál (Math Comput 51:359–373, 1988) we gave a numerical method of solving (4.1), which we briefly explain below. It is interesting to compare these arguments with the relevant arguments for Thue equations and to see how the standard estimates can be modified to cover the inhomogeneous case.

4.1 Elementary Estimates Let α be an algebraic integer of degree n ≥ 3, K = Q(α), and let 0 = m ∈ Z, 0 < ζ < 1. Following Sprindzuk [Sp74] consider the inhomogeneous Thue equation. NK/Q (x + αy + λ) = m in x, y ∈ Z, λ ∈ ZK ,

(4.1)

assuming that |λ| < X1−ζ , where X = max(|x|, |y|). Let (x, y, λ) be a fixed solution of Eq. (4.1). Set β = x + αy + λ. Denote by β (i) the conjugate of β with |β (i) | = min |β (j ) |. Obviously |β (i) | ≤

n

|m|.

(4.2)

© Springer Nature Switzerland AG 2019 I. Gaál, Diophantine Equations and Power Integral Bases, https://doi.org/10.1007/978-3-030-23865-0_4

39

40

4 Inhomogeneous Thue Equations

For j = i we have |β (j ) | ≥ |β (j ) − β (i) | − |β (i) | ≥ |α (j ) − α (i) ||y| − 2X1−ζ −

n

|m|.

On the other hand, by α (i) β (j ) − α (j ) β (i) = (α (i) − α (j ) )x + α (i) λ(j ) − α (j ) λ(i) , whence |β (j ) | ≥

 1  (i) (j ) (i) (j ) 1−ζ (j ) n − α ||x| − (|α | + |α |)X − |α | |m| . |α |α (i) |

The latter two inequalities imply |β (j ) | ≥ c1 X − c2 X1−ζ − c3 > c4 X

(4.3)

if X is large enough (the opposite case is easy to consider), where  |α (i) − α (j ) | c1 = min |α − α |, , |α (i) |   |α (i) | + |α (j ) | c2 = max 2, , |α (i) |   |α (j ) | n c3 = |m| max 1, (i) , |α | 

(j )

c4 =

(i)

c1 , 2

the constants c1 , c2 , c3 , c4 depending of course on i, j . From inequality (4.3) we confer |m| < c5 X1−n (j ) | |β j =i

|β (i) | < 

(4.4)

with |m| . j =i c4 (i, j )

c5 = 

Let η1 , . . . , ηr be the fundamental units of K = Q(α) and consider a full set of nonassociated elements μ of norm ±m in ZK . These data can be computed for

4.2 Baker’s Method

41

example by Kash [DF97] or Magma [BCP97]. The following must be performed for all μ in this finite set. We have β = ξ μ η1a1 · · · ηrar

(4.5)

with a root of unity ξ and a1 , . . . , ar ∈ Z with A = max(|a1 |, . . . , |ar |). Similarly as for Thue equations this equation implies    β (j )    A ≤ c6 max  (j )  ≤ c7 log X, μ 

(4.6) (j )

where c6 is the row norm of the inverse matrix of (log |ηk |) used also in the preceding sections, c7 = 2c6 . We assume that X is large enough.

4.2 Baker’s Method Let i, j, k be distinct indices. Siegel’s identity gets the form (α (i) − α (j ) )(β (k) − λ(k) ) + (α (j ) − α (k) )(β (i) − λ(i) ) +(α (k) − α (i) )(β (j ) − λ(j ) ) = 0, whence (α (i) − α (j ) )β (k) (α (j ) − α (k) )β (i) + δ − 1 = (α (i) − α (k) )β (j ) (α (i) − α (k) )β (j )

(4.7)

with δ = (α (j ) − α (i) )λ(k) + (α (k) − α (j ) )λ(i) + (α (i) − α (k) )λ(j ) having absolute value |δ| < c8 X1−ζ , with c8 = 6|α|. In view of (4.3), (4.4), (4.6), and the last estimate, the right-hand side of Eq. (4.7) can be estimated from above by c9 X−ζ = c9 exp(−ζ log X) < c9 exp(−c10 A),

(4.8)

42

4 Inhomogeneous Thue Equations

with c9 =

|α (j ) − α (k) |c5 + c8 ζ , c10 = . c4 (i, j ) c7

Using the representation (4.5) and applying Theorem 2.1 of Baker and Wüstholz the left-hand side of (4.7) can be estimated from below similarly as for Thue equations by     1   α (i) − α (j ) μ(k)  1  exp(−C log A) < log   (i)  2 2   α − α (k) μ(j )       η(k)   η(k)   1   r  (4.9) +a1 log  (j )  + . . . + ar log  (j )  , η   η r  1

with a large constant C. Comparing (4.8) and (4.9) we get an upper bound A0 for A.

4.3 Reduction, Test The reduction of the bound A0 is based (similarly as for Thue equations) on the upper estimate          α (i) − α (j )  μ(k)   η(k)   η(k)      1   r    + a1 log  (j )  + . . . + ar log  (j )  log   (i)   α − α (k) μ(j )  η   ηr  1

< 2c9 exp(−c10 A), using Lemma 2.4. The possible values of a1 , . . . , ar with absolute values under the reduced bound must be tested directly. For each set of exponents we construct β by (4.5). The corresponding solutions x, y, λ = β − x − αy can be found by solving inequalities. For example, if K is totally real and |y| ≥ |x|, then we must have −|y|1−ζ < β (j ) − x − α (j ) y < |y|1−ζ , for j = 1, . . . , n, that is β (j ) − α (j ) y − |y|1−ζ < x < β (j ) − α (j ) y + |y|1−ζ ,

(4.10)

that means, we only have to consider the values y for which max(β (j ) − α (j ) y − |y|1−ζ ) < min(β (j ) − α (j ) y + |y|1−ζ ). j

j

For the possible y we can use (4.10) to find the corresponding values of x. The case |y| ≤ |x| is similar to consider. An example with a totally real cubic α and ζ = 1/2 is detailed in [Ga88].

4.4 An Analogue of the Bilu-Hanrot Method

43

4.4 An Analogue of the Bilu-Hanrot Method If instead of |λ| < X1−ζ we assume the stronger condition |λ| < |y|1−ζ , then we can use an analogue of the Bilu-Hanrot method. Observe that in our applications in Sects. 11.2.1, 11.2.2, and 14.2.3 this is sufficient, since in those cases the size of λ is bounded by absolute constants. Using (4.2) we can bound x by y: |x| ≤ |β (i) | + |α (i) ||y| + |λ(i) | < c11 |y|. (The positive constants c11 , c12 , . . . can be easily calculated.) Similarly as above we have |β (j ) | > c12 |y| for j = i and |β (i) | < c13 |y|1−n . The estimates (4.6) and (4.8) remain valid similarly, but with |y| instead of X, (4.9) is also valid and we can derive an upper bound for A. If (ukj ) is the inverse matrix (j ) of (log |ηk |), then (4.5) implies ak =

r  j =1

⎛ =⎝

   β (j )    ukj log  (j )  μ 

r  j =1

⎞ ukj ⎠ log |y| +

r  j =1

    β (j )   ukj log    y(α (i) − α (j ) ) 

  r  α (i) − α (j )     + ukj log  .  μ(j )  j =1

We have       β (i) + (α (j ) − α (i) )y + (λ(j ) − λ(i) )   β (j )     log   = log    y(α (i) − α (j ) )    y(α (i) − α (j ) )    β (i) + (λ(j ) − λ(i) )    = log  − 1   y(α (i) − α (j ) )     β (i) + (λ(j ) − λ(i) )    ≤ 2   y(α (i) − α (j ) )  < c14 |y|−ζ < c14 exp(−c15 A).

(4.11)

44

4 Inhomogeneous Thue Equations

Hence by (4.11) we obtain inequalities of type |δk log |y| − ak + νk | < c16 exp(−c15 A) with δk and νk that can be easily calculated. Taking any two of these inequalities and eliminating the term with log |y| we obtain again an inequality of the form |ak ϑ + al − δ| < c17 exp(−c15 A) ≤ c17 exp(−c15 max(|ak |, |al |)) to which the Davenport lemma (Lemma 2.3) can be used. This makes both the reduction and the enumeration processes much faster.

Chapter 5

Relative Thue Equations

Abstract Let M ⊂ K be algebraic number fields, and let K = M(α) with an algebraic integer α. Let 0 = μ ∈ ZM . Consider the relative Thue equation NK/M (X − αY ) = μ in X, Y ∈ ZM . Equations of this type were first considered in effective form by Kotov and Sprindzuk (Dokl Akad Nauk BSSR 17:393–395, 477, 1973). This equation is a direct analogue of (3.1) in the relative case, when the ground ring is ZM instead of Z. The equation given in this form has only finitely many solutions. Relative Thue equations are often considered in the form NK/M (X − αY ) = ημ in X, Y ∈ ZM , where η is an unknown unit in M. In this case the solutions are determined up to a unit factor of M. In our applications index form equations can often be reduced to relative Thue equations for example for sextic fields (Sect. 11.2), octic fields (Sect. 14.2), nonic fields (Sect. 13.2), and for relative cubic (Sect. 13.1) and relative quartic extensions (Sect. 14.1).

5.1 A Fast Algorithm for Finding “Small” Solutions of Relative Thue Equations In this section we give an analogue in the relative case of the algorithm detailed in Sect. 3.3. We follow the arguments of Gaál [Ga15]. Let M ⊂ K be algebraic number fields with m = [M : Q] and n = [K : M] ≥ 3. Let K = M(α) with an algebraic integer α. Let 0 = μ ∈ ZM . Following Kotov and Sprintzuk [KS73] consider the relative Thue equation NK/M (X − αY ) = μ in X, Y ∈ ZM .

© Springer Nature Switzerland AG 2019 I. Gaál, Diophantine Equations and Power Integral Bases, https://doi.org/10.1007/978-3-030-23865-0_5

(5.1)

45

46

5 Relative Thue Equations

Let (1, ω2 , . . . , ωn ) be an integral basis of M. We represent X and Y in the form X = x1 + ω2 x2 + . . . + ωm xm , Y = y1 + ω2 y2 + . . . + ωm ym , with xi , yi ∈ Z (1 ≤ i ≤ m). We set A = max(max |xi |, max |yi |). Our purpose is to determine all solutions of (5.1) with |Y | < C, where C is a large given constant, say 10100 or 10500 . (|Y | denotes the size of Y , that is the maximum absolute value of its conjugates.) Our experience shows that these equations usually have only small solutions, therefore this fast algorithm gives all solutions with a high probability. In our algorithm we shall reduce the bound C in several steps and then enumerate and test the possible values of the solutions under the reduced bound. Denote by γ (j ) , (j = 1, . . . , m) the conjugates of any γ ∈ M. Denote by f (x) the relative defining polynomial of α over M. Denote by α (j k) , (k = 1, . . . , n) the relative conjugates of α over M (j ) , that is the roots of the j -th conjugate of f (x). We also denote by γ (j k) the conjugates of any γ ∈ K corresponding to α (j k) . We assume that n > m if K is not totally real and n > 2m if K is totally real. This condition ensures that our reduction procedure is efficient. Our algorithm is based on the well-known fact that for any fixed solution X, Y ∈ ZM of Eq. (5.1), if Y is large enough, then there is a conjugate of β = X − αY which is very small. We start with two simple lemmas. First we give an upper bound for A in terms of |Y |, then we obtain the crucial inequality (5.9) that we shall use to reduce the bound for A. Let chj i =

 1  (hj )  − α (hi) ) , for 1 ≤ h ≤ m, 1 ≤ i, j ≤ n, i = j, (α 2 chi = 

|μ(h) | 1≤j ≤n,j =i chj i

, for 1 ≤ h ≤ m, 1 ≤ i ≤ n, n

c1 = max h,j,i

|μ(h) | , chj i

where the maximum is taken for 1 ≤ h ≤ m, 1 ≤ i, j ≤ n, i = j . (j ) (j ) Let S be the m × m matrix with entries 1, ω2 , . . . , ωm in the j -th row. Denote by c2 the row norm of S −1 that is the maximum sum of the absolute values of the elements in its rows.

5.1 A Fast Algorithm for Finding “Small” Solutions of Relative Thue Equations

47

Let c3 =

|μ|

1/n

, c4 = max(c1 , c3 ), c5 = 2c2 |α|

|α|

and dhi = chi c5n−1 for 1 ≤ h ≤ m, 1 ≤ i ≤ n. Finally, let C be a given constant. Lemma 5.1 If (X, Y ) ∈ Z2M is a solution of Eq. (5.1) with |Y | > c3 , then A ≤ c5 · |Y |.

(5.2)

Proof Set β = X −αY . For any k let  be the index with |β (k) | = min1≤j ≤n |β (kj ) |. Equation (5.1) implies β (k1) . . . β (kn) = μ(k) . Therefore |β (k) | ≤

n

|μ(k) |, and we obtain 

|X(k) | ≤ |β (k) | + |α (k) | · |Y (k) | ≤

n

|μ| + |α| · |Y |,

whence using |Y | > c3 we get |X| ≤ 2 |α| · |Y |.

(5.3)

By ⎛

⎞ ⎛ (1) ⎞ y1 Y ⎜ .. ⎟ −1 ⎜ .. ⎟ ⎝ . ⎠=S ⎝ . ⎠ yn

Y (m)

we obtain max |yj | ≤ c2 |Y |. Similarly we have max |xj | ≤ c2 |X|, whence by (5.3) we get the assertion (5.2).   Lemma 5.2 If (X, Y ) ∈ Z2M is a solution of Eq. (5.1) with |Y | > c4 , then there exist h, i (1 ≤ h ≤ m, 1 ≤ i ≤ n), such that |β (hi) | ≤ dhi A1−n .

(5.4)

48

5 Relative Thue Equations

Proof Let Y (h) be the conjugate of Y with |Y | = |Y (h) | and let i be determined by |β (hi) | = min |β (hj ) |. 1≤j ≤n

Obviously |β (hi) | ≤

 n |μ(h) |

(5.5)

and for any j = i (1 ≤ j ≤ n) using |Y | > c1 we have |β

(hj )

| ≥ |β

(hj )

−β

(hi)

| − |β

(hi)

| ≥ |(α

(hj )

−α

(hi)

)Y

(h)

|−

 n

|μ(h) | ≥ chj i |Y |. (5.6)

By Eq. (5.1) we get β (h1) . . . β (hn) = μ(h) , whence |β (hi) | ≤ chi · |Y |

1−n

.

By |Y | > c3 Lemma 5.1 applies, therefore (5.2) is satisfied, whence |Y | c5 A−1 , that is |Y |

1−n

≤ c5n−1 A1−n .

(5.7) −1

≤

(5.8)  

Therefore by inequality (5.7) we obtain the assertion (5.4).

5.1.1 The Reduction Procedure In this section we describe the algorithm to reduce the bound C. Our procedure to reduce the bound A < C for A is based on the inequality (h) (h) xm − α (hi) y1 − α (hi) ω2(h) y2 − . . . − α (hi) ωm ym | ≤ |x1 + ω2(h) x2 + . . . + ωm

≤ dhi A1−n ,

(5.9)

which, in view of Lemma 5.2, is satisfied for certain indices i, h with 1 ≤ h ≤ m, 1 ≤ i ≤ n, assumed |Y | > c4 . Our statement is an analogue of Lemma 2.4, applying the extension of Pohst [Po93] of the standard LLL algorithm of Lenstra et al. [LLL82].

5.1 A Fast Algorithm for Finding “Small” Solutions of Relative Thue Equations

49

Let H be a large constant to be given later. Consider now the lattice L generated by the columns of the matrix ⎛

1 0 ⎜ 1 ⎜0 ⎜ . .. ⎜ . ⎜ . . ⎜ ⎜0 0 ⎜ ⎜0 0 ⎜ ⎜ ⎜0 0 ⎜ .. ⎜ .. ⎜ . . ⎜ ⎜0 0 ⎜ ⎜ (h) ⎝ H H Re(ω2 ) (h) H H Im(ω2 )

... 0 0 0 ... 0 0 0 .. .. .. .. . . . . 0 0 ... 1 ... 0 1 0 ... 0 0 1 .. .. .. .. . . . . 0 0 ... 0 (h) (h) . . . H Re(ωm ) H Re(α (hi) ) H Re(α (hi) ω2 ) (h) (h) (hi) (hi) . . . H Im(ωm ) H Im(α ) H Im(α ω2 )

⎞ ... 0 ⎟ ... 0 ⎟ ⎟ .. .. ⎟ . ⎟ . ⎟ ⎟ ... 0 ⎟ ⎟ ... 0 ⎟ ⎟ ⎟ ... 0 ⎟ .. ⎟ .. ⎟ . . ⎟ ⎟ ... 1 ⎟ (h) ⎟ (hi) . . . H Re(α ωm ) ⎠ (h) . . . H Im(α (hi) ωm )

In the totally real case we may omit the last row. Denote by b1 the first vector of an LLL-reduced basis of the lattice L . Lemma 5.3 Assume that x1 , . . . , xm , y1 , . . . , ym are integers with A = max(max |xi |, max |yi |), such that (5.9) is satisfied. If A ≤ A0 for some constant A0 and for the first vector b1 of the LLL reduced basis of L we have |b1 | ≥

(2m + 1)22m−1 · A0 ,

then  A≤

dhi H A0



1 n−1

.

(5.10)

Proof The proof is almost the same as in the proof of Lemma 2.4. Denote by l0 the shortest vector in the lattice L . Assume that the vector l is a linear combination of the lattice vectors with integer coefficients x1 , . . . , xm , y1 , . . . , ym , respectively. Observe that the last two components of l are the real and imaginary parts of β (hi) . Using the inequalities of [Po93] we have |b1 |2 ≤ 22m−1 |l0 |2 . Obviously |l0 | ≤ |l|. The first 2m components of l are in absolute value ≤ A0 , and for the last two components (5.9) is satisfied. Hence we obtain   (2m + 1)A20 = 21−2m (2m + 1) · 22m−1 A20 2 2−2n ≤ 21−2m |b1 |2 ≤ |l0 |2 ≤ |l|2 ≤ 2m · A20 + H 2 dhi A ,

50

5 Relative Thue Equations

whence A0 ≤ dhi H A1−n ,  

which implies the assertion.

Remark 5.4 We made several tests to figure out how one can suitably choose H , for which m and n the procedure is applicable and what is the magnitude of the reduced bound. We summarize our experiences in the following table:

Appropriate value for H

Complex case H = Am 0

Totally real case H = A2m 0

The procedure is efficient for

n>m

n > 2m

The reduced bound is of magnitude

m−1 n−1

A0

2m−1

A0n−1

This phenomenon can be detected in our examples, as well. We start with an initial bound A0 for A, obtained from Lemma 5.1 and |Y | < C, and perform the reduction. In the following steps A0 is the bound obtained in the previous reduction step. In the first reduction steps the new bound is drastically smaller than the original one. Applying Lemma 5.3 in several steps (usually 5–10 steps) the procedure stops by giving (almost) the same bound like the previous one. This reduced bound AR is usually between 10 and 500. Remark 5.5 As it is seen we use the same lattice and our reduction Lemma 5.3 is almost the same as Lemma 2.4. However, the approach and the way of application are completely different and just that is our main goal.

5.1.2 Enumerating Tiny Values of the Variables Let AR denote the reduced bound for A. If AR is small or if m = 2, then it is easy to test all possible Y with |y1 |, |y2 | ≤ AR and using equation n 

(X(h) − α (hj ) Y (h) ) = μ(h)

(5.11)

j =1

(valid for h = 1, 2) we can calculate the values of X(h) corresponding to Y (h) which makes possible to determine x1 , x2 . For m ≥ 3 we proceed as follows. We take a rather small initial value AI (say 10 or 20) such that the above direct procedure can be performed to test y1 , . . . , ym with absolute values ≤ AI within feasible CPU time. Then we only have to consider values of y1 , . . . , ym with max |yj | > AI yielding A = max(max |xj |, max |yj |) > AI .

5.1 A Fast Algorithm for Finding “Small” Solutions of Relative Thue Equations

51

In some steps we construct intervals [As , AS ] the union of which covers the whole interval [AI , AR ]. This means that in the first step we take As = AI and an AS with As ≤ AS ≤ AR . In the following step we set As to be the former AS and take a new AS , etc. We describe now an efficient method to enumerate the variables with As ≤ A ≤ AS . For a given h and i specified in the reduction procedure (1 ≤ h ≤ m, 1 ≤ i ≤ n) by (2.4) we have (h)

(h)

(h) (h) |x1 + ω2 x2 + . . . + ωm xm − α (hi) y1 − α (hi) ω2 y2 − . . . − α (hi) ωm ym | ≤

≤ dhi A1−n ≤ dhi A1−n . s /dhi , then We take H = AS · An−1 s (h)

(h)

(h) (h) H · |x1 + ω2 x2 + . . . + ωm xm − α (hi) y1 − α (hi) ω2 y2 − . . . − α (hi) ωm ym | ≤

≤ AS .

(5.12)

Denote by e1 , . . . , em , f1 , . . . , fm the columns of the matrix defining the lattice L in the preceding section. Using the above H implies that all coordinates of x1 e1 + . . . + xm em + y1 f1 + · · · +m fm are less than or equal to AS yielding |x1 e1 + . . . + xm em + y1 f1 + · · · + ym fm |2 ≤ (n + 1)A2S .

(5.13)

(In the totally real case we omit the last row.) This defines an ellipsoid. The integer points can be enumerated by using the Cholesky decomposition (see Pohst [Po93], Pohst and Zassenhaus [PZ89]). This means to construct an upper triangular matrix R = (rij ) with positive diagonal entries, such that the symmetric matrix of the above quadratic form is written as R T R, that is (denoting here y1 , . . . , ym by xm+1 , . . . , x2m , for simplicity) (5.13) gets the form 2m  i=1

⎛ ⎝rii xi +

2m 

⎞2 rij xj ⎠ ≤ (n + 1)A2S .

j =i+1

By enumerating x2m , x2m−1 , . . . etc. we also use that fact that −AS ≤ xi ≤ AS (1 ≤ i ≤ 2m). Note that the Cholesky decomposition can be improved by using the Fincke-Pohst method [FP85] (see also Pohst [Po93], Pohst and Zassenhaus [PZ89]), involving LLL reduction, but then we loose the above bounds for the xi .

52

5 Relative Thue Equations

5.1.3 The Complete Algorithm In this section we construct the algorithm using the components of the preceding sections. Problem Determine all solutions X, Y ∈ ZM with |Y | < C of the equation NK/M (X − αY ) = μ. We assume that n > m if K is not totally real and n > 2m if K is totally real. Step 1 Calculate the constants chj i , chi (1 ≤ h ≤ m, 1 ≤ i, j ≤ n, i = j ) and c1 , c2 , c3 , c4 , c5 and dhi (1 ≤ h ≤ m, 1 ≤ i ≤ n). Step 2 Set AB = c5 · max(C, c4 ). (This is the initial upper bound for A.) Step 3 For h ∈ {1, . . . , m} and i ∈ {1, . . . , n} perform the reduction procedure. In the first step take A0 = AB , choose a suitable H , perform the LLL basis reduction, and calculate the reduced bound A1 by (5.10). In the next step take A0 = A1 and perform the reduction again. Continue until the reduced bound is not anymore considerably less than the previous bound. Denote by AR,h,i the final reduced bound. Step 4 Set AR = maxh,i AR,h,i . Step 5 If c5 c4 > AR , then set AR = c5 c4 . (Our arguments are only valid for |Y | > c4 that is A > c5 c4 .) Step 6 Enumerate the tiny values of x1 , . . . , xm , y1 , . . . , ym with A ≤ AR using the procedure described above in Sect. 5.1.2. Test all possible vectors by substituting them into the equation.

5.1.4 Computational Aspects All our algorithms were developed in 2014 in Maple [CG88] under Linux and the execution times of our examples refer to a middle category laptop. However, especially at the final enumeration, to find appropriate values As , AS we made several test runs on the supercomputer located in Debrecen, Hungary. The HPC running times were 20–50% shorter even on a single node.

5.1 A Fast Algorithm for Finding “Small” Solutions of Relative Thue Equations

53

5.1.5 Examples Example 1 Let M = Q(i) with integral basis (1, i). Let α be a root of f (x) = x 6 + x + 1 and let K = M(α). Determine all X, Y ∈ ZM with |Y | < 10500 satisfying NK/M (X − αY ) = X6 + XY 5 + Y 6 = 1.

(5.14)

We get A ≤ 0.2252 · 10501 = AB . The reduction process ran as follows: Step 1 2 3 4 5 6

A0 0.2252 · 10501 0.9637 · 10101 0.1809 · 1022 159562 103 17

H 101003 10205 1045 1013 106 105

||b1 || ≥ 0.1424 · 10501 0.6095 · 10102 0.1144 · 1023 0.1009 · 107 651.4291 107.5174

Digits 1150 250 70 30 20 20

New A0 0.9637 · 10101 0.1809 · 1022 159562 103 17 16

CPU time 150 s 7s 2s 1s 1s 1s

The direct method to enumerate the variables with absolute values ≤ AR = 16 took 10 s. Finally all solutions (up to sign) are (x1 , x2 , y1 , y2 ) = (1, 0, 0, 0), (1, 0, −1, 0), (0, 0, 1, 0). √ √ Example 2 Let M = Q(i 2) with integral basis (1, i 2). Let α be a root of f (x) = x 21 − x − 1 and let K = M(α). Determine all X, Y ∈ ZM with |Y | < 10500 satisfying NK/M (X − αY ) = X21 − XY 20 − Y 21 = 1.

(5.15)

We get A ≤ 0.2068 · 10501 = AB . The reduction process ran as follows: Step 1 2 3

A0 0.2068 · 10501 0.5897 · 1026 99

H 101005 1055 107

||b1 || ≥ 0.1307 · 10502 0.3730 · 1027 626.1309

Digits 1200 80 20

New A0 0.5897 · 1026 99 6

CPU time 420 s 5s 2s

For a larger n the reduction is very efficient. In this example |Y | > c4 implies that the main arguments are only valid for |Y | > 9.9271, that is the variables with A ≤ 20 must be considered separately. Therefore we let the direct method run for AR = 20. This took about 5 min and resulted the solutions (x1 , x2 , y1 , y2 ) = (1, 0, 0, 0), (1, 0, −1, 0), (0, 0, −1, 0), (−1, 0, −1, 0).

54

5 Relative Thue Equations

√ Example 3 This √ is an example for the totally real case. Let M = Q( 2) with integral basis (1, 2). Let α be a root of f (x) = x 9 + 3x 8 − 5x 7 + 17x 6 + 7x 5 − 30x 4 − x 3 + 16x 2 − 2x − 1. This totally real nonic polynomial is taken from Voight [Vo08]. Let K = M(α). Determine all X, Y ∈ ZM with |Y | < 10500 satisfying NK/M (X − αY ) = X9 + 3X8 Y − 5X7 Y 2 + 17X6 Y 3 + 7X5 Y 4 − 30X4 Y 5 − X3 Y 6 + 16X2 Y 7 − 2XY 8 − Y 9 = 1.

(5.16)

We get A ≤ 0.5379 · 10501 = AB . The reduction process ran as follows: Step 1 2 3 4 5 6 7 8

A0 0.5379 · 10501 0.8862 · 10189 0.1110 · 1078 0.1439 · 1031 0.3304 · 1013 941870 2612 306

H 102007 10800 10313 10125 1055 1028 1018 1014

||b1 || ≥ 0.3402 · 10502 0.5604 · 10190 0.7022 · 1078 0.9103 · 1031 0.2089 · 1014 0.5956 · 107 16519.5940 1935.2970

Digits 2200 900 400 200 80 80 80 80

New A0 0.8862 · 10189 0.1110 · 1078 0.1439 · 1031 0.3304 · 1013 941870 2612 306 126

CPU time 840 s 120 s 60 s 30 s 30 s 30 s 30 s 30 s

The direct method with AR = 126 executed 21 min. The solutions are (x1 , x2 , y1 , y2 ) = (1, 0, 0, 0), (1, 0, −1, 0), (0, −1, −1, 0), (0, 0, −1, 0), (−1, 0, −1, 0), (0, 1, −1, 0). Example 4 Our last example demonstrates an equation with a cubic base field, m = 3. Let M = Q(ρ), where ρ is defined by the (totally real) polynomial x 3 −x 2 −3x + 1. The field M has integral basis (1, ρ, ρ 2 ). Let α be a root of the (totally complex) polynomial f (x) = x 6 +2x 5 +3x 4 +21. Let K = M(α). Determine all X, Y ∈ ZM with |Y | < 10500 satisfying NK/M (X − αY ) = X6 + 2X5 Y + 3X4 Y 2 + 21Y 6 = 1.

(5.17)

We get A ≤ 0.4268 · 10501 = AB . In this example we used the constants chi and dhi calculated for the given case h, i (while in all other examples we used values valid for all cases). This resulted a better reduction, giving a reduced bound about 15% sharper.

5.1 A Fast Algorithm for Finding “Small” Solutions of Relative Thue Equations

55

The reduction process ran as follows: Step 1 2 3 4 5 6 7 8 9 10

A0 0.4268 · 10501 0.9649 · 10203 0.8196 · 1083 0.8465 · 1035 0.5308 · 1016 0.9237 · 108 52170 2328 598 343

H 101513 10615 10255 10111 1053 1029 1019 5 · 1014 8 · 1012 4 · 1012

||b1 || ≥ 0.1277 · 10503 0.2888 · 10205 0.2453 · 1085 0.2534 · 1037 0.1589 · 1018 0.2764 · 1010 0.1561 · 107 69684.4896 17900.0536 10267.0876

Digits 1700 800 350 200 100 60 50 50 50 50

New A0 0.9649 · 10203 0.8196 · 1083 0.8465 · 1035 0.5308 · 1016 0.9237 · 108 52170 2328 598 343 334

CPU time 600 s 160 s 60 s 30 s 20 s 20 s 20 s 20 s 20 s 20 s

Note that in our example the direct method with AR = 334 yields to test (2 · AR + 1)m nm = (2 · 334 + 1)3 · 63 = 64.674.354.744 cases which is completely impossible. We executed the direct method with AI = 10 taking 7 min. To cover the interval [10,334] we applied the algorithm described for m ≥ 3 in several steps using 200 digits accuracy. It is worthy to choose As and AS with a relatively large difference so that H also becomes large and inequality (5.12) becomes an efficient filter. We performed four steps of the algorithm: Step 1 2 3 4

As 10 50 100 100

AS 50 100 150 334

CPU time 3 min 4 min 10 min 40 min

We had several test runs showing an optimal segmentation of the interval [10,334]. It turned out that the running time for the last step with AS = 334 is not significantly different with As = 100, 150, 200, 250. Therefore it was not worthy to split this interval into further parts. The only solution (up to sign) of Eq. (5.17) is (x1 , x2 , y1 , y2 ) = (1, 0, 0, 0).

56

5 Relative Thue Equations

5.2 Quartic Binomial Thue Equations over Imaginary Quadratic Fields √ Let d > 1 be a square free integer with −d ≡ 1(mod 4), set M √ = Q(i d). Then DM = −d and an integer basis of M is (1, ω) with ω = (1 + i d)/2. Let √ m be a square free positive integer with m = ±1 and m ≡ 2, 3 (mod 4), set ξ = 4 m. In Gaál et al. [GRSz14] we used the algorithm of Sect. 5.1 to solve relative binomial Thue equations over M of type X04 − mY04 = ζ in X0 , Y0 ∈ ZM ,

(5.18)

where ζ is a unit in ZM . We have considered d ∈ {3, 7, 11, 19, 43, 67, 163}. For all these values of d we calculated the solutions of Eq. (5.18) with max(|X0 |, |Y0 |) < 10250 for all m with 1 < m ≤ 5000, m ≡ 2, 3 (mod 4), and (d, m) = 1. In the following statements we list the solutions. √ Theorem 5.6 √ Let d be one of d ∈ {3, 7, 11, 19, 43, 67, 163}, let M = Q(i d), ω = (1 + i d)/2. For 1 < m ≤ 5000, m ≡ 2, 3(mod 4), (d, m) = 1, all solutions with max(|X0 |, |Y0 |) < 10250 of Eq. (5.18) are listed in the table of Sect. 16.2. Remark 5.7 In the application of Theorem 5.6 in Sect. 14.4 we needed unique factorization in M to derive Eq. (5.18). That is why we only consider the above values of d.

5.2.1 Specialties of the Actual Calculations The algorithm for calculating small solutions of relative Thue equations given in Sect. 5.1 consists of two parts. First the bound on the variables (in our case 10250 ) is reduced by using LLL reduction algorithm. In these examples the reduced bound was mostly between 10 and 200. In the second step the tiny solutions under the reduced bound (say 200 in our case) are enumerated. In our present calculation for relative binomial Thue equations we reorganized these two parts to make the procedure more efficient. We proceeded for the given values of d separately. The first part, the reduction was executed for all single m with the fixed d. (This yields about 3000 values of m allowed by the assumptions.) These (about 3000) reduction procedures were executed (all together) within about 3.2–3.6 h of CPU time for each d. The second part of the procedure described in Sect. 5.1, the enumeration of tiny values of the variables (with absolute values under the reduced bound, say 200 in our examples) was performed in a different way. We proceeded for all values of

5.3 Effective Methods for Relative Thue Equations

57

d separately. It turned out to be very CPU time consuming to test the tiny values of x1 , y1 , x2 , y2 for all m if they satisfy the relative Thue equation. On the other hand, we have run the cycles for all x1 , y1 , x2 , y2 under the reduced bound and determined those m for which they yield a solution. Using integer arithmetic and obvious refinements in the enumeration process, to test all x1 , y1 , x2 , y2 and to find the suitable values of m took about 3.5 h. The CPU times refer to an average laptop. The programs were developed in 2014 in Maple [CG88] and were executed under Linux. The same programs were also tested on the high performance computer (supercomputer) of the University of Debrecen where we obtained about 20% better results, calculated for a single processor node.

5.3 Effective Methods for Relative Thue Equations The enumeration method of Wildanger [Wild97], [Wild00] enabled us to construct a feasible algorithm for solving relative Thue equations (cf. Gaál and Pohst [GPo01]) which we detail here. Note that formerly de Weger [We95] and Smart [Sm97] solved some relative Thue equations by using sieve methods, hence their algorithm was not so efficient like ours. Let M ⊂ K be algebraic number fields with m = [M : Q] and n = [K : M] ≥ 3. The rings of integers of K, M will be denoted by ZK , ZM , respectively. Let α ∈ K be an integral generator of K over M, μ ∈ M an algebraic integer, and η an arbitrary unit in M. Consider the relative Thue equation NK/M (X − αY ) = ημ in X, Y ∈ ZM .

(5.19)

According to the effective results by Baker [Ba90], this equation has only finitely many solutions up to multiplication by units in M. We note that Baker’s result was generalized and extended by several authors (for further literature and the latest effective bounds for the sizes of the solutions of (5.19), see Bugeaud and Gy˝ory [BGy96b] and Evertse and Gy˝ory [EGy15].

5.3.1 Baker’s Method and Reduction Let η1 , . . . , ηs be a system of fundamental units in M. Extend this system to a maximal independent system η1 , . . . , ηs , ε1 , . . . , εr of units in K. Then any solution X, Y ∈ ZM of (5.19) can be written as X − αY = δ(η1 )b1 . . . (ηs )bs (ε1 )a1 . . . (εr )ar .

(5.20)

58

5 Relative Thue Equations

Here δ ∈ ZK is an integral element with relative norm μ. Up to unit factors in K there are only finitely many possibilities for δ, which can be determined using the Kash package [DF97], and the following procedure must be performed for each possible value of δ. Assume for simplicity that the possible roots of unity are also contained in δ. The exponents b1 , . . . , bs , a1 , . . . , ar in (5.20) are integers if the above system of independent units is a fundamental system of units. Otherwise, b1 , . . . , bs , a1 , . . . , ar can have a common denominator. In order to make our presentation simpler, we assume that the exponents are integral, otherwise the formulae must be modified in a straightforward way. Setting X =

(η1

)b 1

X , . . . (ηs )bs

Y =

(η1

)b 1

Y . . . (ηs )bs

yields X − αY  = δ(ε1 )a1 . . . (εr )ar .

(5.21)

We will just calculate a1 , . . . , ar , since the solutions of (5.19) are determined only up to unit factors of M. For any γ ∈ K we denote by γ (11) , . . . , γ (1n) , . . . , γ (m1) , . . . , γ (mn) the conjugates of γ , so that for 1 ≤ i ≤ m the elements γ (i1) , . . . , γ (in) are the corresponding relative conjugates of γ over the conjugate field M (i) of M. To simplify our notation, for any i (1 ≤ i ≤ m) and any distinct j1 , j2 , j3 (1 ≤ j1 , j2 , j3 ≤ n) we introduce a symbol I = (ij1 j2 j3 ) and set τ

(I )

=τ

(I )

= νk

νk

(ij1 j2 j3 )

(ij1 j2 )

ρ (I ) = ρ (ij1 j2 )

  (ij ) α 2 − α (ij3 ) δ (ij1 )  =  (ij ) , α 1 − α (ij3 ) δ (ij2 ) (ij1 )

=

εk

(1 ≤ k ≤ r), (ij ) εk 2     (ij j ) a1 (ij j ) ar = ν1 1 2 . . . νr 1 2 ,

and β

(I )

=β

(ij1 j2 j3 )

  (ij ) α 2 − α (ij3 ) · (X − αY  )(ij1 )  =  (ij ) . α 1 − α (ij3 ) · (X − αY  )(ij2 )

Then we have β (I ) = τ (I ) ρ (I ) .

5.3 Effective Methods for Relative Thue Equations

59

Consider the system of linear equations           (I )   a1 log ν1  + . . . + ar log νr(I )  = log ρ (I ) 

(5.22)

in a1 , . . . , ar for any I = (ij1 j2 j3 ) with 1 ≤ i ≤ m and any distinct j1 , j2 , j3 (1 ≤ j1 , j2 , j3 ≤ n) (the equations are independent of j3 ). Since ε1 , . . . , εr are independent over M, the matrix of coefficients on the left side has rank r. Choosing a set of r linearly independent equations and multiplying by the inverse of the coefficient matrix of the system we conclude       A = max(|a1 |, . . . , |ar |) ≤ c1 · log ρ (I ) 

(5.23)

for a certain set I = (ij1 j2 j3 ) of indices, where c1 is the row norm (maximum sum of the absolute values of the elements in a row) of the inverse matrix of the coefficient matrix of (5.22). We choose the set of independent equations so that c1 becomes as small as possible. Now if |ρ (I ) | < 1, then (5.23) implies     A  (I )  , ρ  < exp − c1

(5.24)



and if |ρ (I ) | > 1, then the same holds for |ρ (I ) | = 1/|ρ (I ) | < 1 with I  = (ij2 j1 j3 ). From now on we assume that (5.24) is valid. The following procedure (application of Baker’s method, reduction) must be performed for each possible values of i, j1 , j2 since we cannot predict which of the |ρ (I ) | satisfies the crucial inequality (5.24). Let 1 ≤ j3 ≤ n be any index distinct from j1 , j2 . Using Siegel’s identity we have (α (ij1 ) − α (ij2 ) )(X − α (ij3 ) Y  ) + (α (ij2 ) − α (ij3 ) )(X − α (ij1 ) Y  ) +(α (ij3 ) − α (ij1 ) )(X − α (ij2 ) Y  ) = 0. For I = (ij1 j2 j3 ) and I  = (ij3 j2 j1 ) we obtain 

β (I ) + β (I ) = 1.

(5.25)

Using | log x| < 2|x − 1| holding for all |x − 1| < 0.795 and applying (5.24), from (5.25) we get            A  (I )  (I )    (I )  , − 1 = 2 · β  ≤ c2 exp −  ≤ 2 · β log β c1

(5.26)

60

5 Relative Thue Equations

where c2 = 2 · |τ (I ) |. On the other hand,              (I ) + log β (I )  = log τ (I ) + a1 · log ν1     . . . + ar · log νr(I ) + a0 · log(−1) ,

(5.27)

where log denotes the principal value of the logarithm and a0 ∈ Z with |a0 | ≤ |a1 | + . . . + |ar | + 1. Set A = max(|a1 |, . . . , |ar |, |a0 |), then A ≤ A ≤ rA + 1. Note that (5.26) implies       A − 1   . log β (I )  ≤ c2 exp − rc1

(5.28)

In case the terms in (5.27) are linearly independent, then we can directly apply Theorem 2.1 of Baker and Wüstholz to the linear form in (5.27) to derive a lower bound of type       log β (I )  ≥ exp(−C · log A ),  which, compared to (5.28),  implies an upper bound for A and A.  Note that if log τ (I ) in (5.27) is dependent on the other terms, we can reduce the number of variables in the linear form. The variable a0 can be omitted for totally real fields K. The bounds obtained by Baker’s method are about 1020 for r = 2, 3 and go up to about 10500 for r = 7, 8. Using the estimate (5.28) we can apply Lemma 2.4 to reduce the bound. After about four to five steps, the reduction procedure stabilizes, i.e., the new bound is not any smaller than the previous bound. Then we stop the procedure. The final reduced bound is usually between 100 and 1000.

5.3.2 Enumeration Since we usually have r ≥ 3 for relative Thue equations, in the whole algorithm the most critical step is the test of all possible values of the exponents a1 , . . . , ar below the reduced bounds. We use the version of Wildanger’s enumeration method described in Sect. 2.3. Equation (5.25) is just what we had in (2.5). Here the multi-indices I I = (ij1 j2 j3 ) range for 1 ≤ i ≤ m and distinct 1 ≤ j1 , j2 , j3 ≤ n. The structure of β (I ) ensures the property (2.6).

5.3 Effective Methods for Relative Thue Equations

61

We let I ∗ = {I1 , . . . , It } be a set of tuples I = (ij1 j2 j3 ) with the following properties: 1. if (ij1 j2 j3 ) ∈ I ∗ , then either (ij2 j3 j1 ) ∈ I ∗ or (ij3 j2 j1 ) ∈ I ∗ , 2. if (ij1 j2 j3 ) ∈ I ∗ , then either (ij1 j3 j2 ) ∈ I ∗ or (ij3 j1 j2 ) ∈ I ∗ , 3. the vectors ⎞  ⎛  (I )  log νk 1  ⎟ ⎜ ⎟ ⎜ .. ek = ⎜ ⎟ (1 ≤ k ≤ r) .  ⎠ ⎝  (It )  log νk  are linearly independent. These are tantamount with the conditions in Sect. 2.3. Since ε1 , . . . , εr are multiplicatively independent over M, the last condition can be satisfied if we take sufficiently many tuples. Note that choosing a minimal set of tuples satisfying those conditions reduces the amount of necessary computations considerably. We are now ready to apply the method in Sect. 2.3.

5.3.3 An Example √ √ √ Let M = Q( 2) with integral basis (1, ω = 2) and fundamental unit η = 1+ 2. Consider the equation X4 − 2X3 Y + (−2 − ω)X2 Y 2 + (3 + ω)XY 3 + (1 + ω)Y 4 = ±ηk in X, Y ∈ ZM , k ∈ Z . The corresponding octic field K is totally real with seven fundamental units, among them η. Hence we had six unknown exponents. The term involving the α-s was independent from the others in the logarithmic linear forms. Baker’s method gave a bound 1053 , which was reduced in three steps to 1097, 121, and 85, respectively. In the first reduction step we took H = 10350 and used a precision of 420 digits. The next steps were much easier, and the whole reduction procedure required about 5 min. In the final enumeration procedure we had to consider 18 ellipsoids (that is, we had to test 18 tuples ij1 j2 j3 ). The vector g was independent from the vectors e1 , . . . , e6 (cf. Sect. 2.3). This means that in fact we enumerated quadratic forms in seven variables, one of them restricted to 1. The reduced bound 85 implied an initial constant S0 = 10269 for the final enumeration. We summarize the enumeration procedure in the following table. In the second and third columns S > s denote the subsequent values Sk > Sk+1 . In the fourth column Digits is the precision we used, the fifth column contains the number of tuples found (in the 18 ellipsoids together), in the last column we display the running time (for the 18 ellipsoids together).

62

5 Relative Thue Equations

The last line, Step 26 corresponds to the single ellipsoid (2.17). The possible exponents were all tested if there were corresponding solutions (X, Y ) of the equation, this took some seconds. These calculations were performed in 2000, and the total CPU time for this example took about 1 h. The solutions of the equation are (X, Y ) = (−ω, 1 − ω), (−1 + ω, −2 + ω), (ω, −1), (−1, −1), (0, −1), (1, 0), (−1, −1 + ω), (2 − ω, −2 + ω), (−2, −1), (1, −ω), (1 − ω, −1), (−4 + ω, −6 + 2ω), (−1 − ω, ω), (ω, −2 − 2ω), (−1, 2 − ω) and of course all multiplies of them by units of M. Step 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26

S 10269 1050 1020 1012 1010 108 107 106 105 104.5 10000 6000 3000 1500 1000 500 250 150 100 50 40 30 20 10 5 3

s 1050 1020 1012 1010 108 107 106 105 104.5 104 6000 3000 1500 1000 500 250 150 100 50 40 30 20 10 5 3

Digits 150 70 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50

Tuples 0 0 0 0 4 42 195 2081 2185 4957 5005 7274 8178 7306 9113 10907 10077 9265 11431 6249 6297 6287 7039 4459 1306 5399

CPU time 5s 5s 5s 30 s 60 s 60 s 60 s 180 s 180 s 180 s 210 s 240 s 240 s 180 s 240 s 240 s 240 s 180 s 180 s 120 s 120 s 120 s 120 s 120 s 70 s 60 s

5.4 Totally Real Thue Equations over Imaginary Quadratic Fields

63

5.4 Totally Real Thue Equations over Imaginary Quadratic Fields In this section we consider special relative Thue equations that can be reduced to absolute Thue equations, which makes the resolution much easier. We follow the arguments of Gaál et al. [GJR18a]. A similar idea is used for solving the relative Thue equations when we consider index from equations in sextic fields with an imaginary quadratic and a real cubic subfield, cf. Sect. 11.2.4. Let F (x, y) be a binary form of degree n ≥ 3 with rational integer coefficients. Assume that f (x) = F (x, 1) has leading coefficient 1 and distinct real roots α1 , . . . , αn . Let 0 < ε < 1, 0 < η < 1 and let K ≥ 1. Set A = min |αi − αj |, i=j



B = min i



|αj − αi |,

j =i

 K , 1 , (1 − ε)n−1 B    1/n  1/n K K , (2C)1/(n−2) , C2 = max , C 1/(n−2) , C1 = max εA εA  1/n K (1 + η)n−1 K D= , E= . n−1 η(1 − ε) AB (1 − ε)n−1 C = max

√ Let m > 1 be a square-free positive integer, and set M = Q(i m). Consider the relative Thue inequality |F (x, y)| ≤ K in x, y ∈ ZM .

(5.29)

If −m ≡ 1 (mod 4), then x, y ∈ ZM can be written as √ √ (2x1 + x2 ) + x2 i m 1+i m = , x = x1 + x2 2 2 √ √ (2y1 + y2 ) + y2 i m 1+i m y = y1 + y2 = 2 2 with x1 , x2 , y1 , y2 ∈ Z. If −m ≡ 2, 3 (mod 4), then x, y ∈ ZM can be written as √ √ x = x1 + x2 i m, y = y1 + y2 i m with x1 , x2 , y1 , y2 ∈ Z. In the proof of our main result, Theorem 5.9, we shall use the following lemma.

64

5 Relative Thue Equations

Lemma 5.8 Let x, y ∈ Z, y = 0. Assume that     αi − x  ≤ d  0 y  |y|n for some i0 (1 ≤ i0 ≤ n) and d > 0. If  |y| ≥

d ηA

1/n ,

then |F (x, y)| ≤ d(1 + η)n−1



|αj − αi0 |.

j =i0

Proof By our assumption, we have         αj − x  ≤ |αj − αi | + αi − x  ≤ (1 + η)|αj − αi | 0 0  0 y  y for j = i0 . Therefore       n       n    αj − x  = αi − x  · αj − x  ≤ d · (1 + η)n−1 · |αj − αi0 |, 0       n y y y |y| j =i0

j =1

j =i0

 

which implies our assertion. Now we present our main result in this section: Theorem 5.9 Let (x, y) ∈ Z2M be a solution of (5.29). Assume that |y| > C1 if −m ≡ 1 (mod 4),

(5.30)

|y| > C2 if −m ≡ 2, 3 (mod 4).

(5.31)

x2 y1 = x1 y2 .

(5.32)

Then

I. Further, if −m ≡ 1 (mod 4), then the following holds: 2n K IA1. If 2y1 + y2 = 0, then 2x1 + x2 = 0 and |F (x2 , y2 )| ≤ √ n . (5.33) ( m) IA2.

If |2y1 + y2 | ≥ 2D, then |F (2x1 + x2 , 2y1 + y2 )| ≤ 2n E. (5.34)

5.4 Totally Real Thue Equations over Imaginary Quadratic Fields

65

If y2 = 0, then x2 = 0 and |F (x1 , y1 )| ≤ K.

IB1.

2n

2 If |y2 | ≥ √ D, then |F (x2 , y2 )| ≤ √ n E. m ( m)

IB2.

(5.35) (5.36)

II. If −m ≡ 2, 3 (mod 4), then the following holds: K IIA1. If y1 = 0, then x1 = 0 and |F (x2 , y2 )| ≤ √ n . ( m)

(5.37)

IIA2.

If |y1 | ≥ D, then |F (x1 , y1 )| ≤ E.

(5.38)

IIB1.

If y2 = 0, then x2 = 0 and |F (x1 , y1 )| ≤ K.

(5.39)

IIB2.

D E If |y2 | ≥ √ , then |F (x2 , y2 )| ≤ √ n . m ( m)

(5.40)

Proof Let (x, y) ∈ Z2M be an arbitrary solution of (5.29) with y = 0. Let βj = x − αj y, j = 1, . . . , n, then the inequality (5.29) can be written as |β1 · · · βn | ≤ K.

(5.41)

Let i0 be the index with |βi0 | = min |βj |. j

1

Then |βi0 | ≤ K n and together with (5.30) and (5.31) we get 1

|βj | ≥ |βj − βi0 | − |βi0 | ≥ |αj − αi0 | · |y| − K n ≥ (1 − ε) · |αj − αi0 | · |y| for j = i0 . From the previous inequality and (5.41), we have |βi0 | ≤ 

K

j =i0

|βj |

≤

c , |y|n−1

(5.42)

with c=

(1 − ε)n−1

K 

j =i0

|αj − αi0 |

.

By (5.42) we obtain         αi − xy  = αi − x  ≤ c ,  0 |y|2   0 y  |y|n

66

5 Relative Thue Equations

hence     αi0 |y|2 − xy  ≤

c , |y|n−2

which implies |Im(xy)| ≤ Note that

c |y|n−2


C1 . (a) If |2y1 + y2 | < 2D, then √ (i) If |y2 | < 2D/ m, then we have only finitely many values for y1 , y2 , we proceed as in √ 1. (ii) If |y2 | ≥ 2D/ m,√ then we use IB2. We solve F (x2 , y2 ) = k for all k ∈ Z with |k| ≤ 2n E/( m)n . We determine the possible values of y1 which satisfy |2y1 + y2 | < 2D. We substitute x2 , y1 , y2 into x2 y1 = x1 y2 to see if there exist corresponding integer x1 . (b) If |2y1 + y2 | ≥ 2D, then we use IA2. We calculate the solutions X = 2x1 + x2 , Y = 2y1 + y2 of F (X, Y ) = k for all k ∈ Z with |k| ≤ 2n E. √ (i) If |y2 | < 2D/ m, then there are only finitely many possible values for y2 . We determine y1 from Y . Using X = 2x1 + x2 we set x2 = X − 2x1 , substitute x2 = X − 2x1 , y1 , y2 into x2 y1 = x1 y2 and test if there is a corresponding x√ 1 in Z. (ii) If |y2 |√≥ 2D/ m we use IB2. We solve F (x2 , y2 ) = k for |k| ≤ 2n E/( m)n . We determine x1 , y1 from x2 , y2 and X, Y . For solving absolute Thue equations F (x, y) = k for certain values k ∈ Z one can efficiently apply Kash [DF97] and Magma [BCP97]. Remark 5.10 An appropriate choice of the parameters ε, η of Theorem 5.9 makes the resolution much easier. It is worthy to keep C1 , C2 and also D small, to avoid

70

5 Relative Thue Equations

extensive tests of small possible solutions. On the other hand, if E is small, then there are fewer Thue equations (over Z) to be solved. Of course we cannot make all these constants simultaneously small, therefore we need to make a compromise, taking into consideration also the value of K (which also determines the number of Thue equations to be solved). Usually it is worthy to try several values of ε, η before we start solving (5.29).

5.4.2 An Example √ Let M = Q(i 5), and let F (x, y) = x 4 − 9x 3 y − 21x 2 y 2 + 88xy 3 + 48y 4 and consider the solutions of |F (x, y)| ≤ 20 in x, y ∈ ZM .

(5.47)

The polynomial F (x, y) is irreducible and the roots of F (x, 1) are approximately −3.4271, −0.49938, 2.7581, 10.1684. We may set A = 2.9278, B = 101.7426. Further, let ε = 0.1 and η = 0.1. We are in case II. Calculating the constants, Theorem 5.9 gives: Assume |y| > 7.2229. Then: IIA1. If y1 = 0, then x1 = 0 and |F (x2 , y2 )| ≤ 0.8000. IIA2.

If |y1 | ≥ 0.9796, then |F (x1 , y1 )| ≤ 36.5157.

IIB1.

If y2 = 0, then x2 = 0 and |F (x1 , y1 )| ≤ 20.

IIB2.

If |y2 | ≥ 0.4381, then |F (x2 , y2 )| ≤ 1.4606. 1

First we consider the values with |y| ≤ C2 = 7.2229. We have |x| √ ≤ 20 4 + max |αj | · C√2 = 75.64. Enumerating and testing all possible x = x1 + i 5x2 and y = y1 + i 5y2 satisfying these bounds we obtain the solutions (x1 , x2 , y1 , y2 ) = (0, 0, 0, 0), (1, 0, 0, 0), (2, 0, 0, 0), (1, 0, −2, 0), (2, 0, −4, 0), up to sign. If y1 = 0, then by IIA1 we have x1 = 0 and |F (x2 , y2 )| ≤ 0.8, whence |F (x2 , y2 )| = 0, x2 = 0, y2 = 0. If y2 = 0, then by IIB1 we have x2 = 0 and |F (x1 , y1 )| ≤ 20. Using Magma we solve F (x1 , y1 ) = k for −20 ≤ k ≤ 20. We obtain the solutions (x1 , y1 ) = (0, 0), (1, 0), (1, −2), (2, 0), (2, −4), up to sign. These bring the above solutions (x1 , x2 , y1 , y2 ) again.

5.5 Simplest Quartic and Simplest Sextic Thue Equations over Imaginary. . .

71

From now on we assume that y1 = 0 and y2 = 0. If |y1 | ≤ 0.9796 and |y2 | ≤ 0.4381, then by IIA2 we have |F (x1 , y1 )| ≤ 36.5157 and by IIB2 we have |F (x2 , y2 )| ≤ 1.4606. In addition to the above calculation we solve F (x1 , y1 ) = k for 21 ≤ |k| ≤ 36 but we do not get any further solutions. Hence the solutions of |F (x1 , y1 )| ≤ 36.5157 are (x1 , y1 ) = (0, 0), (1, 0), (1, −2), (2, 0), (2, −4), up to sign. Also the solutions of |F (x2 , y2 )| ≤ 1.4606 are (x1 , y1 ) = (0, 0), (1, 0), (1, −2), up to sign. Testing these possible (x1 , x2 , y1 , y2 ) we do not get any new solutions. If either |y1 | < 0.9796 or |y2 | < 0.4381, then y1 = 0 or y2 = 0 which cases we have already considered. Hence all solutions of (5.47) are (x, y) = (0, 0), (1, 0), (2, 0), (1, −2), (2, −4), up to sign. The calculation takes just a few seconds.

5.5 Simplest Quartic and Simplest Sextic Thue Equations over Imaginary Quadratic Fields The most important application of Theorem 5.9 of Sect. 5.4 is to the infinite parametric families of simplest quartic and simplest sextic Thue equations. Here we recall the results of Gaál et al. [GJR18b]. Let t be an integer parameter. The infinite parametric families of number fields generated by the roots of the polynomials (3)

ft (x) = x 3 − (t − 1)x 2 − (t + 2)x − 1, (4) ft (x) = x 4 − tx 3 − 6x 2 + tx + 1, (6) ft (x) = x 6 − 2tx 5 − (5t + 15)x 4 −20x 3 + 5tx 2 + (2t + 6)x + 1,

(t ∈ Z), (t ∈ Z \ {−3, 0, 3}), (t ∈ Z \ {−8, −3, 0, 5}),

are called simplest cubic, simplest quartic, and simplest sextic fields, respectively. They are extensively studied in algebraic number theory, starting with Shanks [Sh74], in the cubic case. It was shown by Lettl et al. [LPV99] that these are all parametric families of number fields which are totally real cyclic with Galois group generated by a mapping of type x → ax+b cx+d with a, b, c, d ∈ Z. In 1990 Thomas [Tho90] considered an infinite parametric family of Thue equations, corresponding to the simplest cubic fields. For t ∈ Z, let Ft(3) (x, y) = x 3 − (t − 1)x 2 y − (t + 2)xy 2 − y 3 and consider (3)

Ft (x, y) = ±1 in x, y ∈ Z.

72

5 Relative Thue Equations

Thomas described the solutions for large enough parameters t, and later on the solutions were found for all parameters by Mignotte [Mi93] (cf. Lemma 8.2). This was the first infinite parametric family of Thue equations that was completely solved. Instead of a single equation, the solutions were given for infinitely many equations, for all values of the parameter t. These equations are also called the infinite parametric family of the simplest cubic Thue equations. A couple of other infinite parametric families of Thue equations were completely solved, see Sect. 3.8, among others the parametric family of simplest quartic Thue equations by Lettl and Peth˝o [LP95], Chen and Voutier [CV97], and the parametric family of simplest sextic Thue equations by Lettl et al. [LPV98], Hoshi [Ho12]. Further, the simplest cubic Thue equations and some other families of Thue equations were considered over imaginary quadratic fields, see Sect. 5.6. In this section we consider simplest quartic and simplest sextic Thue equations over imaginary quadratic fields. Let t be√an integer parameter, let m ≥ 1 be a square-free positive integer, and set M = Q(i m) with ring of integers ZM . Let (4)

Ft (x, y) = x 4 − tx 3 y − 6x 2 y 2 + txy 3 + y 4 , and let (6)

Ft (x, y) = x 6 − 2tx 5 y − (5t + 15)x 4 y 2 − 20x 3 y 3 + 5tx 2 y 4 + (2t + 6)xy 5 + y 6 . We give all solutions of the infinite parametric families of simplest quartic and simplest sextic relative Thue equations. More precisely we give all solutions of the simplest quartic relative Thue inequalities |Ft(4) (x, y)| ≤ 1 in x, y ∈ ZM and of the simplest sextic relative Thue inequalities |Ft(6) (x, y)| ≤ 1 in x, y ∈ ZM . We formulate now our main results. In both theorems we exclude the parameters t ∈ Z for which the binary form involved is reducible over Z. Theorem 5.11 Let t ∈ Z with t = −3, 0, 3. All solutions of (4)

|Ft (x, y)| ≤ 1 in x, y ∈ ZM are up to sign given by the following: for any m and any t: (x, y) = (0, 0), (0, 1), (1, 0), for any m and any t = 1: (x, y) = (1, 2), (2, −1), for any m and any t = −1: (x, y) = (2, 1), (−1, 2),

(5.48)

5.5 Simplest Quartic and Simplest Sextic Thue Equations over Imaginary. . .

73

for any m and any t = 4: (x, y) = (2, 3), (3, −2), for any m and any t = −4: (x, y) = (3, 2), (−2, 3), for m = 1 and any t: (x, y) = (0, i), (i, 0), for m = 3 and any t: (x, y) = (ω, 0), (0, ω), (1 − ω, 0), (0, 1 − ω), for m = 1 and t = 1: (x, y) = (i, 2i), (2i, −i), for m = 1 and t = −1: (x, y) = (2i, i), (−i, 2i), for m = 1 and t = 4: (x, y) = (2i, 3i), (3i, −2i), for m = 1 and t = −4: (x, y) = (3i, 2i), (−2i, 3i), for m = 3 and t = 1: (x, y) = (2ω−2, −ω+1), (ω−1, 2ω−2), (−2ω, ω), (ω, 2ω), for m = 3 and t = −1: (x, y) = (−ω + 1, 2ω − 2), (2ω − 2, ω − 1), (ω, −2ω), (2ω, ω), for m = 3 and t = 4: (x, y) = (3ω − 3, −2ω + 2), (2ω − 2, 3ω − 3), (2ω, 3ω), (3ω, −2ω), for m = 3 and t = −4: (x, y) = (−2ω + 2, 3ω − 3), (3ω − 3, 2ω − 2), (3ω, 2ω), (−2ω, 3ω), √ where ω = (1 + i 3)/2. Theorem 5.12 Let t ∈ Z, t = −8, −3, 0, 5. All solutions of (6)

|Ft (x, y)| ≤ 1 in x, y ∈ ZM

(5.49)

are up to sign given by the following: for any m and any t: (x, y) = (0, 0), (0, 1), (1, 0), (1, −1), for m = 1 and any t: (x, y) = (0, i), (i, 0), (i, −i), for m = 3 and any t: (x, y) = (ω, 0), (0, ω), (ω, −ω), (1−ω, 0), (0, ω−1), (ω−1, −ω + 1).

5.5.1 Proof of Theorem 5.11 In this section we turn to the proof of Theorem 5.11. In our proof we shall use Theorem 5.9 of Sect. 5.4 and the corresponding results in the absolute case. For right-hand sides ±1 Chen and Voutier [CV97] gave all solutions of simplest quartic Thue equations. Lemma 5.13 Let t ∈ Z with t ≥ 1, t = 3. All solutions of Ft(4) (x, y) = ±1 in x, y ∈ Z are given by (x, y) = (±1, 0), (0, ±1). Further, for t = 1 we have (x, y) = (1, 2), (−1, −2), (2, −1), (−2, 1) and for t = 4 we have (x, y) = (2, 3), (−2, −3), (3, −2), (−3, 2). For larger right-hand sides we can use the statement of Lettl et al. [LPV99].

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5 Relative Thue Equations

Lemma 5.14 Let t ∈ Z, t ≥ 58 and consider the primitive solutions (i.e., solutions with (x, y) = 1) of (4)

|Ft (x, y)| ≤ 6t + 7 in x, y ∈ Z.

(5.50)

If (x, y) is a solution of (5.50), then every pair in the orbit {(x, y), (y, −x), (−x, −y), (−y, x)} is also a solution. Every orbit has a solution with y > 0, −y ≤ x ≤ y. If an orbit contains a primitive solution, then all solutions in this orbit are primitive. All solutions of the above inequality with y > 0, −y ≤ x ≤ y are (0, 1), (±1, 1), (±1, 2). Remark 5.15 Since (4)

(4)

Ft (x, y) = F−t (y, x), it is enough to solve the inequality (5.48) only for t > 0. Also, we have (4)

(4)

(4)

(4)

Ft (x, y) = Ft (−x, −y) = Ft (y, −x) = Ft (−y, x). Therefore, if (x, y) ∈ Z2M is solution, then (y, −x), (−y, x), (−x, −y) are solutions, too. Proof of Theorem 5.11 for m = 1, 3 Using the estimates of [LPV99] for the roots (4) of the polynomial Ft (x, 1) we obtain A > 0.9833 and B > 58.1 for t ≥ 58. Calculating the roots for 0 < t < 58 we obtain A > 0.8284, B > 4.6114 for any t > 0, t = 3. Set ε = 0.1924,

η = 0.169.

For t > 0, t = 3 and a square-free m with m = 1, 3 our Theorem 5.9 implies: Corollary 5.16 Let (x, y) ∈ Z2M be solutions of (5.48). Assume that |y| > 6.2741. Then x2 y1 = x1 y2 .

5.5 Simplest Quartic and Simplest Sextic Thue Equations over Imaginary. . .

75

I. Let −m ≡ 1 (mod 4). (4)

IA1. If 2y1 + y2 = 0, then 2x1 + x2 = 0 and |Ft (x2 , y2 )| ≤ 0.326. IA2. If |2y1 + y2 | ≥ 2.618, then |Ft(4) (2x1 + x2 , 2y1 + y2 )| ≤ 48.526. (4)

IB1.

If y2 = 0, then x2 = 0 and |Ft (x1 , y1 )| ≤ 1.

IB2.

If |y2 | ≥ 0.989, then |Ft (x2 , y2 )| ≤ 0.990.

(4)

II. Let −m ≡ 2, 3 (mod 4). IIA1. If y1 = 0, then x1 = 0 and |Ft(4) (x2 , y2 )| ≤ 0.25. (4)

IIA2.

If |y1 | ≥ 1.309, then |Ft (x1 , y1 )| ≤ 3.032.

IIB1.

If y2 = 0, then x2 = 0 and |Ft (x1 , y1 )| ≤ 1.

IIB2.

If |y2 | ≥ 0.925, then |Ft (x2 , y2 )| ≤ 0.7582.

(4)

(4)

Case I −m ≡ 1 (mod 4) (a) Assume that |y| > 6.2741. By Corollary 5.16 we have: If y2 = 0, then by IB1 we have x2 = 0 and using Lemma 5.13, |Ft(4) (x1 , y1 )| ≤ 1 implies |x1 |, |y1 | ≤ 3. This contradicts to |y| > 6.2741. (4) If y2 = 0, then IB2 implies |Ft (x2 , y2 )| ≤ 0.990, whence y2 = 0, a contradiction. Therefore |y| > 6.2741 is not possible. (b) Consider now |y| ≤ 6.2741. By Remark 5.15 if (x, y) is a solution then so also is (y, −x). As above we obtain that |x| > 6.2741 is not possible, hence |x| ≤ 6.2741. We enumerate all x, y with |x| ≤ 6.2741 and |y| ≤ 6.2741 and we obtain the solutions (x, y) = (0, 0), (0, ±1), (±1, 0). Additionally we have up to sign for t = 1: (x, y) = (1, 2), (2, −1) and for t = 4: (x, y) = (2, 3), (3, −2). Case II −m ≡ 2, 3 (mod 4) Similar to Case I, we obtain the same solutions. According to Remark 5.15 we proved Theorem 5.11 for all t with t = −3, 0 − 3 and for m = 1, 3.

76

5 Relative Thue Equations

Proof of Theorem 5.11 for m = 1 Set ε = 0.1792,

η = 0.0308.

For t > 0, t = 3 and m = 1 Theorem 5.9 implies: Corollary 5.17 Let (x, y) ∈ Z2M be a solution of (5.48). Assume |y| > 6.736. Then x2 y1 = x1 y2 . Further, (4)

IIA1. if y1 = 0, then x1 = 0 and |Ft (x2 , y2 )| ≤ 1, IIA2.

(4)

if |y1 | ≥ 1.98, then |Ft (x1 , y1 )| ≤ 1.981, (4)

IIB1. if y2 = 0, then x2 = 0 and |Ft (x1 , y1 )| ≤ 1, IIB2.

if |y2 | ≥ 1.98, then |Ft(4) (x2 , y2 )| ≤ 1.981.

(a) Assume |y| > 6.736. By the above Corollary we deduce: If y1 = 0, then by IIA1 we have |Ft(4) (x2 , y2 )| ≤ 1, whence by Lemma 5.13 |y2 | ≤ 3, contradicting |y| > 6.736. (4) If |y1 | > 3, then by IIA2 we have |Ft (x1 , y1 )| ≤ 1, whence by Lemma 5.13 |y1 | ≤ 3, a contradiction. Therefore only |y1 | = 1, 2, 3 is possible. Using IIB1 and IIB2 we similarly obtain that only |y2 | = 1, 2, 3 is possible. But |y1 | = 1, 2, 3, |y2 | = 1, 2, 3 contradicts |y| > 6.736. (b) Hence only |y| ≤ 6.736 is possible. If (x, y) is a solution, then so also is (y, −x), therefore we must also have |x| ≤ 6.736. Enumerating the set (x, y) ∈ Z2M with |x|, |y| ≤ 6.736 we obtain (x, y) = (0, 0), (0, ±1), (±1, 0), (0, ±i), (±i, 0). Additionally we have up to sign for t = 1 (x, y) = (1, 2), (i, 2i), (2, −1), (2i, −i) and for t = 4 (x, y) = (2, 3), (2i, 3i), (3, −2), (3i, −2i). According to Remark 1 we have proved Theorem 5.11 for all t with t = −3, 0, 3 and for m = 1.

5.5 Simplest Quartic and Simplest Sextic Thue Equations over Imaginary. . .

77

Proof of Theorem 5.11 for m = 3 First we assume t ≥ 58. Then A > 0.9833 and B > 58.1. Set ε = 0.6273,

η = 0.0361.

2 be solutions of (5.48) and let m = 3. Assume Corollary 5.18 Let (x, y) ∈ ZM t ≥ 58 and

|y| > 1.621. Then x2 y1 = x1 y2 . Further IA1.

(4)

if 2y1 + y2 = 0, then 2x1 + x2 = 0 and |Ft (x2 , y2 )| ≤ 1.778, (4)

IA2. if |2y1 + y2 | ≥ 3.497, then |Ft (2x1 + x2 , 2y1 + y2 )| ≤ 343.753, IB1.

if y2 = 0, then x2 = 0 and |Ft(4) (x1 , y1 )| ≤ 1,

IB2.

if |y2 | ≥ 2.019, then |Ft (x2 , y2 )| ≤ 38.195.

(4)

(a) Assume |y| > 1.621. Then by the above corollary: (4)

If 2y1 + y2 = 0, then by IA1 2x1 + x2 = 0 and |Ft (x2 , y2 )| ≤ 1.778. By Lemma 5.13 this later inequality implies (x2 , y2 ) = (0, 0), (0, ±1), (±1, 0). However, for (x2 , y2 ) = (0, ±1), (±1, 0) one of the the equations 2y1 + y2 = 0 and 2x1 + x2 = 0 have no integer solutions in y1 , resp. x1 . If (x2 , y2 ) = (0, 0), then 2y1 + y2 = 0 implies y1 = 0, but (y1 , y2 ) = (0, 0) contradicts |y| > 1.621. (4) If |2y1 + y2 | > 3.497, then IA2 implies |Ft (2x1 + x2 , 2y1 + y2 )| ≤ 343.753. Using Lemma 5.14 we can easily list all primitive and non-primitive solutions of this inequality and we always have |2y1 + y2 | ≤ 4. Therefore only |2y1 + y2 | = 1, 2, 3, 4 is possible. Using IB1 and IB2 we similarly obtain that only |y2 | = 1, 2 is possible. The equations |2y1 + y2 | = 1, 2, 3, 4, |y2 | = 1, 2 leave only a few possible values for (y1 , y2 ). (b) If |x| > 1.621, then we similarly obtain |2x1 + x2 | = 1, 2, 3, 4, |x2 | = 1, 2, since if (x, y) is a solution, then so also is (y, −x). (c1) If |x| > 1.621 and |y| > 1.621, then we test the finite set |2x1 + x2 | = 1, 2, 3, 4, |x2 | = 1, 2, |2y1 + y2 | = 1, 2, 3, |y2 | = 1, 2. (c2) If |x| > 1.621 and |y| ≤ 1.621, then we test the finite set |2x1 + x2 | = 1, 2, 3, 4, |x2 | = 1, 2, |y| ≤ 1.621.

78

5 Relative Thue Equations

(c3) If |x| ≤ 1.621 and |y| > 1.621, then we test the finite set |x| ≤ 1.621, |2y1 + y2 | = 1, 2, 3, 4, |y2 | = 1, 2. (c4) Finally, if |x| ≤ 1.621 and |y| ≤ 1.621, then we test this finite set. All together up to sign we get the following solutions for arbitrary t ≥ 58: (x, y) = (0, 0), (1, 0), (0, 1), (ω, 0), (0, ω), (1 − ω, 0), (0, 1 − ω). (4)

Let now 0 < t < 58 Considering the roots of the polynomial Ft (x, 1) = 0 for these parameters we obtain A > 0.8284, B > 4.6114. Set ε = 0.0348,

η = 0.0005.

Corollary 5.19 Let m = 3 and 0 < t < 58. Let (x, y) ∈ Z2M be a solution of (5.48) and assume |y| > 34.688. Then x2 y1 = x1 y2 . Further (4)

IA1. if 2y1 + y2 = 0, then 2x1 + x2 = 0 and |Ft (x2 , y2 )| ≤ 1.778, IA2. if |2y1 + y2 | ≥ 9.824, then |Ft(4) (2x1 + x2 , 2y1 + y2 )| ≤ 17.825, (4)

IB1.

if y2 = 0, then x2 = 0 and |Ft (x1 , y1 )| ≤ 1,

IB2.

if |y2 | ≥ 5.672, then |Ft (x2 , y2 )| ≤ 1.981.

(4)

(a) Assume |y| > 34.688. Then by the above corollary we have: (4)

If y2 = 0, then by IB1 |Ft (x1 , y1 )| ≤ 1. By Lemma 5.13 we know the possible solutions y1 . These, together with y2 = 0 contradict |y| > 34.688. (4) If |y2 | ≥ 5.672, then by IB2 |Ft (x2 , y2 )| ≤ 1, whence by Lemma 5.13 |y2 | ≤ 3, a contradiction. Therefore only |y2 | = 1, 2, 3, 4, 5 is possible. (4) If 2y1 + y2 = 0, then by IA1 we have 2x1 + x2 = 0 and |Ft (x2 , y2 )| ≤ 1. From Lemma 5.13 we get the possible values of y2 and we calculate y1 from 2y1 +y2 = 0. These are in contradiction with |y| > 34.688. (4) If |2y1 +y2 | ≥ 9.824, then by IA2 we have |Ft (2x1 +x2 , 2y1 +y2 )| ≤ 17. Using (4) Magma [BCP97] we solve the equation Ft (2x1 + x2 , 2y1 + y2 ) = d for all t ≤ 58 and |d| ≤ 17 and list the solutions. All these solutions contradict |2y1 +y2 | > 9.824. Therefore only |2y1 + y2 | = 1, . . . , 9 is possible.

5.6 Further Results on Relative Thue Equations

79

In the set |y2 | = 1, 2, 3, 4, 5, |2y1 + y2 | = 1, . . . , 9 all y = y1 + ωy2 have absolute values less than 34.688 which is in contradiction with |y| > 34.688. (b) Therefore only |y| ≤ 34.688 is possible. If (x, y) is a solution, then so also is (y, −x), therefore we similarly obtain |x| ≤ 34.688. Enumerating all x, y with these properties we obtain up to sign the following solutions: for arbitrary t: (1, 0), (0, 1), (ω, 0), (0, ω), (1 − ω, 0), (0, 1 − ω), for t = 1: (1, 2), (2, −1), (2ω−2, −ω+1), (ω−1, 2ω−2), (−2ω, ω), (ω, 2ω), for t = 4: (2, 3), (3, −2), (3ω − 3, −2ω + 2), (2ω − 2, 3ω − 3), (2ω, 3ω), (3ω, −2ω). According to Remark 5.15 we have proved Theorem 5.11 for all t with t = −3, 0, 3 and m = 3.

5.5.2 Proof of Theorem 5.12 This proof is similar to the proof of Theorem 5.11, see Gaál et al. [GJR18b]. In the proof we apply the results of Lettl et al. [LPV98] and Hoshi [Ho12] on the rational (6) integer solutions of equation Ft (x, y) = ±1 and the results of Lettl et al. [LPV99] (6) on the rational integer solutions of the inequality |Ft (x, y)| ≤ 120t + 323.

5.5.3 Computational Aspects All auxiliary calculations were made in 2018 by using Maple [CG88]. The resolution of Thue equations was performed by using Magma [BCP97]. The running time all together took a few hours.

5.6 Further Results on Relative Thue Equations For relative Thue equations see also Evertse and Gy˝ory [EGy15]. Infinite parametric families of relative Thue equations were considered among others by Heuberger et al. [HPT02], Ziegler [Zie05], [Zie06], Heuberger et al. [HPT06], Jadrijevi´c and Ziegler [JZ06], Heuberger [He06], and Kirschenhofer and Lampl [KL07]. All these families are considered over imaginary quadratic fields.

Chapter 6

The Resolution of Norm Form Equations

Abstract Although there is an extensive literature of the explicit resolution of Thue equations (see for example Chap. 3), the problem of algorithmic resolution of norm form equations was not investigated formerly. Our purpose is now to fill this gap and to give an efficient method for solving norm form equations under general conditions. The reason to include this algorithm in this book is that we use the same tools of Chap. 2 as for the above types of Thue equations.

6.1 Preliminaries Let α1 = 1, α2 , . . . , αm be algebraic integers, linearly independent over Q, let K = Q(α2 , . . . , αm ), L = Q(α1 , . . . , αm−1 ), and assume that [K : L] ≥ 3.

(6.1)

Let 0 = b ∈ Z and consider the norm form equation NK/Q (x1 + α2 x2 + . . . + αm xm ) = d in x1 , x2 , . . . , xm ∈ Z, xm = 0.

(6.2)

Using Baker’s method Gy˝ory gave effective upper bounds for the solutions of norm form equations of the above type, see for example [Gy81] (see [BGy96b] for improved bounds), reducing the equation to unit equations in two variables. Some of his ideas are used here, as well. In order to apply Baker’s method it was necessary to make assumptions on the coefficients: (6.1) was the most general assumption of these. Note that Smart [Sm95] gave a method for solving triangularly connected decomposable form equations, which involve also some special norm form equations, but his purpose was not to consider norm form equations utilizing their special properties, hence his general method is not feasible for norm form equations of the above type. © Springer Nature Switzerland AG 2019 I. Gaál, Diophantine Equations and Power Integral Bases, https://doi.org/10.1007/978-3-030-23865-0_6

81

82

6 The Resolution of Norm Form Equations

Our purpose is to give an efficient algorithm (cf. Gaál [Ga00c]) for the explicit resolution of Eq. (6.2). In the course of our method we need to use Baker’s method, hence we also have to assume (6.1). In fact we reduce the problem to solving a special type of relative Thue equation over L. One of our goals is to show that by solving Eq. (6.2) it is sufficient to deal with linear forms in r(K) − r(L) variables, where r(K) resp. r(L) denote the unit rank of K resp. L. The second goal is to show that the enumeration method described in Sect. 2.3 can be applied in its original form. This way we become an efficient method for the enumeration of small exponents in the corresponding unit equation for reasonable values of r(K) − r(L) (up to about 11). Let l = [L : Q], k = [K : L] and denote by γ (ij ) (1 ≤ i ≤ l, 1 ≤ j ≤ k) the conjugates of any γ ∈ K so that γ (i1) , . . . , γ (ik) are just corresponding relative conjugates of γ over the conjugate field L(i) of L. For elements μ of L we write μ(i) for μ(i1) = . . . = μ(ik) . Assume that η1 , . . . , ηs is a set of fundamental units in L. Let us extend this system to a system of independent units η1 , . . . , ηs , ε1 , . . . , εr of full rank in K. Denote by q the index of the unit group generated by these units in the whole unit group of K. Calculate a full set of nonassociated integers ν of K of norm ±d. The algorithm must be performed for each element ν of this set. Assume that x = (x1 , . . . , xm ) ∈ Zm is a solution of (6.2). Let l(x) = x1 + α2 x2 + . . . + αm xm . For 1 ≤ i ≤ l, 1 ≤ j ≤ k we have (ij )

(i) l (ij ) (x) = x1 + α2(i) x2 + . . . + αm−1 xm−1 + αm xm

  b1  a1  ar   bs   q (ij ) q (ij ) q (i) q ε1 = ζ (ij ) ν (ij ) η1 · · · ηs(i) · · · εr

(6.3)

with some integers b1 , . . . , bs , a1 , . . . , ar ∈ Z, where ζ is a root of unity and we use throughout a fixed determination of the q-th root of the numbers involved. For any i (1 ≤ i ≤ l) and distinct j1 , j2 , j3 (1 ≤ j1 , j2 , j3 ≤ k) we have     (ij ) (ij ) (ij ) (ij ) αm 1 − αm 2 l (ij3 ) (x) + αm 2 − αm 3 l (ij1 ) (x)   (ij ) (ij ) + αm 3 − αm 1 l (ij2 ) (x) = 0, whence (ij )

(ij )

αm 2 − αm 3 (ij )

(ij )

αm 1 − αm 3

(ij )

·

(ij )

l (ij1 ) (x) αm 2 − αm 1 l (ij3 ) (x) + (ij ) = 1, · (ij ) (ij ) 3 l (ij2 ) (x) αm − αm 1 l 2 (x)

6.2 Solving the Unit Equation

83

that is  

(ij )

(ij )

αm 2 − αm 3

 

ζ (ij1 ) ν (ij1 )



(ij1 )

ε1

(ij )



 aq1 ···

(ij1 )

εr

 ar q

(ij )

(ij ) (ij ) 2 εr 2 αm 1 − αm 3 ζ (ij2 ) ν (ij2 ) ε1    (ij )  ar   a1 (ij ) (ij ) αm 2 − αm 1 ζ (ij3 ) ν (ij3 ) ε(ij3 ) q εr 3 q 1  + · · · = 1. (ij2 ) (ij ) (ij ) (ij ) εr 2 αm 3 − αm 1 ζ (ij2 ) ν (ij2 ) ε1

(6.4)

This is the unit equation we are going to solve. Note that the denominator q in the exponents will not cause any troubles. Introduce

γ

(ij1 j2 j3 )

  (ij ) α 2 − α (ij3 ) ζ (ij1 ) ν (ij1 )  =  (ij ) , α 1 − α (ij3 ) ζ (ij2 ) ν (ij2 )

 (ij j ) ρk 1 2

=

(ij1 )

εk

 q1

(ij2 )

εk

(1 ≤ k ≤ r)

and     (ij j ) a1 (ij j ) ar τ (ij1 j2 ) = ρ1 1 2 . . . ρr 1 2 , then we have (ij )

β (ij1 j2 j3 ) =

(ij )

αm 2 − αm 3 (ij )

(ij )

αm 1 − αm 3

·

l (ij1 ) (x) = γ (ij1 j2 j3 ) τ (ij1 j2 ) . l (ij2 ) (x)

for any i (1 ≤ i ≤ l) and any distinct j1 , j2 , j3 (1 ≤ j1 , j2 , j3 ≤ k). Equation (6.4) can be written in the form β (ij1 j2 j3 ) + β (ij3 j2 j1 ) = 1.

(6.5)

In the following we consider Eq. (6.5).

6.2 Solving the Unit Equation By solving the system of linear equations          (ij j )   (ij j )  a1 log ρ1 1 2  + . . . + ar log ρr 1 2  = log τ (ij1 j2 ) 

(6.6)

84

6 The Resolution of Norm Form Equations

in a1 , . . . , ar (1 ≤ i ≤ l, 1 ≤ j1 , j2 ≤ k, j1 = j2 ), we obtain       A = max(|a1 |, . . . , |ar |) ≤ c1 · log τ (ij1 j2 ) 

(6.7)

for a certain set i, j1 , j2 of indices, where c1 (and in the following c2 , c3 . . .) can be calculated easily (cf. (5.23)). Exchanging j1 and j2 if necessary (6.7) implies that there are indices i, j1 , j2 with     A  (ij1 j2 )  . τ  < exp − c1

(6.8)

The following steps must be performed for all possible values of i, j1 , j2 . Let 1 ≤ j3 ≤ k be any index distinct from j1 , j2 . Applying (6.8), from (6.5) we get          log β (ij3 j2 j1 )  ≤ 2 · β (ij3 j2 j1 ) − 1     A   = 2 · β (ij1 j2 j3 )  ≤ c2 exp − . c1 On the other hand,            (ij j ) log β (ij3 j2 j1 )  = log γ (ij3 j2 j1 ) + a1 · log ρ1 3 2 +     (ij j ) . . . + ar · log ρr 3 2 + a0 · log(−1) ,

(6.9)

(6.10)

where log denotes the principal value of the logarithm, and a0 ∈ Z with |a0 | ≤ |a1 | + . . . + |ar | + 1. Set A = max(|a1 |, . . . , |ar |, |a0 |), then A ≤ A ≤ rA + 1. In case the terms in the above linear form are independent (otherwise we can reduce the number of variables), using Theorem 2.1 of Baker and Wüstholz and (6.9) we conclude      A + 1   (ij3 j2 j1 )  , (6.11) exp(−C · log A ) ≤ log β  ≤ c2 exp − rc1 which implies an upper bound AB for A of magnitude 1020 up to 10500 for r = 2 up to 8. Using (6.10) and (6.11) we have an estimate of type |ξ + a1 ξ1 + . . . + ar ξr + a0 ξ0 | < c2 exp(−c3 A − c4 ),

(6.12)

where       (ij j ) (ij j ) ξ = log γ (ij3 j2 j1 ) , ξ1 = log ρ1 3 2 , . . . , ξr = log ρr 3 2 , ξ0 = log(−1).

6.3 Calculating the Solutions of the Norm Form Equation

85

Using inequality (6.12) we can reduce the bound AB for A by applying Lemma 2.4 of Sect. 2.2.2. The final reduced bound AR is usually between 100 and 1000. Let I be the set of all tuples I = (i, j1 , j2 , j3 ) with 1 ≤ i ≤ l, 1 ≤ j1 , j2 , j3 ≤ k so that j1 , j2 , j3 are distinct. Introduce (ij1 j2 )

(I )

β (I ) = β (ij1 j2 j3 ) , γ (I ) = γ (ij1 j2 j3 ) , ρh = ρh

(1 ≤ h ≤ r).

Then we have   ar  (I ) a1 . . . ρr(I ) β (I ) = γ (I ) ρ1 and setting I  = (i, j3 , j2 , j1 ) Eq. (6.5) can be written as 

β (I ) + β (I ) = 1. The set I satisfies (2.6) (cf. Sect. 2.3). Let I ∗ = {I1 , . . . , It } be a set of tuples I with the following properties: 1. if (ij1 j2 j3 ) ∈ I ∗ , then either (ij2 j3 j1 ) ∈ I ∗ or (ij3 j2 j1 ) ∈ I ∗ , ∗ ∗ 2. if (ij1 j2 j3 ) ∈ I ∗ , then (ij  1 j3 j2 ) ∈ I or(ij3 j1 j2 ) ∈ I ,   either  (It )   (I1 )  (1 ≤ h ≤ r) are linearly 3. the vectors eh = log ρh  , . . . , log ρh  independent. These conditions are tantamount to the ones in Sect. 2.3. We can apply the method of Sect. 2.3 to enumerate all tuples of exponent vectors satisfying (6.5) with coordinates less than AR in absolute value.

6.3 Calculating the Solutions of the Norm Form Equation The algorithm of the preceding section gives us all possible tuples (a1 , . . . , ar ) of exponents in (6.3). For any i, j (1 ≤ i ≤ l, 1 ≤ j ≤ k) set   b1   bs q (i) q · · · ηs(i) , δ (i) = η1 and   a1  ar  (ij ) q (ij ) q γ (ij ) = ζ (ij ) ν (ij ) ε1 · · · εr . The δ (i) are not yet known, but the γ (ij ) are determined by the exponents (a1 , . . . , ar ). Then we have (ij )

(i) l (ij ) (x) = x1 + α2(i) x2 + . . . + αm−1 xm−1 + αm xm = δ (i) γ (ij ) .

(6.13)

86

6 The Resolution of Norm Form Equations

For any 1 < i ≤ l we obtain     (i1) (i2) xm , − αm δ (i) γ (i1) − γ (i2) = l (i1) (x) − l (i2) (x) = αm hence 0 = xm = δ (i)

γ (i1) − γ (i2) (i1)

(i2)

αm − αm

,

that is δ (i) = κi δ (1) ,

(6.14)

with (i1)

κ1 = 1,

κi =

(i2)

γ (11) − γ (12) αm − αm (11) (12) γ (i1) − γ (i2) αm − αm

for i = 2, . . . , l.

Substituting our expressions into the original equation (6.2) it can be written in the form k l  

l (ij ) (x) = d,

i=1 j =1

whence we obtain k  l    κi δ (1) γ (ij ) = d, i=1 j =1

that is ⎛ ⎞−1  −k k l  l  kl   (1) (ij ) ⎠ ⎝ δ =d γ κi , i=1 j =1

(6.15)

i=1

from which we can calculate the value of δ (1) . This gives at once the value of δ (i) by (6.14). Finally, solving the system of linear equations (6.13) (1 ≤ i ≤ l, 1 ≤ j ≤ k) in x1 , . . . , xm we get the solutions of Eq. (6.2).

6.4 Examples We illustrate our algorithm by two detailed examples. The calculations were performed in 1999. The basic number field data were calculated by using Kash [DF97].

6.4 Examples

87

Example 1 Consider first the field K generated by a root ξ of the polynomial f (x) = x 9 − x 8 − 31x 7 + 8x 6 + 200x 5 − 87x 4 − 97x 3 + 27x 2 + 12x − 1. This field is totally real, has integral basis (1, ξ, ξ 2 , ξ 3 , ξ 4 , ξ 5 , ξ 6 , ξ 7 , ω9 ), with ω9 = (14800 + 24483ξ + 15778ξ 2 + 15468ξ 3 +19731ξ 4 + 4153ξ 5 + 1420ξ 6 + 4197ξ 7 + ξ 8 )/25349. The discriminant of the field is DK = 107226034120512 = 26 · 33 · 373 · 1073 . The field K has a totally real cubic subfield L generated by α defined by the polynomial g(x) = x 3 − x 2 − 3x + 1 with discriminant DL = 148 = 22 · 37. (Note that K has also another cubic subfield generated by the root of x 3 − x 2 − 4x + 1 with discriminant 321 = 3 · 107 but this does not play any roles in our arguments.) The field L has integral basis (1, α, α 2 ) and fundamental units η1 = α,

η2 = 2α − 1.

These elements have the following coefficients in the integral basis of K: η1 = (−430, −703, −454, −472, −568, −117, −42, −122, 736), η2 = (−6383, −10561, −6838, −6694, −8428, −1791, −626, −1811, 10936). These units together with ε1 = (328, 539, 346, 360, 433, 89, 32, 93, −561), ε2 = (758, 1242, 800, 832, 1001, 206, 74, 215, −1297), ε3 = (3590, 5940, 3838, 3746, 4739, 1010, 352, 1018, −6148), ε4 = (6055, 10022, 6492, 6334, 7995, 1702, 594, 1718, −10375), ε5 = (103, 164, 108, 112, 135, 28, 10, 29, −175), ε6 = (6225, 10295, 6682, 6551, 8218, 1745, 611, 1767, −10670) form a system of fundamental units in K. (Hence s = 2, r = 6, q = 1.) Consider the norm form equation NK/Q (x1 + αx2 + α 2 x3 + ξ x4 ) = ±1 in x1 , x2 , x3 , x4 ∈ Z with x4 = 0.

(6.16)

88

6 The Resolution of Norm Form Equations

We had c1 = 0.763 and c2 = 4.291 for all possible i, j1 , j2 . Since our example is a totally real one, we did not have to use a0 . Baker’s method gave A = max(|a1 |, . . . , |a6 |) ≤ 1036 = AB . In the reduction procedure we had dimension 7, c3 = 1/c1 , c4 = 0. The following table summarizes the steps of the reduction procedure. Note that in each step we had to perform nine reductions.

Step I Step II Step III Step IV

A< 1036 429 49 36

|b1 | > 1039 9709 1109 815

H = 10280 1030 1022 1021

Precision 700 digits 100 digits 60 digits 60 digits

New bound for A 429 49 36 35

CPU time 20 min 8s 5s 5s

Hence our algorithm gave the reduced bound AR = 35. In the enumeration process we used I ∗ = {(i123), (i231), (i312)|i = 1, 2, 3}, that is, we had t = 9 ellipsoids to consider. The initial bound was S = 0.4116 · 10153 that we got using the reduced bound for A. Note that in this example the vector g is linearly dependent on e1 , . . . , e6 . The following table is a summary of the enumeration process.

Step I Step II Step III Step IV Step V Step VI Step VII Step VIII Step IX Step X Step XI Step XII

S 10153 1020 1010 108 106 105 104 103 102 10 5 3

s 1020 1010 108 106 105 104 103 102 10 5 3

Precision 100 digits 50 digits 50 digits 50 digits 50 digits 50 digits 50 digits 50 digits 50 digits 50 digits 50 digits 50 digits

CPU time 10 s 5s 4s 4s 3s 3s 3s 3s 3s 5s 5s 3s

Tuples found 0 0 0 2 8 16 34 96 133 15 15 34

The last line refers to the enumeration of the ellipsoid (2.17) (Sect. 2.3) with S = 3. We tested all tuples we found in the enumeration process if they are solutions of (6.4). We found 14 solutions of (6.4), and the components were all ≤2 in absolute value. For these tuples we calculated the corresponding solutions of Eq. (6.16). We obtained the following solutions:

6.4 Examples

89 x1 0 0 1 0 1 −1 0 −1

x2 0 0 −1 −1 −1 0 2 −2

x3 0 1 −1 1 0 1 −1 0

x4 −1 −1 1 1 −1 1 1 1

If (x1 , x2 , x3 , x4 ) is a solution, then so also is (−x1 , −x2 , −x3 , −x4 ), but we list only one of them. Example 2 Our second example refers to a more complicated situation. Consider the field K generated by a root ξ of the polynomial f (x) = x 12 − 80x 10 − 85x 9 + 568x 8 + 184x 7 − 1041x 6 + 40x 5 + 432x 4 − 19x 3 − 52x 2 − 2x + 1. This field is totally real, has integral basis (1, ξ, ξ 2 , ξ 3 , ξ 4 , ξ 5 , ξ 6 , ξ 7 , ξ 8 , ω10 , ω11 , ω12 ) with ω10 = (1 + ξ 3 + 2ξ 4 + ξ 6 + ξ 7 + ξ 8 + ξ 9 )/3, ω11 = (2 + ξ + 2ξ 3 + 2ξ 4 + 2ξ 5 + 2ξ 6 ξ 10 )/3, ω12 = (107761264539 + 9245049222ξ + 31097752879ξ 2 + 40137945519ξ 3 +34157911107ξ 4 + 93111405784ξ 5 + 51616938926ξ 6 +54389034027ξ 7 + 110416671757ξ 8 + 1812369088ξ 9 +25415148001ξ 10 + ξ 11 )/113333753409. The field K has a totally real quartic subfield L generated by α defined by the polynomial g(x) = x 4 −4x 2 +x+1. (Note that K has also a cubic subfield generated by the root of x 3 + 4x 2 − 2x − 1 but it does not play a role in our arguments.) The field L has integral basis (1, α, α 2 , α 3 ) and fundamental units η1 = α, η2 = 1 − α,

η3 = −2α + α 2 + α 3 .

The units η1 , η2 , η3 together with ε1 , . . . , ε8 form a system of fundamental units in K, where the coefficients of ε1 , . . . , ε8 in the integral basis of K are the following:

90

6 The Resolution of Norm Form Equations

ε1 = (−10895130684, 3196295645, −6147000968, 2471771060, 4012072493, −8357582270, 202752252, −10392673880, −21467491622, −1074736976, −15071211644, 22402348184), ε2 = (761572045, −223421774, 429676837, −172777352, −280445134, 584196946, −14172204, 726450169, 1500582390, 75124353, 1053481036, −1565929104), ε3 = (97534039010, −28613481877, 55028420322, −22127482530, −35916377883, 74817712333, −1815053823, 93036007507, 192178618710, 9621127196, 134918633553, −200547525759), ε4 = (53135222443, −15588237092, 29978737346, −12054752441, −19566755165, 40759675309, −988817143, 50684755933, 104696306673, 5241459687, 73501842816, −109255573732), ε5 = (−137633535407, 40377438576, −77652439063, 31224828557, 50682797994, −105577768977, 2561283431, −131286213220, −271189658479, −13576693470, −190388183616, 282999302268), ε6 = (22062864796, −6472564754, 12447804013, −5005387339, −8124529892, 16924276765, −410577392, 21045379385, 43472114179, 2176364587, 30519515049, −45365218051), ε7 = (−62200893641, 18247825609, −35093562605, 14111475601, 22905139427, −47713891702, 1157523982, −59332340433, −122559077246, −6135731847, −86042366935, 127896224154), ε8 = (465526096893, −136571013123, 262648465963, −105613596395, −171427445547, 357101971563, −8663180146, 444057171072, 917260919255, 45921258230, 643961282807, −957205380390). Hence s = 3, r = 8, q = 1. Consider the norm form equation NK/Q (x1 + αx2 + α 2 x3 + α 3 x4 + ξ x5 ) = ±1 in x1 , x2 , x3 , x4 , x5 ∈ Z with x5 = 0.

(6.17)

6.4 Examples

91

We had c1 = 0.5187 and c2 = 3.9495 for all possible i, j1 , j2 . Since our example is a totally real one, we did not have to use a0 . Baker’s method gave A = max(|a1 |, . . . , |a8 |) ≤ 1046 = AB . In the reduction procedure we had dimension 9, c3 = 1/c1 , c4 = 0. The following table summarizes the steps of the reduction procedure. Note that in each step we had to perform 12 reductions.

Step I Step II Step III Step IV

|b1 |> 1048 23884 2277 2024

A< 1046 472 45 40

H = 10440 1040 1035 1030

Precision 1100 digits 100 digits 80 digits 80 digits

New bound for A 472 45 40 34

CPU time 98 min 60 s 60 s 60 s

Hence our algorithm gave the reduced bound AR = 34. In the enumeration process we used I ∗ = {(i123), (i231), (i312)|i = 1, 2, 3, 4}, that is we had t = 12 ellipsoids to consider. The initial bound was S = 0.128 · 10174 that we got using the reduced bound for A. Note that also in this example the vector g is linearly dependent on e1 , . . . , e8 . The following table is a summary of the enumeration process.

Step I Step II Step III Step IV Step V Step VI Step VII Step VIII Step IX Step X Step XI Step XII Step XIII Step XIV Step XV Step XVI

S 10174 1020 1010 108 106 105 104 1000 500 250 120 60 30 15 7 4

s 1020 1010 108 106 105 104 103 500 250 120 60 30 15 7 4

Precision 100 digits 50 digits 50 digits 50 digits 50 digits 50 digits 50 digits 50 digits 50 digits 50 digits 50 digits 50 digits 50 digits 50 digits 50 digits 50 digits

CPU time 15 s 10 s 10 s 8s 5s 15 s 15 s 15 s 10 s 10 s 12 s 10 s 10 s 10 s 10 s 3s

Tuples found 0 2 2 24 26 91 178 57 45 37 60 24 17 18 16 125

92

6 The Resolution of Norm Form Equations

The last line refers to the enumeration of the ellipsoid (2.17) (Sect. 2.3) with S = 4. We tested all tuples we found in the enumeration process if they are solutions of (6.4). We found five solutions of (6.4), and the components were all ≤1 in absolute value. For these tuples we calculated the corresponding solutions of Eq. (6.17). We obtained the following solutions: x1 −2 1 0 0

x2 4 3 1 0

x3 0 0 −1 0

x4 −1 −1 0 0

x5 −1 1 1 1

If (x1 , x2 , x3 , x4 , x5 ) is a solution, then so also is (−x1 , −x2 , −x3 , −x4 , −x5 ), but we list only one of them.

6.5 Further Results on Norm Form Equations There are general criteria on the finiteness of the number of solutions of norm form equations by Schmidt [Schm72], that were extended to the so-called p-adic case by Schlickewei [Schl77]. These celebrated results are ineffective. Effective upper bounds for the absolute values of the solutions of norm form equations were given by Gy˝ory (see for example [Gy81]) under some restrictions on the coefficients.

Chapter 7

Index Form Equations in General

Abstract In this chapter we investigate the structure of the index form equation (1.7). Discovering special properties of the index form, especially factorization properties, makes the resolution of index form equations much easier. In the numerical examples the field K is often the composite of its subfields. This special case is considered in Sect. 7.5. The general results on composite fields have several applications, see for example Sects. 11.4, 13.2, 15.1, and 15.3.

7.1 The Structure of the Index Form Let K be an algebraic number field of degree n with integral basis (1, ω2 , . . . , ωn ) and set (i)

L(i) (X) = X1 + ω2 X2 + . . . + ωn(i) Xn , where γ (i) (1 ≤ i ≤ n) denote the conjugates of any γ ∈ K. Let DK be the discriminant of K. According to (1.6) we have  1 I (X2 , . . . , Xn ) = ± √ Lij (X), |DK | 1≤i