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Thue equations and related topics Akhtari, Shabnam 2008

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Thue Equations and Related Topics by Shabnam Akhtari B.Sc., Sharif University of Technology, 2002 M.Sc., Simon Fraser University, 2004 Ph.D., The University of British Columbia, 2008 A THESIS SUBMITTED IN PARTIAL FULFILMENT OF THE REQUIREMENTS FOR THE DEGREE OF Doctor of Philosophy in The Faculty of Graduate Studies (Mathematics) The University Of British Columbia (Vancouver) August, 2008 c© Shabnam Akhtari 2008 Abstract Using a classical result of Thue, we give an upper bound for the number of solutions to a family of quartic Thue equations. We also give an upper bound upon the number of solutions to a family of quartic Thue inequalities. Using the Thue-Siegel principle and the theory of linear forms in logarithms, an upper bound is given for general quartic Thue equations. As an application of the method of Thue-Siegel, we will resolve a con- jecture of Walsh to the effect that the Diophantine equation aX4 − bY 2 = 1, for fixed positive integers a and b, possesses at most two solutions in positive integers X and Y . Since there are infinitely many pairs (a, b) for which two such solutions exist, this result is sharp. It is also effectively proved that for fixed positive integers a and b, there are at most two positive integer solutions to the quartic Diophantine equation aX4 − bY 2 = 2. We will also study cubic and quartic Thue equations by combining some classical methods from Diophantine analysis with modern geometric ideas. ii Table of Contents Abstract . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ii Table of Contents . . . . . . . . . . . . . . . . . . . . . . . . . . . . iii Acknowledgements . . . . . . . . . . . . . . . . . . . . . . . . . . . vi Dedication . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . vii Statement of Co-Authorship . . . . . . . . . . . . . . . . . . . . . viii 1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 Bibliography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6 2 The Method Of Thue-Siegel For Binary Quartic Forms . 8 2.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . 8 2.2 The Method Of Thue-Siegel . . . . . . . . . . . . . . . . . . 10 2.3 Equivalent Forms . . . . . . . . . . . . . . . . . . . . . . . . 13 2.4 Reduction To A Diagonal Form . . . . . . . . . . . . . . . . 16 2.5 Resolvent Forms . . . . . . . . . . . . . . . . . . . . . . . . . 21 2.6 Gap Principles For The Thue Equation . . . . . . . . . . . . 23 2.7 Gap Principles For The Thue Inequality . . . . . . . . . . . . 25 2.8 Some Algebraic Numbers . . . . . . . . . . . . . . . . . . . . 27 2.9 Approximating Polynomials . . . . . . . . . . . . . . . . . . 32 2.10 An Auxiliary Lemma . . . . . . . . . . . . . . . . . . . . . . 38 2.11 Proof of Theorem 2.1.2 . . . . . . . . . . . . . . . . . . . . . 41 2.12 Representation Of Unity By F (x, y) . . . . . . . . . . . . . . 44 2.12.1 The Proof Of Theorem 2.1.1 for I > 1536 . . . . . . . 44 2.12.2 Forms With Small Discriminant . . . . . . . . . . . . 44 Bibliography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47 iii Table of Contents 3 The Diophantine equation aX4 − bY 2 = 1 . . . . . . . . . . . 49 3.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . 49 3.2 An Equivalent Problem . . . . . . . . . . . . . . . . . . . . . 50 3.3 Reduction To A Family Of Thue Equations . . . . . . . . . . 52 3.4 Padé Approximation . . . . . . . . . . . . . . . . . . . . . . . 56 3.5 Some Algebraic Numbers . . . . . . . . . . . . . . . . . . . . 61 3.6 Three Auxiliary Lemmas . . . . . . . . . . . . . . . . . . . . 64 3.7 The Proof Of Theorem 3.1.1 . . . . . . . . . . . . . . . . . . 72 Bibliography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 76 4 The Diophantine Equation aX4 − bY 2 = 2 . . . . . . . . . . . 79 4.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . 79 4.2 Linear Recurrences . . . . . . . . . . . . . . . . . . . . . . . 80 4.3 Reduction To Thue Equations . . . . . . . . . . . . . . . . . 81 4.4 Lower Bounds For k, t And |y| . . . . . . . . . . . . . . . . . 84 4.5 Associated Fourth Roots Of Unity . . . . . . . . . . . . . . . 87 4.6 The Main Lemma . . . . . . . . . . . . . . . . . . . . . . . . 89 4.6.1 Approximating Polynomials . . . . . . . . . . . . . . 90 4.6.2 Gap Principles . . . . . . . . . . . . . . . . . . . . . . 93 4.6.3 The Proof of Lemma 4.6.1 . . . . . . . . . . . . . . . 96 4.7 An Effective Measure Of Approximation . . . . . . . . . . . 98 4.8 Completion Of The Proof Of Theorem 4.1.1 . . . . . . . . . 105 Bibliography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 108 5 Addendum on the equation aX4 − bY 2 = 2 . . . . . . . . . . 110 Bibliography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 113 6 Geometry Of Quartic Thue Equations . . . . . . . . . . . . . 114 6.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . 114 6.2 Quartic Forms And Elliptic Curves . . . . . . . . . . . . . . 116 6.3 Heights . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 118 6.4 The Thue-Siegel Principle . . . . . . . . . . . . . . . . . . . . 120 6.5 Large Solutions . . . . . . . . . . . . . . . . . . . . . . . . . 121 6.6 Small Solutions . . . . . . . . . . . . . . . . . . . . . . . . . 123 6.7 Forms With Real Roots . . . . . . . . . . . . . . . . . . . . . 125 6.7.1 Exponential Gap Principle . . . . . . . . . . . . . . . 126 6.7.2 Geometry Of The Curve φ(t) . . . . . . . . . . . . . . 130 6.7.3 Linear Forms In Logarithms . . . . . . . . . . . . . . 135 iv Table of Contents Bibliography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 138 7 Geometry of Binary Cubic Forms . . . . . . . . . . . . . . . 140 7.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . 140 7.2 The Covariants of Binary Cubic Forms . . . . . . . . . . . . 142 7.3 Some Functions In The Number Field Q( √−3D) . . . . . . . 144 7.4 Geometric Gap Principles . . . . . . . . . . . . . . . . . . . . 148 7.5 Linear Forms In Logarithms . . . . . . . . . . . . . . . . . . 156 7.6 Proof Of The Main Results . . . . . . . . . . . . . . . . . . 163 Bibliography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 166 8 Conclusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 168 Bibliography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 170 v Acknowledgements It is difficult to overstate my gratitude to my Ph.D supervisor, Professor Michael Bennett. Throughout my thesis-writing period, he provided en- couragement, support and advice. I would have been lost without his help and stimulating suggestions. I was delighted to interact with Professor Greg Martin. I enjoyed all graduate courses he offered during my time in UBC. I would like to thank him for being always so kind and helpful. I was fortunate in having Professor Stephen Choi as my M.Sc supervisor in Simon Fraser University. He is the one who warmly welcomed me in this country and helped me find my way. I am specially thankful to him for his help and advice in finding the right place for my Ph.D study. I am grateful to Professor David Boyd for spending his valuable time reading my thesis. I am also grateful to Professor Ebad Mahmoudian, Professor Siavash Shahshahani and Dr. Yahya Tabesh who kindly and unforgettably made my immigration to Canada possible. I am indebted to Dr. Hadi Jorati for his understanding, for his company and for his honest friendship. I am very thankful to Dr. Julia Gordon for being such a good listener whenever I needed one. Words fail me to express my appreciation to my mother, Zahra Hosseini. She is my first math teacher and the creator of my passion for Mathematics. Selfishly, I left her alone in the most difficult time of her life to pursue my graduate study. My absence was not easy for her but she went through it patiently. I owe her everything. My lovely sisters, Shaghayegh and Nazli Akhtari, exceptionally inspire and enrich my life. I feel a deep sense of gratitude for my father, Mansour Akhtari. His memory still provides strength and persistent courage. It is to him that I dedicate this work. vi Dedication To the memory of Baba. vii Statement of Co-Authorship Chapters 3 and 4 of this thesis is joint work with Dr. Togbe and Dr. Walsh. I used the method explained in Chapter 2 to solve a family of quartic equation which rises from the equation aX4−bY 2 = 2. This project was suggested to me by Dr. Walsh who had been working with Dr. Togbe on finding an upper bound for the number of integral solutions to the equation aX4 − bY 2 = 2. We give an upper bound upon the number of integral solutions of aX4 − bY 2 = 2 for the first time, combining the result of their previous study on the subject with my method of solving Thue equations. viii Chapter 1 Introduction In 1909, Thue [25] derived the first general sharpening of Liouville’s theorem on rational approximation to algebraic numbers, proving, if θ is algebraic of degree n ≥ 3 and  > 0, that there exists a constant c(θ, ) such that∣∣∣∣θ − pq ∣∣∣∣ > c(θ, )q n2+1+ for all p ∈ Z and q ∈ N. In the intervening years, Thue’s result has been improved and generalized by several mathematicians. In 1921, Siegel [21] showed that∣∣∣∣θ − pq ∣∣∣∣ < 1qκ (1.1) has at most finitely many solutions if κ > 2n1/2. In 1940s, Dyson [10] and Gelfond [13] independently improved this to κ > (2n)1/2. Finally in 1955, Roth reduced this condition to κ > 2. Roth’s result is best possible, in the sense that we cannot replace the inequality κ > 2 with κ = 2. The following theorem is an almost immediate consequence of Thue’s result Theorem 1.0.1 (Thue). Let F (x, y) be an irreducible binary form in Z[x, y] of degree at least three and m a nonzero integer, then the equation F (x, y) = m (1.2) has only finitely many integer solutions (x, y). To prove this theorem, apply Thue’s inequality to the roots of F (x, 1) = 0 (see chapter III of [22] for a complete proof). Equations (1.2) are often called Thue equations, in honour of Axel Thue. Another proof of Theorem 1.0.1 was given by Skolem in 1935 [23] using p-adic power series, under certain weak restrictions imposed on F . Nevertheless all of these results, 1 Chapter 1. Introduction starting from that of Thue, are ineffective. In general they do not supply an algorithm to compute all solutions of (1.2). This means that the constant c(θ, ) in the Thue theorem, or constants in any other of its improvements, can not be determined explicitly from the proof as can be done in Liouville’s theorem. Baker’s method (1966) for bounding linear forms in logarithms of algebraic numbers provides an effective, but slight improvement over the Liouville’s theorem and thus an explicit upper bound for the solutions of (1.2). In this thesis, we are mainly interested in upper bounds for the number of solutions to equation (1.2) with degree 4. We will apply ineffective meth- ods similar to those of Thue and Siegel, since they seem to lead to far better estimates in this context than those corresponding to, for instance, the the- orem of Roth. In Thue and Siegel’s papers, the construction of auxiliary polynomials, at least for some algebraic numbers α, was completely explicit, based on the method of Padé approximations and hypergeometric polyno- mials. The method of Padé approximations and evaluation of asymptotics of auxiliary polynomials gives precise values for all constants, and allows one to start with a “good” approximation of a relatively small height. In 1983, J.H. Evertse showed that indeed the number of primitive so- lutions to (1.2), for irreducible F of degree n ≥ 3, could be bounded by a function depending only on n and m, but otherwise independent of F . Evertse [12] obtains the bound 715(( n 3)m+1) 2 + 6× 72(n3)m(ν+1) for the number of primitive solutions of (1.2), where ν is the number of prime factors of the constant term m. This is a special case of Evertse’s results as he also treats equations in number fields. There has developed an extensive body of literature devoted to explicitly solving Thue equations, or bounding the number of such integral solutions; in the latter regard, we mention a result of Bombieri and Schmidt [6]: Theorem 1.0.2. If F is an irreducible binary form of degree n and m is a nonzero integer, then the number of primitive solutions to the Diophantine equation (1.2) is not greater than c0n 1+ν , where c0 is an absolute constant, ν is the number of distinct prime factors of m. 2 Chapter 1. Introduction Further advances on the number of solutions to Thue equations and inequalities were made by Stewart [24]. Theorem 1.0.3. Let F be an irreducible binary form of degree n and m be a given integer. The number of solutions of the Thue inequality |F (x, y)| ≤ m is at most nm2/n(1 + logm1/n). The argument of Bombieri and Schmidt [6] implies that if N is the cor- responding upper bound in the special case m = 1, then Nnν is an upper bound for the number of solutions to (1.2). This makes the Thue equation F (x, y) = 1 of special interest. If F is a linear form, F (x, y) = ax + by say, then (1.2) has solutions in integer (x, y) if and only if the greatest common denominator d of a and b divides m. Moreover, if (x0, y0) is one solution of (1.2), then the other solutions of (1.2) are given by x = x0 + tb d , y = y0 − ta d , where t runs through the non-zero integers. If F is a quadratic form of positive discriminant D, then (1.2) has either no solutions or infinitely many. To decide whether (1.2) has solutions or not, one may use the continued fraction expansion of √ D (see [14] for details). If F is a quadratic form of negative discriminant D then (1.2) has at most finitely many solutions. This is because each solution (x, y) to ax2 + bxy + cy2 = m will satisfy (2ax+ by)2 −Dy2 = 4am, thus |y| ≤ ∣∣∣∣4amD ∣∣∣∣1/2 . Dirichlet gave an upper bound for the number of solutions to (1.2), when F is a quadratic form with negative discriminant (see [9]). 3 Chapter 1. Introduction The first important result on the number of solutions of cubic Thue equations was obtained by Siegel. Nagell [17] and Delone [8] independently showed that (1.2) has at most five solutions in integers x, y where F is a cubic irreducible binary form with negative discriminant. This bound is sharp, for if F (x, y) = x3 − xy2 + y3 then F has discriminant −23 and the equation F (x, y) = 1 has 5 solutions, namely (1, 0), (0, 1), (−1, 1), (1, 1) and (4,−3). Cubic Thue equations with positive discriminant have been treated in several papers including [11], [5] and [18]. We will adjust and use the ideas and techniques from these three papers to obtain new results for quartic Thue equations. The problem of determining upper bound for the number of integer points on elliptic curves has received considerable attention. Ljunggren de- rived remarkable sharp bounds for the number of solutions to various quartic Diophantine equations, particularly those of the shape aX4 − bY 2 = ±1. However, for general a and b, there had previously been no absolute upper bound for the number of integral solutions to aX4 − bY 2 = 1. In Chapter 2, we will reduce this equation to a quartic Thue inequality and will apply the method of Chapter 1 to solve the inequality. Mordell [16], shows that all the rational points of a cubic curve f(x, y, z) = 0 of genus one, can be found from a finite number by the chord and tangent process. To prove this, he first explains how the problem is equivalent to that of finding the integer solutions in x, y, z of an equation with integer coefficients: ax4 + bx3y + cx2y2 + dxY 3 + ey4 = z2. This fact motivates Chapters 1 and 4 of this thesis. Here we will give a survey of the contents of this thesis. Chapter 2 [3] is inspired by a paper of J.H. Evertse [11] and its improve- ment by M.A. Bennett [5], where the method of Thue-Siegel is used to give 4 Chapter 1. Introduction an upper bound for the number of solutions to cubic Thue equations. In Chapter 1, we give an upper bound for the number of solutions to a family of quartic Thue equations. We also give an upper bound upon the number of solutions to a family of quartic Thue inequalities. Chapter 3 [2] resolves a conjecture of P.G. Walsh and sharpens classical work of W. Ljunggren. The main result is Theorem 1.0.4. Let a and b be positive integers. Then equation aX4 − bY 2 = 1 (1.3) has at most two solutions in positive integers (X,Y ). Since there are infinitely many pairs (a, b) for which two such solutions exist, this result is best possible.To obtain the main result of [2] , we have appealed to classical results of Thue from the theory of Diophantine approx- imation, together with modern refinements, particularly those of Evertse. We apply these techniques to particular families of quartic inequalities (the general machinery is developed in [3]). Similar arguments lead to the result of Chapter 4 [4], on the equation aX4− bY 2 = 2, which arose from commu- nication with P.G. Walsh. Chapter 5 is a short chapter explaining how the result of Chapter 4 can be extended to a more general family of equations. In Chapter 6 [1], we combine techniques from Algebraic Number Theory with the theory of linear forms in logarithms (Baker’s method), to obtain a clearer view of the geometry of quartic Thue equations. Chapter 4 presents an extension of some geometric ideas of Okazaki [18] and an application of Stewart’s method [24] for Thue equations with arbitrary degree n to quartic ones. In Chapter 7, we will give a survey of the know results for cubic Thue equations and we’ll give slight improvment to them. We will study different methods used to give an upper bound for the number of integral solutions to cubic Thue equations. In Chapter 8, our concluding chapter, we discuss our future plans for research along the lines of this thesis. 5 Bibliography [1] S. Akhtari. Geometry of quartic Thue equations. preprint. [2] S. Akhtari. The Diophantine equation aX4− bY 2 = 1. to appear in J. Reine Angew. Math. (2008). [3] S. Akhtari, The method of Thue-Siegel for binary quartic forms. sub- mitted (2007). [4] S. Akhtari, A. Togbe and G. Walsh. The Diophantine equation aX4− bY 2 = 2. to appear in Acta. Arith.(2008). [5] M.A. Bennett. On the representation of unity by binary cubic forms. Trans. Amer. Math. Soc. 353 (2001), 1507-1534. [6] E. Bombieri and W.M. Schmidt. On Thue’s equation. Invent. Math. 88 (1987), 69-81. [7] G.V. Chudnovsky. On the method of Thue-Siegel. Ann. of Math. II Ser. 117 (1983), 325-382. [8] B.N. Delone. Über die Darstellung der Zahlen durch die binären kubis- chen Formen von negativer Diskriminante, Math. Zeitschr. 31 (1930), 1-26. [9] P.G.L. Dirichlet. Vorlesungen über zahlentheorie, 4th ed., Vieweg & Sohn, Braunschweig (1894). [10] F.J. Dyson. The approximation of algebraic numbers by rationals, Acta Math.79 (1947), 225-240. [11] J.H. Evertse. On the representation of integers by binary cubic forms of positive discriminant. Invent. Math.73(1983), 117-138. [12] J.H. Evertse. Upper bounds for the number of solutions of Diophantine equations. Math. Centrum. Amsterdam( 1983). 6 Bibliography [13] A.O. Gelfond. Transcendental and Algebraic Numbers. Dover pub. New York (1960). [14] L.K. Hua. Introduction to Number Theory, Springer Verlag, Berlin, Heidelberg, New York (1982). [15] W. Ljunggren. On the representation of integers by certain binary cubic and biquadratic forms. Acta. Arith. XVII (1971). [16] L.J. Mordell. On the rational solutions of indeterminate equations of the 3rd and 4th degrees. Proc. Camb. Phil. Soc.21 (1922), 179-192. [17] T. Nagell. Darstellung ganzer Zahlen durch bind̈re kubischr formen mit negativer Diskriminante, Math. Zeitscher. 28 (1928), 10-29. [18] R. Okazaki. Geometry of a cubic Thue equation, Publ. Math. Debre- cen. 61 (2002),267-314. [19] K.F. Roth. Rational approximation to algebraic numbers, Mathe- matika 2 (1955) 1-20. [20] C.L. Siegel. Über einige Anwendungen diophantischer Approximatio- nen, Abh. Preuss. Akad. Wiss. 1 (1929). [21] C.L. Siegel. Approximation algebraischer Zahlen, Math. Zeitschr. 10 (1921), 173-213. [22] W.M. Schmidt, Diophantine Approximations and Diophantine Equa- tions, Lecture Notes in Mathematics, Springer Verlag (2000). [23] Th. Skolem, Einige Sätze über p-adische Potenzreihen mit Anwendung auf gewisse exponentielle Gleichungen, Math. Ann. 111 (1935), 399- 424. [24] C. L. Stewart, On the number of solutions of polynomial congruences and Thue equations, J. Amer. Math. Soc., 4 (1991), 793-838. [25] A. Thue. Über Annäherungenswerte algebraischen Zahlen. J. reine angew. Math. 135 (1909), 284-305. 7 Chapter 2 The Method Of Thue-Siegel For Binary Quartic Forms 1 2.1 Introduction In 1909, Thue [17] proved that if F (x, y) is an irreducible binary form of degree at least 3 with integer coefficients, and h a nonzero integer, then the equation F (x, y) = h has only finitely many solutions in integers x and y . In this paper we will consider irreducible binary quartic forms with in- teger coefficients, i.e. polynomials of the shape F (x, y) = a0x4 + a1x3y + a2x2y2 + a3xy3 + a4y4. The discriminant D of F is given by D = a60(α1 − α2)2(α1 − α3)2(α1 − α4)2(α2 − α3)2(α2 − α4)2(α3 − α4)2, where α1 , α2, α3 and α4 are the roots of F (x, 1) = a0x4 + a1x3 + a2x2 + a3x+ a4. The invariants of F form a ring, generated by two invariants of weights 4 and 6, namely I = IF = a22 − 3a1a3 + 12a0a4 and J = JF = 2a32 − 9a1a2a3 + 27a21a4 − 72a0a2a4 + 27a0a23. These are algebraically independent and every invariant is a polynomial in I and J . For the invariant D, we have 27D = 4I3 − J2. 1A version of this paper has been submitted for publication. Akhtari. S. The Method Of Thue-Siegel For Binary Quartic Forms. 8 Chapter 2. The Method Of Thue-Siegel For Binary Quartic Forms In what follows, we will just consider the forms F for which the quantity JF is 0; i.e. for which we have 27D = 4I3. Let h be a positive integer. The number of solutions in integers x and y of the equation |F (x, y)| = h. (2.1) will be the focus of our study in this paper. Theorem 2.1.1. Let F (x, y) be an irreducible binary quartic form with integer coefficients and positive discriminant that splits in R. If JF = 0, then the Diophantine equation |F (x, y)| = 1 possesses at most 12 solutions in integers x and y (with (x, y) and (−x,−y) regarded as the same) . Each positive definite form is equivalent to a reduced form (see [8]). We give the definition of reduced form in section 2.7. Theorem 2.1.2. Let F (x, y) be a reduced irreducible binary quartic form with integer coefficients and positive discriminant that splits in R. If JF = 0, then the inequality |F (x, y)| ≤ h possesses at most 12 solutions (x, y), with |y| ≥ 2h3/4 (3I)1/8 . In section 2.2, we will show that to apply a classical theorem of Thue [17] from Diophantine approximation to a quartic form F , one needs to assume JF = 0. Another reason for us to be interested in these results, despite what are apparently quite serious restriction upon F , is that we know important families of quartic forms with these properties. For example a solution to the equation aX4 − bY 2 = 1 gives rise to a solution to the Thue equation x4 + 4tx3y − 6tx2y2 − 4t2xy3 + t2y4 = t21, where t1|t. We have applied the methods of this paper to treat the above Thue equation in [1]. The method of Thue and Siegel based on Padé approximation to bi- nomial functions has been used to study binary cubic forms with positive discriminant, for decades. In 1939, Krechmar [10] showed that when the discriminant of quartic form F (x, y) is sufficiently large ( DF  h216/5), the equation F (x, y) = h has at most 20 solutions in integers x and y, provided that JF = 0 and all roots of F (x, 1) are real numbers. 9 Chapter 2. The Method Of Thue-Siegel For Binary Quartic Forms 2.2 The Method Of Thue-Siegel The relationship between a system of approximations to an arbitrary cubic irrationality and Padé approximations to 3 √ 1− x was first established by Thue [19]. Siegel [12], [13] extended Thue’s result [19], via hypergeometric polynomials to sharpen the bounds for the number of solutions of Diophan- tine equation f(x, y) = k, for certain binary forms f(x, y) of degree r. He also established bounds for the number of solutions to axn − byn = c, where n ≥ 3 [14]. To find an upper bound for the number of solutions to F (x, y) = 1, we use the method of Thue-Siegel which is based on the following result of Thue. Lemma 2.2.1. Suppose that P (x) is a polynomial of degree n and there is a quadratic polynomial U(x) such that U(x)P ′′(x)− (n− 1)U ′(x)P ′(x) + n(n− 1) 2 U ′′(x)P (x) = 0. (2.2) Let Y (x) = 2U(x)P ′(x)− nU ′(x)P (x) and h = n2 − 1 4 ( U ′(x)2 − 2U(x)U ′′(x)). Consider the recurrences Pr+1(x) = krY (x)Pr(x)− P (x)2Pr−1(x); Qr+1(x) = krY (x)Qr(x)− P (x)2Qr−1(x), with the initial conditions P0(x) = Q0(x) = 2 3 h, P1(x) = U(x)P ′(x)− n− 12 U ′(x)P (x), Q1(x) = xP1(x)− U(x)P (x), where c1 = 3 2 , c2 = 2(2n− 1)(2n+ 1) 3(n− 1)(n+ 1) h, krcr = 2r + 1 2 10 Chapter 2. The Method Of Thue-Siegel For Binary Quartic Forms and cr+1 − cr−1 kr = 2h n2 (n− 1)(n+ 1) . Then polynomials Pr(x), Qr(x) are of degree rn+ 1 and satisfy equation αPr(x)−Qr(x) = (x− α)2r+1Rr(x) for a polynomial Rr(x) . To apply Theorem 2.2.1 to the polynomial P (x) = a0x4+ a1x3+ a2x2+ a3x+ a4, suppose that for a quadratic polynomial U(x) = u2x2 + u1x+ u0, we have 0 = U(x)P ′′(x)− 3U ′(x)P ′(x) + 6U ′′(x)P (x) = (12a0u0 − 3a1u1 + 2a2u2)x2 + (6a1u0 − 4a2u1 + 6a3u2)x +2a2u0 − 3a3u1 + 12a4u2. This implies that  12a0 −3a1 2a23a1 −2a2 3a3 2a2 −3a3 12a4  u0u1 u2  = 0. Therefore, det  12a0 −3a1 2a23a1 −2a2 3a3 2a2 −3a3 12a4  = 4(2a32 − 9a1a2a3 + 27a21a4 − 72a0a2a4 + 27a0a23) = 4J = 0. In this paper, we always suppose that J = 0. In section 2.5, we will show that if JF = 0 then there are linear forms ξ = ξ(x, y) and η = η(x, y) so that F (x, y) = −1 8 √ 3IA4 ( ξ4 − η4) . We will use Padé approximation via hypergeometric polynomials to approx- imate ηξ with rational numbers. The main idea here is to replace the con- struction of a family of dense approximations to ξη , by a family of rational approximations to the function (1 − z)1/4. Consider the system of linear 11 Chapter 2. The Method Of Thue-Siegel For Binary Quartic Forms forms Rr(z) = −Qr(z) + (1 − z)1/4Pr(z) that approximate (1 − z)1/4 at z = 0, such that Rr(z) = z2r+1R̄r(z), R̄r(z) is regular at z = 0, and Pr(z) and Qr(z) are polynomials of degree r. Thue [16], [18] explicitly found poly- nomials Pr(z) and Qr(z) (see Lemma 2.2.1) and Siegel [12] identified them in terms of hypergeometric polynomials. Refining the method of Siegel, Ev- ertse [9] used the theory of hypergeometric functions to give an upper bound for the number of solutions to the equation f(x, y) = 1, where f is a cubic binary form with positive discriminant. Here we adjust Lemma 4 of [9] for quartic forms. Lemma 2.2.2. Let r, g be integers with r ≥ 1, g ∈ {0, 1}. Put Ar,g(z) = r∑ m=0 ( r − g + 14 m )( 2r − g −m r − g ) (−z)m, Br,g(z) = r−g∑ m=0 ( r − 14 m )( 2r − g −m r ) (−z)m. (2.3) (i) There exists a power series Fr,g(z) such that for all complex numbers z with |z| < 1 Ar,g(z)− (1− z)1/4Br,g(z) = z2r+1−gFr,g(z) (2.4) and |Fr,g(z)| ≤ ( r−g+1/4 r+1−g )( r−1/4 r )( 2r+1−g r ) (1− |z|)− 12 (2r+1−g). (2.5) (ii) For all complex numbers z with |1− z| ≤ 1 we have |Ar,g(z)| ≤ ( 2r − g r ) . (2.6) (iii) For all complex numbers z 6= 0 and for h ∈ {1, 0} we have Ar,0(z)Br+h,1(z) 6= Ar+h,1(z)Br,0(z). (2.7) Proof. This lemma has been proven in [1]. 12 Chapter 2. The Method Of Thue-Siegel For Binary Quartic Forms 2.3 Equivalent Forms We will call forms F1 and F2 equivalent if they are equivalent under SL2(Z)- action (i.e. if there exist integers b, c, d and e such that F1(bx+ cy, dx+ ey) = F2(x, y) for all x and y, where be− cd = ±1). Denote by NF the number of solutions in integers x and y of the Diophantine equation |F (x, y)| = h. Note that if F1 and F2 are equivalent, then NF1 = NF2 , IF1 = IF2 and JF1 = JF2 . Let us define, for a quartic form F , an associated quartic form, the Hessian H, by H(x, y) = δ2F δx2 δ2F δy2 − ( δ2F δxδy )2 . Then H(x, y) = A0x4 +A1x3y +A2x2y2 +A3xy3 +A4y4, where A0 = 3(8a0a2 − 3a21), A1 = 12(6a0a3 − a1a2), A2 = 6(3a1a3 + 24a0a4 − 2a22), (2.8) A3 = 12(6a1a4 − a2a3), A4 = 3(8a2a4 − 3a23). We have the following identities (see Proposition 5 of [8]): IH = 122I2F , (2.9) JH = 123(2I3F − J2F ) and DH = 126J2FDF , where H is the Hessian of F and DF , DH are the discriminants of F and H, respectively . We note that A0A 2 3 −A4A21 = 123(a0a23 − a4a21)JF 13 Chapter 2. The Method Of Thue-Siegel For Binary Quartic Forms and A33 + 8A1A 2 4 − 4A2A3A4 = 123(a33 + 8a1a24 − 4a2a3a4)JF . When JF = 0, we obtain A0A 2 3 = A4A 2 1, (2.10) A33 + 8A1A 2 4 = 4A2A3A4. Therefore, H(x, y) = A0x4 +A1x3y +A2x2y2 +A3xy3 +A4y4 = 1 4A23A4 (2A1A4x2 +A23xy + 2A4A3y 2)2, when A3A4 6= 0. Whereby, from (2.9),∣∣A43 − 16A1A24A3∣∣ = ∣∣48A23A4IF ∣∣ . (2.11) In order to make good use of the above identities, we prove the following lemma: Lemma 2.3.1. Let F (x, y) be a quartic form with JF = 0. There exists a form equivalent to F (x, y), for which A3A4 6= 0. Proof. If A4 = 0, then by (2.10) we have A4 = A3 = 0 and therefore, H(x, y) = x2(A0x2 +A1xy +A2y2). Let x = mX + lY and y = pX + qY, where m, l, p and q are integers satisfying mq − lp = ±1. Suppose that φ(X,Y ) is equivalent to F (x, y) under this substitution with Hessian Hφ(X,Y ) = A′0X 4 +A′1X 3Y +A′2X 2Y 2 +A′3XY 3 +A′4Y 4. We have, A′4 = Hφ(0, 1) = HF (l, q) = l 2(A0l2 +A1lq +A2q2). We have assumed that F (x, y) is irreducible, henceHF (x, y) is not identically zero. Therefore, the integers l and q can be chosen so that A′4 6= 0. 14 Chapter 2. The Method Of Thue-Siegel For Binary Quartic Forms Let t ∈ Z and put M = m+ lt, P = p+ qt. Let φ(X,Y ) be the equivalent form to F (x, y) under the substitution x =MX + lY and y = PX + qY. Then A′3 = 4Ml 3A0 + (l3P + 3Ml2q)A1 + (2l2Pq + 2Mlq2)A2 +(q3M + 3Pq2l)A3 + 4Pq3A4 = (m+ lt)(4l3A0 + 3l2qA1 + 2lq2A2 + q3A3) +(p+ qt)(l3A1 + 2l2qA2 + 3lq2A3 + 4q3A4) = K + 4t(l4A0 + l3qA1 + l2q2A2 + lq3A3 + q4A4) = K + 4tA′4. Since A′4 6= 0, the integer t can be chosen so that A′3 6= 0. In the following, we will show that F (x, y) or one of its equivalences (under GL2(Z)-action) satisfies |A4| < 4I. From now on, we will suppose that A3A4 6= 0. Let x = mX + lY and y = pX + qY, where m, l, p and q are integers satisfying mq − lp = ±1. Let φ(X,Y ) be equivalent to F (x, y) under this substitution and φ(X,Y ) = a′0X 4 + a′1X 3Y + a′2X 2Y 2 + a′3XY 3 + a′4Y 4. We observe that A′4 = Hφ(0, 1) = HF (l, q), where Hφ(X,Y ) = A′0X4 +A′1X3Y +A′2X2Y 2 +A′3XY 3 +A′4Y 4. To continue, we will be in need of the following Proposition due to Her- mite. 15 Chapter 2. The Method Of Thue-Siegel For Binary Quartic Forms Proposition 2.3.2. Suppose that f11x2 + 2f12xy + f22y2 is a binary form with D = f11f22 − f212 6= 0. Then there is an integer pair (u1, u2) 6= (0, 0) for which 0 < |f11u21 + 2f12u1u2 + f22u21| < √ 4 3 |D|. Proof. See [5], page 31. Proposition 2.3.2 implies that we can choose l and q, such that 0 < |A′4| = 1 |4A23A4| (2A1A4l2 +A23lq + 2A4A3q 2)2 < 1 |4A23A4| ∣∣∣∣13(A43 − 16A1A24A3) ∣∣∣∣ = 4I, where the last equality comes from (2.11). 2.4 Reduction To A Diagonal Form Our goal in this section will be to reduce the problem at hand to considera- tion of diagonal forms over a suitable imaginary quadratic field. The method of Thue-Siegel is particularly well suited for application to such forms. We will show that Lemma 2.4.1. Let F be the binary form in Theorem 2.1.1. Then F (x, y) = 1 96A23A4 √−3I ( ξ4(x, y)− η4(x, y)) , where ξ and η are complex conjugate linear forms in x and y. Let H(x, y) = A0x4 + A1x3y + A2x2y2 + A3xy3 + A4y4 with A3A4 6= 0, be the Hessian of F (x, y). Suppose that x = mξ + lη and y = pξ + qη, with ξ(x, y)η(x, y) = 2A1A4x2 +A23xy + 2A4A3y 2, 16 Chapter 2. The Method Of Thue-Siegel For Binary Quartic Forms for some m, l, p, q ∈ C. Let ∆ = mq − lp. Therefore, F (x, y) = F (mξ + lη, pξ + qη) = a′0ξ 4 + a′1ξ 3η + a′2ξ 2η2 + a′3ξη 3 + a′4η 4 = Φ(ξ, η) and ξ = λ(αx+ βy), η = µ(γx+ δy) and λµ = 1 (the values of λ and µ will be determined later). The Hessian H ′(ξ, η) of Φ(ξ, η) satisfies H ′(ξ, η) = A′0ξ 4 +A′1ξ 3η +A′2ξ 2η2 +A′3ξη 3 +A′4η 4 = ∆2H(x, y) = ∆2 4A23A4 ξ2η2. Hence, A′0 = A ′ 1 = A ′ 3 = A ′ 4 = 0; A ′ 2 = ∆ 2 1 4A23A4 . (2.12) On the other hand, A′0 = 3(8a ′ 0a ′ 2 − 3a ′2 1 ) and A′1 = 12(6a ′ 0a ′ 3 − a′1a′2). It is easy to check that for any form F (x, y), −10a4A0 + 2a3A1 − a2A2 + a1A3 − 2a0A4 = 6J. So, for Φ(ξ, η), we obtain −10a′4A′0 + 2a′3A′1 − a′2A′2 + a′1A′3 − 2a′0A′4 = 6JΦ = 6∆6JF = 0, where a′i are the coefficients of Φ and A ′ i are the coefficients of its Hessian. Therefore, by (2.12), a′2 = 0 and by (2.8), a′1 = a ′ 3 = 0, whereby, F (x, y) = Φ(ξ, η) = a′0ξ 4 + a′4η 4. Our goal now is to determine the value of λ and µ = 1λ in ξ = λ(αx+βy) and η = µ(γx+ δy), so that a′4 = −a′0. We have a′0 = F (m, p) = F (−δp γ , p ) = p4 γ4 F (−δ, γ) 17 Chapter 2. The Method Of Thue-Siegel For Binary Quartic Forms and a′4 = F (l, q) = F (−βq α , q ) = q4 α4 F (−β, α). Also ( m l p q )( λα λβ µγ µδ ) = ( 1 0 0 1 ) , whereby, q = −λpα µγ . Therefore, to obtain −a′0 = a′4, we may take λ and µ so that µ8 = µ4 λ4 = −F (−β, α) F (δ,−γ) . We have shown that F (x, y) can be written as a′0 ( ξ4(x, y)− η4(x, y)), where ξ = λ(αx+ βy), η = µ(γx+ δy) (2.13) and λµ = 1. It remains to calculate the value of a′0. Using (2.12) and (2.8), we get A′2 = ∆ 2 1 4A23A4 = 6(3a′1a ′ 3 + 24a ′ 0a ′ 4 − 2a′22 ) = 144a′0a′4. Substituting a′4 by −a′0, we obtain a ′2 0 = − ∆2 242A23A4 = − 1 (αδ − βγ)2242A23A4 . To calculate (αδ − βγ)2, we recall that 2A1A4x2 +A23xy + 2A3A4y 2 = (αx+ βy)(γx+ δy), consequently, by (2.11), (αδ − βγ)2 = A43 − 16A1A24A3 = 48A23A4I (2.14) and therefore, a′0 = ± 1 96A23A4 √−3I , where I = IF . 18 Chapter 2. The Method Of Thue-Siegel For Binary Quartic Forms We will assume, without loss of generality, that a′0 = 1 96A23A4 √−3I . (2.15) For the binary form F (x, y) with Hessian H(x, y), the sextic covariant Q(x, y) is defined by Q(x, y) = δF δx . δH δy − δF δy . δH δx . Since we have taken H(x, y) = 1 4A23A4 (2A1A4x2+A23xy+2A4A3y 2)2, we may write Q(x, y) = 1 2A23A4 (2A1A4x2 +A23xy + 2A4A3y 2).ψ(x, y), where ψ(x, y) = (A23x+ 4A3A4y) δF δx − (4A1A4x+A23y) δF δy . We have (see equation (25) of [8]) 16H3 + 9Q2 = 44 × 33IHF 2. Let us set W (x, y) = 2A1A4x2 +A23xy + 2A4A3y 2 to get Q(x, y) = 1 2A23A4 W (x, y)ψ(x, y) and W 4(x, y) + 9A23A4ψ 2(x, y) = 44 × 33IA43A24F 2(x, y). (2.16) Since W (x, y) = ξη and F (x, y) = a′0(ξ4 − η4), (2.16) implies that (ξ4 + η4)2 = −36A23A4ψ2(x, y) (2.17) and therefore, by (2.10) ξ4 + η4 = ±6A 2 3 A1 √ −A0ψ(x, y). Note that if all roots of F (x, 1) are real then I > 0 and A0 < 0 ( see [7], Exercise 1 on page 217). So for every x, y ∈ Z, there is b ∈ Q, such that ξ4(x, y) + η4(x, y) = b √ −A0. 19 Chapter 2. The Method Of Thue-Siegel For Binary Quartic Forms We have also seen that for integers x, y ∈ Z, there is a ∈ Z such that ξ4(x, y)− η4(x, y) = i a √ 3I. Therefore, for integers x, y, the quantities ξ4(x, y) and η4(x, y) are complex conjugates and integral in Q (√−A0,√−3I). Moreover, √−A0 ξ4(x, y) and√−A0 η4(x, y) are algebraic integers in Q(√A0I/3). We will work in the number field K = Q (√ A0I/3 ) . We also have, for every pair of integers (x, y), ξ4(x, y) η4(x, y) = b √−A0 + ia √ 3I b √−A0 − ia √ 3I = −A0b2 − 3a2I + i6ab √−A0I/3 −A0b2 + 3a2I , where b ∈ Q and a ∈ Z depend on (x, y). Therefore, ξ4(x, y) η4(x, y) ∈ Q( √ A0I/3), for integers x and y. Note that, in (2.13), we started with two linear forms and continued with their fourth powers. Let the linear form ξ = ξ(x, y) be a fourth root of ξ4(x, y) and define η(x, y) = ξ̄(x, y). Indeed, η(x, y) is a fourth root of η4. Hence, when F (x, 1) splits in R, we can define the complex conjugate linear forms ξ(x, y) and η(x, y), so that ξ4 − η4 = 96A23A4 √−3IF (x, y) and |ξη| = ∣∣2A1A4x2 +A23xy + 2A4A3y2∣∣ . Observe that if the pair (ξ, η) satisfies the above identities, then there are precisely three others satisfying these identities, given by (iξ,−iη) , (−ξ,−η) and (−iξ, iη), where i = √−1. We will, however, work with (ξ, η), a fixed choice of complex conjugate forms. Let (x1, y1) and (x2, y2) be two distinct pairs of integers. Then we have( λα λβ µγ µδ )( x1 x2 y1 y2 ) = ( ξ1 ξ2 η1 η2 ) . If x1y2 − x2y1 is a nonzero integer then by (2.14), we get |ξ1η2 − ξ2η1| = |(αδ − βγ)(x1y2 − x2y1)| ≥ 12A 2 3 |A1| √ |IA0| 3 . (2.18) 20 Chapter 2. The Method Of Thue-Siegel For Binary Quartic Forms 2.5 Resolvent Forms Suppose that ξ and η are linear forms in Lemma 2.4.1. Let us define ξ′ = ξ (12A23)1/4|A4|1/8 and η′ = η (12A23)1/4|A4|1/8 , so that F (x, y) = −1 8 √ 3IA4 ( ξ ′4(x, y)− η′4(x, y) ) . From (2.8), it is easy to see that all coefficients ofW 2(x, y) = (4A23A4)H(x, y) are multiples of 12A23|A4|. By (2.16), we conclude that all coefficients of ψ2(x, y) are multiples of 16A23|A4|. Using (2.17), we conclude that the real part of ξ4 has the factor 12A23A4 . Since ξ 4 − η4 = a′0F , by (2.15), the imaginary part of ξ4 has also the factor 12A23A4. So, for x, y ∈ Z, we have ξ4 |12A23A4| , η4 |12A23A4| ∈ Q (√ −A0, √−3I ) . By (2.10), √−A4 ξ4 |12A23A4| , √−A4η4 |12A23A4| ∈ Q (√ A0I/3 ) , for x, y ∈ Z. From now on, we will consider the following diagonal form representation for F : F (x, y) = −1 8 √ 3IA4 ( ξ4(x, y)− η4(x, y)) . (2.19) For this new choice of ξ and η, we have |ξη| = ∣∣∣∣∣∣2A1A4x 2 +A23xy + 2A4A3y 2√ 12A23 √|A4| ∣∣∣∣∣∣ = √ H √|A4| 3 . (2.20) From (2.18), we have |ξ1η2 − ξ2η1| ≥ 2 √ I |A4|1/4 . (2.21) We also have ξ4(x, y) = ±ψ(x, y) 4A4 ± 4F (x, y) √ 3IA4. (2.22) Lemma 2.4.1 can hence be restated as follows: 21 Chapter 2. The Method Of Thue-Siegel For Binary Quartic Forms Lemma 2.5.1. Let F be the binary form in Theorem 2.1.1. Then F (x, y) = 1 8 √ 3IA4 ( ξ4(x, y)− η4(x, y)) , where ξ and η are complex conjugate linear forms in x and y. Furthermore, for integers x, y we have ξ4, η4 ∈ Q (√ A0I/3 ) . Note that by (2.10), Q( √ A0I/3) = Q( √ A4I/3). We call a pair of complex conjugates ξ and η satisfying the above iden- tities a pair of resolvent forms, and note that if (ξ, η) is one pair, there are precisely three others, given by (iξ,−iη) , (−ξ,−η) and (−iξ, iη), where i = √−1. We will, however, work with (ξ, η), a fixed pair of resolvent forms. Let ω be a fourth root of unity (for some j ∈ {1, 2, 3, 4}, let ω = e 2jpii4 ). We say that the integer pair (x, y) is related to ω if∣∣∣∣ω − η(x, y)ξ(x, y) ∣∣∣∣ = min0≤k≤3 ∣∣∣∣e2kpii/4 − η(x, y)ξ(x, y) ∣∣∣∣ . Let us define z = 1− ( η(x,y) ξ(x,y) )4 , where (ξ, η) is a fixed pair of resolvent forms ( in other words, ηξ is a fourth root of (1− z)). We have |1− z| = 1 , |z| < 2. Lemma 2.5.2. Let ω be a fourth root of unity and the integral pair (x, y) satisfies F (x, y) = −1 8 √ 3IA4 (ξ4(x, y)− η4(x, y)) = h, with ∣∣∣∣ω − η(x, y)ξ(x, y) ∣∣∣∣ = min0≤k≤3 ∣∣∣∣e2kpii/4 − η(x, y)ξ(x, y) ∣∣∣∣ . If |z| ≥ 1 then ∣∣∣∣ω − η(x, y)ξ(x, y) ∣∣∣∣ ≤ |z|. (2.23) If |z| < 1 then ∣∣∣∣ω − η(x, y)ξ(x, y) ∣∣∣∣ < pi12 |z|. (2.24) 22 Chapter 2. The Method Of Thue-Siegel For Binary Quartic Forms Proof. If 1 ≤ z = ∏ 0≤k≤3 ∣∣∣∣e2kpii/4 − η(x, y)ξ(x, y) ∣∣∣∣ , then ∣∣∣∣ω − η(x, y)ξ(x, y) ∣∣∣∣ = min0≤k≤3 ∣∣∣∣e2kpii/4 − η(x, y)ξ(x, y) ∣∣∣∣ ≤ z. Now suppose that |z| < 1. Let 4θ = arg ( η(x, y)4 ξ(x, y)4 ) . We have √ 2− 2 cos(4θ) = |z| < 1. Therefore, |θ| < pi12 . Since ∣∣∣∣ω − η(x, y)ξ(x, y) ∣∣∣∣ ≤ |θ|, we obtain ∣∣∣∣ω − η(x, y)ξ(x, y) ∣∣∣∣ ≤ 14 |4θ|√2− 2 cos(4θ) ∣∣∣∣1− η(x, y)4ξ(x, y)4 ∣∣∣∣ . From the fact that |4θ|√ 2−2 cos(4θ) < pi 3 whenever 0 < |θ| < pi12 , we conclude∣∣∣∣ω − η(x, y)ξ(x, y) ∣∣∣∣ < pi12 |z|, as desired. 2.6 Gap Principles For The Thue Equation Suppose that we have distinct solutions to |F (x, y)| ≤ h indexed by i, say (xi, yi), related to ω with |ξ(xi+1, yi+1)| ≥ |ξ(xi, yi)|. For concision, we will write ηi = η(xi, yi) and ξi = ξ(xi, yi). By (2.21), |ξiηi+1 − ξi+1ηi| ≥ 2 √ I |A4|1/4 . 23 Chapter 2. The Method Of Thue-Siegel For Binary Quartic Forms On the other hand, by (2.23) and (2.24), we have |ξiηi+1 − ξi+1ηi| = |ξi(ηi+1 − ωξi+1)− ξi+1(ηi − ωξi)| ≤ 8h √ |3I A4| ( |ξi| |ξ3i+1| + |ξi+1| |ξ3i | ) ≤ 8h √ |3I A4| ( 2 |ξi+1| |ξ3i | ) , since we assumed |ξi| ≤ |ξi+1|. Combining this with (2.21), we conclude |ξi+1| ≥ 1 8 √ 3h |A4|1/4 |ξi|3. (2.25) Let us now assume that there are 4 distinct solutions to |F (x, y)| = h related to a fixed choice of ω, corresponding to ξ−1, ξ0, ξ1 and ξ2, where |ξ−1| < |ξ0| < |ξ1| < |ξ2|. We will deduce a contradiction, which shows that at most 3 such solutions can exist under certain assumptions. By (2.25), |zj+1| ≤ |zj | 3192h2 I , where zi = 1 − η 4 i ξ4i = ±8h √ 3I A4 ξ4i . Since |z−1| ≤ 2, if I > 1536h2 then |z0| , |z1| , |z2| < 1. Suppose that |z0| < 1. By (2.23) and (2.24), |ξ−1η0 − ξ0η−1| = |ξ−1(ωη0 − ξ0)− ξ0(ωη−1 − ξ−1)| < 8h(1 + pi 12 ) √ |3I A4| ( |ξ0| |ξ3−1| ) . Combining this with (2.21), we conclude |ξ0| > √ 3 (12 + pi)h |A4|1/4 |ξ−1|3. Similarly, we get |ξ0η1 − ξ1η0| = |ξ0(ωη1 − ξ1)− ξ1(ωη0 − ξ0)| < 8h √ |3I A4| pi12 ( |ξ0| |ξ31 | + |ξ1| |ξ30 | ) ≤ 4pi 3 h √ |3I A4| ( |ξ1| |ξ30 | ) , which leads to |ξ1| > √ 3 2pih |A4|1/4 |ξ0|3 > 92pih4(12 + pi)3 |A4| |ξ−1| 9. (2.26) 24 Chapter 2. The Method Of Thue-Siegel For Binary Quartic Forms Note that ∣∣∣∣8h√|3I A4|ξ4−1 ∣∣∣∣ = |z−1| = ∣∣∣∣1− (η−1ξ−1 )4 ∣∣∣∣ ≤ 2 and therefore, |ξ−1|4 > 4h √ |3I A4|. (2.27) Thus, when I > 1536h2 we have |ξ1| > I 98 9 ( 4 √ 3 )9/4 |A4|9/8 2pi(12 + pi)3h7/4 |A4| > 0.06 I 9 8 |A4|1/8 h7/4 . (2.28) 2.7 Gap Principles For The Thue Inequality In Section 2.3, we have shown that the Hessian of F satisfies the following formula. H(x, y) = A0x4 +A1x3y +A2x2y2 +A3xy3 +A4y4 = 1 4A23A4 (2A1A4x2 +A23xy + 2A4A3y 2)2. In fact, it is known (see page 78 of [8]) that for any quartic form F (x, y) with J = 0, the algebraic covariant −19 H(x, y) is the square of a quadratic form, say −1 9 H(x, y) = m2(x, y). The polynomial m(x) = m(x, 1) is a positive definite quadratic. We say that the quartic form F (x, y) = a0x4 + a1x3y + a2x2y2 + a3xy3 + a4y4 with positive discriminant is reduced if and only if the positive definite quadratic form m(x, y) is reduced. Here, we remark that the real quadratic form f(x, y) = ax2 + bxy + cy2 is called reduced if |b| ≤ a ≤ c. Suppose that our quartic form F (x, y) is reduced. We assume that y 6= 0. Put m(x, y) = y2m(z) = y2 ( Az2 +Bz + C ) , where z = xy . Note that m(z) assumes a minimum equal to 4AC−B2 4A at z = −B2A . Since m(x, y) = 1 |36A23A4|(2A1A4x 2+A23xy+2A4A3y 2)2, by (2.11), we get 4AC −B2 = 16A1A3A 2 4 −A43 36A23A4 = 4 3 I. 25 Chapter 2. The Method Of Thue-Siegel For Binary Quartic Forms Since m(x, y) is reduced, we have A2 ≤ AC ≤ 1 3 (4AC −B2) = 4 9 I. Therefore, m(x, y) ≥ √ I 2 y2, Hence, we have shown that Lemma 2.7.1. If the quartic form F (x, y) in Theorem 2.1.2 is reduced, then for its Hessian H(x, y) we have |H(x, y)| ≥ 9 4 Iy4. So we can assume that |H(x, y)| ≥ 12h3√3I when looking for pairs of solutions (x, y) with |y| ≥ 2h3/4 (3I)1/8 . Suppose that we have distinct solutions to |F (x, y)| ≤ h indexed by i, say (xi, yi), related to ω with |ξ(xi+1, yi+1)| ≥ |ξ(xi, yi)|. Let F (xi, yi) = hi, F (xi+1, yi+1) = hi+1. Since |hi| , |hi+1| < h, (2.25) still holds and we have |ξi+1| ≥ 1 8 √ 3h |A4|1/4 |ξi|3. Let us now assume that there are 4 distinct solutions to |F (x, y)| ≤ h related to a fixed choice of ω, corresponding to ξ−1, ξ0, ξ1 and ξ2, where |ξ−1| < |ξ0| < |ξ1| < |ξ2| and F (xj , yj) = hj . We will deduce a contradiction. By (2.25) and since |hj | < h, |zj+1| ≤ |zj | 3192h5 I , where zi = 1 − η 4 i ξ4i = −8hi √ 3IA4 ξ4i . Since |z−1| ≤ 2, if I > 1536h5 then |z0| , |z1| , |z2| < 1. Suppose that I > 1536h5 then similar to section 2.6, we can prove that inequalities (2.26) hold and we have |ξ1| > 3 2pih |A4|1/4 |ξ0|3 > 9 √ 3 2pih4(12 + pi)3 |A4| |ξ−1| 9. 26 Chapter 2. The Method Of Thue-Siegel For Binary Quartic Forms Let us now assume that |H(x−1, y−1)| ≥ 12 h 3 √ 3I∣∣A23A4∣∣ . Then by (2.20), |ξ−1|4 ≥ 4h3 √ |3IA4|. Note that under this assumption, we have |z−1| < 1 and |ξ1| > (4 √ 3)9/4I 9 8h11/4 |A4|1/8 ( 3 2pi )4 > 0.1h11/4I 9 8 |A4|1/8 . (2.29) 2.8 Some Algebraic Numbers Combining the polynomials Ar,g and Br,g in Lemma 2.2.2 with the resolvent forms, we will consider the complex sequences Σr,g given by Σr,g = η2 ξ2 Ar,g(z1)− η1 ξ1 Br,g(z1) where z1 = 1− η41/ξ41 . For any pair of integers (x, y), ξ4(x, y) and η4(x, y) are algebraic integers in Q( √ A0I/3)( see Lemma 2.5.1). We have seen that A0 < 0 and one can assume A3A4 6= 0. Therefore from (2.10), we have A1 6= 0. Define Λr,g = (9|A4|) 1−g 4 ξ4r1 ξ 1−g 1 ξ2Σr,g. We will show that Λr,g is either an integer in Q( √ A0I 3 ) or a fourth root of such an integer. If Λr,g 6= 0, this provides a lower bound upon |Λr,g|. Lemma 2.8.1. The linear forms ξ ξ(1, 0) , η η(1, 0) have their coefficients in Q( √ A0I/3). Proof. By (2.10) and (2.14), we have αδ − βγ = ± √ A43 − 16A1A24A3 = ± 4A23 A1 √ 3IA0 = 12A23 A1 √ IA0 3 . 27 Chapter 2. The Method Of Thue-Siegel For Binary Quartic Forms Since 2A1A4x2 +A23xy + 2A3A4y 2 = (αx+ βy)(γx+ δy) = √ 12A23 √ |A4|ξ(x, y)η(x, y), we conclude that αγ , βδ , αδ + βγ ∈ Z. Thus, for integral pair (s, t), we obtain ξ(s, t) ξ(1, 0) , η(s, t) η(1, 0) ∈ Q( √ A0I/3)[s, t]. Lemma 2.8.2. If (x1, y1) and (x2, y2) are two pairs of rational integers then the numbers √ 3|A4|1/2ξ(x1, y1)η(x2, y2), ξ(x1, y1)3ξ(x2, y2) and η(x1, y1)3η(x2, y2) are integers in Q( √ A0I/3). Proof. For any pair of integers (x, y), Lemma 2.8.1 implies that ξ(x, y) ξ(1, 0) ∈ Q( √ A0I/3). Thus, ξ(x1, y1) ξ(x2, y2) ∈ Q( √ A0I/3). Since √ 3|A4|1/2ξ(x2, y2)η(x2, y2) = W (x2, y2)2|A3| ∈ Q, the algebraic integer √ 3|A4|ξ(x1, y1)η(x2, y2) belongs to Q( √ A0I/3). Let ξ(x, y) = 1x+ 2y. Clearly, 1 and 2 are algebraic integers and so are 41,  3 12,  2 1 2 2, 1 3 2 and  4 2. Since ξ 4 is an integer in Q( √ A0I/3), we conclude that 41,  3 12,  2 1 2 2, 1 3 2 and  4 2 are all algebraic integers in Q( √ A0I/3). ξ(x1, y1)3ξ(x2, y2) is an integer in Q( √ A0I/3) , because it can be written as a linear combination with rational integer coefficients in 41 ,  3 12,  2 1 2 2, 1 3 2 and 42. We can similarly show that that η(x1, y1)3η(x2, y2) is also an integer in Q( √ A0I/3). 28 Chapter 2. The Method Of Thue-Siegel For Binary Quartic Forms LetK = Q( √ A4I/3) and OK be its ring of integers. Note that by (2.10) Q( √ A4I/3) = Q( √ A0I/3). We put O = {m+ n √ A4I/3 2 ∈ OK | m,n ∈ Z}. It is easy to check thatO is a subring ofOK. Let d be the largest square-free divisor of A0I/3. From the well-known characterization of algebraic integers in quadratic fields, we have OK = {a+ b 1 + √ d 2 | a, b ∈ Z} if d ≡ 1 mod 4 and OK = {a+ b √ d | a, b ∈ Z} if d ≡ 2, 3 mod 4. Therefore, θ ∈ O if and only if θ ∈ OK, θ − θ̄ ∈ Z( √ A4I/3), (2.30) where θ̄ is the complex conjugate of θ. This implies that |θ| ≥ | Im θ| ≥ 1 2 ( √ A4I/3). (2.31) Lemma 2.8.3. If (x1, y1) and (x2, y2) are two pairs of rational integers then the numbers √ 3|A4|1/2ξ(x1, y1)η(x2, y2), ξ(x1, y1)3ξ(x2, y2) and η(x1, y1)3η(x2, y2) belong to O. Proof. For pairs of integers (x1, y1) and (x2, y2), we know√ 3|A4|1/2ξ(x1, y1)η(x2, y2), and √ 3|A4|1/2ξ(x2, y2)η(x1, y1), 29 Chapter 2. The Method Of Thue-Siegel For Binary Quartic Forms are complex conjugate elements of OK. By the definition of the resolvent forms ξ and η in Section 2.5 and (2.14), the transformation (x, y) → (ξ, η) has determinant ± 4A3 √ 3IA4 2 √ 3A3 |A4|1/4 . Therefore√ 3|A4|1/2ξ(x1, y1)η(x2, y2)− √ 3|A4|1/2ξ(x2, y2)η(x1, y1) = ±2 √ 3IA4. By (2.30), this implies√ 3|A4|1/2ξ(x1, y1)η(x2, y2) ∈ O. Let ξ(x, y) = 1x+ 2y. Since 41x 4 + 4312x 3y + 621 2 2x 2y2 + 4132xy 3 + 42y 4 = ξ4(x, y), and by (2.22), ξ4(x, y) = ±ψ(x, y) 4A4 ± 4F (x, y) √ 3IA4, we have 41, 4 3 12, 6 2 1 2 2, 41 3 2 and  4 2 are elements of O and their imaginary parts have factor 12. From the proof of Lemma 2.8.2, 41 ,  3 12,  2 1 2 2, 1 3 2 and 42 are algebraic integers. Therefore, by (2.30),  4 1 ,  3 12,  2 1 2 2, 1 3 2 and 42 are elements of O. So ξ(x1, y1) 3ξ(x2, y2) belongs to O, because it can be written as a linear combination with rational integer coefficients in 41 ,  3 12, 21 2 2, 1 3 2 and  4 2. We can similarly prove the lemma for η(x1, y1)3η(x2, y2). For every polynomial P (z) = anzn+an−1zn−1+ . . .+a1z+a0, we define P ∗(x, y) = xnP (y/x) = a0xn + a1xn−1y + . . .+ an−1xyn−1 + anyn. Let Ar,g and Br,g be as in (2.3) and Cr,g(z) = Ar,g(1− z), Dr,g(z) = Br,g(1− z), where Ar,g and Br,g are the polynomials in Lemma 2.2.2. For z 6= 0, we have Dr,0(z) = zrCr,0(z−1), hence 30 Chapter 2. The Method Of Thue-Siegel For Binary Quartic Forms A∗r,0(z, z − z̄) = zrAr,0(1− z̄ z ) = zrCr,0( z̄ z ) = z̄rDr,0( z z̄ ) = z̄rBr,0(1− z z̄ ) (2.32) = B∗r,0(z̄, z̄ − z) = B̄∗r,0(z, z − z̄). Lemma 2.8.4. For any pair of integers (x, y), the numbers A∗r,g(ξ 4(x, y), ξ4(x, y)− η4(x, y)) and B∗r,g(ξ 4(x, y), ξ4(x, y)− η4(x, y)) belong to O. Proof. It is clear that A∗r,g(ξ 4(x, y), ξ4(x, y)− η4(x, y)) and B∗r,g(ξ 4(x, y), ξ4(x, y)− η4(x, y)) belong to Q( √ A0I/3). So we need only show that they are integers in Q( √ A0I/3). This follows immediately from Lemma 4.1 of [6] since ξ4(x, y)− η4(x, y) = −8h √ 3IA4F (x, y). We now proceed to show that for any r ∈ Z, Λr,0 and Λ4r,1 are in O. Λr,g = (9|A4|) 1−g 4 ξ4r1 ξ 1−g 1 ξ2Σr,g = (9|A4|) 1−g 4 ( ξ1−g1 η2A ∗ r,g(ξ 4 1 , ξ 4 1 − η41)− ξ3g1 ξ2η1B∗r,g(ξ41 , ξ41 − η41) ) . By Lemma 2.8.2 and (2.32), Λr,0 ∈ Z. √ A0I/3. By Lemma 2.8.3 and Lemma 2.8.4, Λ4r,1 belongs to O. Next we will show that Λ 4 r,1 is not a rational integer when Σr,1 is nonzero. We have Σr,g = η2 ξ2 Ar,g(z1)− η1 ξ1 Br,g(z1) = η ξ [ η2/η ξ2/ξ Ar,g(z1)− η1/η ξ1/ξ Br,g(z1) ] , 31 Chapter 2. The Method Of Thue-Siegel For Binary Quartic Forms where η = η(1, 0) and ξ = ξ(1, 0). By Lemmas 2.8.1 and 2.8.4, η2/ηξ2/ξAr,g(z1)− η1/η ξ1/ξ Br,g(z1) ∈ Q( √ A0I/3) . Hence f = Q( √ A0I/3,Σr,g) = Q( √ A0I/3, ξ η ). If we choose complex number X so that ξ(X, 1) = η(X, 1) then by Lemma 2.8.1, X ∈ f. We have F (X, 1) = −1 8 √ 3IA4 (ξ4(X, 1)− η4(X, 1)) = 0. Since we have assumed that F is irreducible and has only real roots, the number A0 is negative, and therefore, X has degree 4 over Q( √ A0I/3).We also know that X ∈ f = Q( √ A0I/3)(Σr,g). Thus, Σr,g must have degree 4 over Q( √ A0I/3). Now suppose Λ4r,1 ∈ Z. Then we have for some ρ ∈ {±1,±i}, that Λr,1 = ρΛ̄r,1. Hence by Lemma 2.8.2 Σr,1 = ξ−4r1 ξ −1 2 ρΛ̄r,1 = ξ−4r1 ξ −1 2 η 4r 1 η2ρ ( ξ2 η2 Ar,1 ( 1− ξ 4 1 η41 ) − ξ1 η1 Br,1 ( 1− ξ 4 1 η41 )) = ξ−4r1 ξ −1 2 η 4r 1 η2ρ ( ξ2 η2 Ar,1 ( 1− ξ 4 1 η41 ) − ξ1 η1 Br,1 ( 1− ξ 4 1 η41 )) = ρ η4r1 ξ4r1 ( Ar,1 ( 1− ξ 4 1 η41 ) − ξ1η2 ξ2η1 Br,1 ( 1− ξ 4 1 η41 )) . This, together with Lemmas 2.8.2 and 2.8.4, implies that Σr,1 ∈ Q( √ A0I/3, ρ), which contradicts the fact that Σr,1 has degree 4 over Q( √ A0I/3) (ρ ∈ {±1,±i} ). We conclude that Λr,1 can not be a rational integer. Therefore, by (2.31), we may conclude that if Λr,g 6= 0, then for g ∈ {0, 1} |Λr,g| ≥ 2 −g 4 (−A4I/3) 12− 3g 8 . (2.33) 2.9 Approximating Polynomials In order to apply (2.21), we must make sure that Λr,g or equivalently Σr,g does not vanish. First we will show that for small r, Σr,0 6= 0. 32 Chapter 2. The Method Of Thue-Siegel For Binary Quartic Forms Lemma 2.9.1. Suppose that (x, y) is a pair of solutions to F (x, y) = ±1 with I > 1536 or a pair of solutions to |F (x, y)| ≤ h with |y| > 2h3/4 (3I)1/8 . For this pair of solutions and r ∈ {1, 2, 3, 4, 5}, we have Σr,0 6= 0. Proof. Let r ∈ {1, 2, 3, 4, 5}. Suppose that Σr,0 = 0. From (2.4), we can find for each r, a polynomial Fr(z) ∈ Z[z], satisfying Ar,0(z)4 − (1− z)B4r,0 = z2r+1Fr(z). In fact, we have A1(z) = 4A1,0(z) = 8− 5z, B1(z) = 4B1,0(z) = 8− 3z, F1(z) = 320− 320z + 81z2, A2(z) = 32 3 A2,0(z) = 64− 72z + 15z2, B2(z) = 32 3 B2,0(z) = 64− 56z + 7z2, F2(z) = 86016− 172032z + 114624z2 − 28608z3 + 2401z4, A3(z) = 128A3,0(z) = 2560− 4160z + 1872z2 − 195z3, B3(z) = 128B3,0(z) = 2560− 3520z + 1232z2 − 77z3, F3(z) = 14057472000− 42172416000z +48483635200z2 − 26679910400z3 +7150266240z4 − 839047040z5 +35153041z6, A4(z) = 2048 5 A4,0(z) = 28672− 60928z + 42432z2 − 10608z3 + 663z4, B4(z) = 2048 5 B4,0(z) = 28672− 53760z + 31680z2 − 6160z3 + 231z4, 33 Chapter 2. The Method Of Thue-Siegel For Binary Quartic Forms F4(z) = 13989396348928− 55957585395712z +91916125077504z2 − 79896826347520z3 +39463764078592z4 − 11050000539648z5 +1648475542656z6 − 113348764800z7 +2847396321z8, A5(z) = 8192 21 A5,0(z) = 98304− 258048z + 243712z2 −99008z3 + 15912z4 − 663z5, B5(z) = 8192 21 B5,0(z) = 98304− 233472z + 194560z2 −66880z3 + 8360z4 − 209z5. and F5(z) = 121733331812352− 608666659061760z +1301756554248192z2 − 1555026262622208z3 +1136607561252864z4 − 523630732640256z5 +151029162176512z6 − 26204424888320z7 +2515441608384z8 − 113971885760z9 +1908029761z10. We also define A∗r and B∗r via A∗r(x, y) = x rAr(y/x), and B∗r (x, y) = x rBr(y/x). Since Σr,0 is assumed to be zero, η42 ξ42 = η41(B ∗ r (ξ 4 1 , ξ 4 1 − η41))4 ξ41(A∗r(ξ41 , ξ41 − η41))4 . Let Ir be the integral ideal in Q( √ IA0/3) generated by ξ41(A ∗(ξ41 , ξ41−η41))4 and η41(B ∗(ξ41 , ξ41 − η41))4 and N(Ir) be the absolute norm of Ir. Since the 34 Chapter 2. The Method Of Thue-Siegel For Binary Quartic Forms ideal generated by ξ41(A ∗ r(ξ 4 1 , ξ 4 1 − η41))4 − η41(B∗r (ξ41 , ξ41 − η41))4 divides (ξ42 − η42).Ir, we obtain |ξ1|4(4r+1)|A4r(z1)− (1− z1)B4r (z1)| = |ξ41(A∗r(ξ41 , ξ41 − η41))4 − η41(B∗r (ξ41 , ξ41 − η41))4. Since Ir is contained in an imaginary quadratic field, by (2.15), we get |ξ1|4(4r+1)|A4r(z1)− (1− z1)B4r (z1)| ≤ N(Ir)1/2|ξ42 − η42| By (2.4), A4r(z1)− (1− z1)B4r (z1) = z2r+11 Fr(z1), and so we conclude |z1|2r+1|Fr(z1)| ≤ N(Ir)1/2|ξ42 − η42||ξ1|−4(4r+1); i.e. 1 ≤ N(Ir) 1/2|ξ42 − η42||ξ1|−4(4r+1) |z1|2r+1|Fr(z1)| . Since ξ41 = (ξ 4 1 − η41)(1− η 4 1 ξ41 )−1 = (ξ41 − η41)z−11 we obtain 1 ≤ N(Ir) 1/2|ξ42 − η42||ξ41 − η41|−4r−1|z1|2r |Fr(z1)| . Noting that |z1| = ∣∣ξ−41 ∣∣ ∣∣ξ41 − η41∣∣ and ∣∣ξ4i − η4i ∣∣ = ∣∣8h√3IA4F (x, y)∣∣, we obtain for r ∈ {1, 2, 3, 4, 5}, |ξ1|8r ≤ (N(Ir) 1/2|ξ41 − η41|−4r−1)|8h √ 3IA4|2r+1 |Fr(z1)| . (2.34) To estimate N(Ir)1/2, we choose a finite extension M of Q( √ A0I/3) so that the ideal generated by ξ41 and ξ 4 1 − η41 in M is a principal ideal, with generator p, say. We denote the extension of Ir to M, by I′r. Let rr be the ideal inM generated by A∗r(u, v) and B∗r (u, v), where u = ξ41 p and v = ξ41−η41 p . Since A∗r(x, x− y) = B∗r (y, y − x), p4r+1r4rB ∗ r (0, 1) 4 ⊂ p4r+1r4r(u,B∗r (0, v)4)(u− v,B∗r (0, v)4) (2.35) ⊂ p4r+1r4r(u,B∗r (0, v)4)(u− v,A∗r(v, v)4) ⊂ p4r+1r4r(u, u− v)(u,B∗r (u, v)4)(u− v,A∗r(u, v)4) ⊂ p4r+1(uA∗(u, v)4, (u− v)B∗r (u, v)4) = I′r, 35 Chapter 2. The Method Of Thue-Siegel For Binary Quartic Forms where (m1, . . . ,mn) denote the ideal in M generated by m1, . . . ,mn. We have A∗1(x, y)−B∗1(x, y) = −2y. Therefore, 2(v) ⊂ (A∗1(u, v), B∗1(u, v)) ⊂ r1, where (v) is the ideal generated by v in M. Since B∗1(0, 1) = −3, it follows from (2.35) that 1296(ξ41 − η41)5 ⊂ 1296p(ξ41 − η41)4 = p516v4B∗1(0, 1)4 ⊂ I′1. For r = 2, we first observe that B∗1(x, y)A ∗ 2(x, y)−A∗1(x, y)B∗2(x, y) = −10y3 and (−32x+ 7y)A∗2(x, y)− (−32x+ 15y)B∗2(x, y) = 80xy2. Therefore, by (2.35) we have 80(v)2 ⊂ (−10v3, 80uv2) ⊂ (A∗2(u, v), B∗2(u, v)) ⊂ r2. Since B∗2(0, 1) = 7, we have 804 × 74(ξ41 − η41)9 ⊂ 804 × 74p(ξ41 − η41)8 = 804p9v8B∗2(0, 1)4 ⊂ I′2. When r = 3, we have B∗2(x, y)A ∗ 3(x, y)−A∗2(x, y)B∗3(x, y) = −210y5 (1616x2 − 1078xy + 77y2)A∗3(x, y) − (1616x2 − 1482xy + 195y2)B∗3(x, y) = −16800x2y3. Substituting 77 for B∗3(0, 1), we conclude 168004 × 774(ξ41 − η41)13 ⊂ 168004 × 774p(ξ41 − η41)12 = 168004p13v12B∗3(0, 1) 4 ⊂ I′3. For r = 4, setting G4(x, y) = 14178304x3 − 15889280x2y + 4071760xy2 − 162393y3, H4(x, y) = 14178304x3 − 19433856x2y + 6714864xy2 − 466089y3, 36 Chapter 2. The Method Of Thue-Siegel For Binary Quartic Forms we may verify that B∗3(x, y)A ∗ 4(x, y)−A∗3(x, y)B∗4(x, y) = −6006y7 and G4(x, y)A∗4(x, y)−H4(x, y)B∗4(x, y) = −150678528y4x3. This implies that 1506785284 × 2314(ξ41 − η41)17 ⊂ 1506785284 × 2314p(ξ41 − η41)16. Since this latter quantity is equal to 1506785284p17v16B∗4(0, 1)4, it follows that 1506785284 × 2314(ξ41 − η41)17 ⊂ I′4. Finally, for r = 5, we have B∗4(x, y)A ∗ 5(x, y)−A∗4(x, y)B∗5(x, y) = −14586y7 and G5(x, y)A∗5(x, y)−H5(x, y)B∗5(x, y) = −134424576y5x4, where G5(x, y) = 43706368x4 − 69346048x3y + 32767856x2y2 −4764782xy3 + 123519y4, H5(x, y) = 43706368x4 − 80272640x3y + 46006896x2y2 −8845746xy3 + 391833y4. This implies that 1344245764 × 2094(ξ41 − η41)21 ⊂ 1344245764 × 2094p(ξ41 − η41)20 whereby 1344245764 × 2094(ξ41 − η41)21 ⊂ 1344245764p21v20B∗5(0, 1)4 ⊂ I′5. From the preceding arguments, we are thus able to deduce the following series of inequalities : N(I1)1/2|ξ41 − η41|−5 ≤ 1296, 37 Chapter 2. The Method Of Thue-Siegel For Binary Quartic Forms N(I2)1/2|ξ41 − η41|−9 ≤ 5604, N(I3)1/2|ξ41 − η41|−13 ≤ (77× 16800)4, N(I4)1/2|ξ41 − η41|−17 ≤ (231× 150678528)4 and N(I5)1/2|ξ41 − η41|−21 ≤ (134424576× 209)4. Substituting any of these in (2.34) provides a contradiction to inequality (2.28) when I > 1536 and a contradition to (2.29) when |y| > 2h3/4 (3I)1/8 . Note that under both assumptions I > 1536 and |y| > 2h3/4 (3I)1/8 , the function z is small. This makes |Fr(z)| large enough for our contradictions. Lemma 2.9.2. If r ∈ N and h ∈ {0, 1}, then at most one of {Σr,0,Σr+h,1} can vanish. Proof. Let r be a positive integer and h ∈ {0, 1} . Following an argument of Bennett [3], we define the matrix M: M =  Ar,0(z1) Ar+h,1(z1) η1 ξ1 Ar,0(z1) Ar+h,1(z1) η1 ξ1 Br,0(z1) Br+h,1(z1) η2 ξ2  . The determinant of M is zero because it has two identical rows. Expanding along the first row, we get 0 = Ar,0(z1)Σr+h,1 −Ar+h,1(z1)Σr,0 + η2 ξ2 (Ar,0(z1)Br+h,1(z1)−Ar+h,1(z1)Br,0(z1)). If Σr,0 = 0 and Σr+h,1 = 0 then Ar,0(z1)Br+h,1(z1)−Ar+h,1(z1)Br,0(z1) = 0 which contradicts part (iii) of Lemma 2.2.2. 2.10 An Auxiliary Lemma We now combine the upper bound for Λr,g obtained in (2.33) with the lower bounds from Lemma 2.2.2 to prove the following lemma. Lemma 2.10.1. If Σr,g 6= 0, then c1(r, g)|ξ1|4r+1−g|ξ2|−3 + c2(r, g)|ξ1|−4r−3(1−g)|ξ2| > 1, 38 Chapter 2. The Method Of Thue-Siegel For Binary Quartic Forms where we may take c1(1, 0) = 4 √ 3pih |A4|1/4 and c2(1, 0) = 27h3 ( 3 |A4|1/2 )1/2 (9 √ 3I |A4|)2 5128 and for (r, g) 6= (1, 0), c1(r, g) = 2 √ 3pi h |A4|1/4 ( |A4|1/2 3 )g/4 4r√ r and c2(r, g) = 27 √ 3h2r+1−g |A4|1/4 ( |A4|1/2 3 )g/4 (9 √ 3I |A4|)2r−g √ 2√ rpi4r . Proof. By the definition of Λr,g and (2.4), we can write |Λr,g| = (9|A4|)(1−g)/4|ξ1|4r+1−g|ξ2| ∣∣∣∣(η2ξ2 − ω)Ar,g(z1) + ωz2r+1−g1 Fr,g(z1) ∣∣∣∣ . Since |1− z1| = 1 , |z1| ≤ 1 and |zi| = 8h √ 3I |ξ4i | , by (2.5), (2.6) and inequality (2.24), we have |Λr,g| ≤ (9|A4|)(1−g)/4|ξ1|4r+1−g|ξ2|L, (2.36) where L is equal to( 2r − g r ) 2pih √ 3I |A4| 3|ξ42 | + ( r−g+1/4 r+1−g )( r−1/4 r )( 2r+1−g r ) (9h√3I |A4||ξ41 | )2r+1−g . Comparing this with (2.33), we obtain c1(r, g)|ξ1|4r+1−g|ξ2|−3 + c2(r, g)|ξ1|−4r−3(1−g)|ξ2| > 1, where we may take c1 and c2 so that c1(r, g) ≥ 2 √ 3pi h |A4|1/4 ( |A4|1/2 3 )g/4( 2r r ) 39 Chapter 2. The Method Of Thue-Siegel For Binary Quartic Forms and c2(r, g) ≥ 27 √ 3h2r+1−g |A4|1/4 ( |A4|1/2 3 )g/4 (9 √ 3I |A4|)2r−g ( r−g+1/4 r+1−g )( r−1/4 r )( 2r+1−g r ) . Substituting r = 1 and g = 0, we get the desired values for c1(1, 0) and c2(1, 0). Let us apply the following version of Stirling’s formula (see Theorem (5.44) of [15]): 1 2 √ k 4k ≤ ( 2k k ) < 1√ pik 4k, for k ∈ N. This leads to the stated choice of c1 immediately. To evaluate c2(r, g), we first note that( 2r + 1− g r ) ≥ ( 2r r ) ≥ 4 r 2 √ r . Next we will show that( r − g + 1/4 r + 1− g )( r − 1/4 r ) < 1√ 2pir , (2.37) for r ∈ N and g ∈ {0, 1}, whence we may conclude that( r−g+1/4 r+1−g )( r−1/4 r )( 2r+1−g r ) < √2√ rpi4r . This leads immediately to the stated choice of c2. It remains to show (2.37). Let us set Xr = ( r − 3/4 r )( r − 1/4 r ) = yr r , whereby Xr+1 = ( r + 1/4 r + 1 )( r + 3/4 r + 1 ) = ( r2 + r + 2/9 r2 + r ) yr r + 1 . This implies y1 = 3/16 , yr = 3 16 r−1∏ k=1 k2 + k + 3/16 k2 + k . Since ∞∏ k=1 k2 + k + 3/16 k2 + k = 16 3Γ(1/4)Γ(3/4) = 16 3 √ 2pi , 40 Chapter 2. The Method Of Thue-Siegel For Binary Quartic Forms we obtain Xr < 1√ 2pir . For r ∈ N, we have ( r − 3/4 r ) > ( r + 1/4 r + 1 ) . So when g ∈ {0, 1}, ( r − g + 1/4 r + 1− g )( r − 1/4 r ) ≤ Xr, which completes the proof. 2.11 Proof of Theorem 2.1.2 Let us now assume that there are 4 distinct solutions (xi, yi) to reduced form |F (x, y)| ≤ h related to ω with |yi| > h3/4(3I)1/8 , corresponding to ξ−1, ξ0, ξ1 and ξ2, where we have ordered these in nondecreasing modulus. From (2.29) and Lemma 2.7.1, we have |ξ1| > 0.1 |A4|1/8 I9/8h11/4. We will deduce a contradiction, implying that at most 3 such solutions can exist. Then Theorem 2.1.1 will be proven, since there are 4 choices of ω. We will show that |ξ2| is arbitrarily large in relation to |ξ1|. This con- tradicts (2.29). Lemma 2.11.1. Let F (x, y) be the quartic form in Theorem 2.1.2. Suppose that (x1, y1) and (x2, y2) are 2 pairs of solutions to |F (x, y)| ≤ h, both related to ω, a fixed fourth root unity, with 0.1 |A4|1/8 h11/4I9/8 < |ξ1| < |ξ2|, where ξj = ξ(xj , yj). Then, for each positive integer r, |ξ2| > 4 r√r 27 √ 3 |A4|1/8 h2r+1 (9 √ 3I |A4|)−2r|ξ1|4r+3. 41 Chapter 2. The Method Of Thue-Siegel For Binary Quartic Forms Proof. By (2.26), |ξ2| ≥ 3|ξ1| 3 2pih|A4|1/4 . This implies c1(1, 0)|ξ1|5|ξ2|−3 ≤ 4h4pi × 123/2 √ 3 |A4|1/2 |ξ1|−4 Therefore, by (2.29) and from the fact that |A4| < 4I, we obtain c1(1, 0)|ξ1|5|ξ2|−3 < 0.01 Lemma 2.9.1 implies that Σ1,0 6= 0. So we may apply Lemma 2.10.1 to get c2(1, 0)I3|ξ1|−7|ξ2| > 0.99. One may now conclude that |ξ2| > 0.99 c2(1, 0) |ξ1|7 > 0.93h−3 ( 3 |A4|1/2 )−1/2 (9 √ 3I |A4|)−2|ξ1|7. This proves the lemma for r = 1. Moreover, we may conclude that c1(2, 0)|ξ1|9|ξ2|−3 < 18h 10√pi × 16× (5× 27)3√ 2|A4|1273 ( 9 √ 3I |A4| )6 |ξ1|−12 . Since |A4| ≤ 4I , by (2.29) we have c1(2, 0)|ξ1|9|ξ2|−3 < 0.1. Via Lemmas 2.10.1 and 2.9.1, we obtain |ξ2| > 0.9 c2(1, 0) |ξ1|11. This leads to the proof of the Lemma for r = 2, after substituting the value of c2(2, 0). To complete the proof, we use induction on r. Suppose that for some r ≥ 2, |ξ2| > 4 r√r 27 √ 3 |A4|1/8 h2r+1 (9 √ 3I |A4|)−2r|ξ1|4r+3. Then c1(r + 1, 0)|ξ1|4r+5|ξ2|−3 < 18 √ pi × 273 |A4|1/8 h6r+4 42r−1 √ (r + 1)r √ r ( 9 √ 3I |A4| )6r |ξ1|−8r−4 . 42 Chapter 2. The Method Of Thue-Siegel For Binary Quartic Forms By (2.29), we have c1(r + 1, 0)|ξ1|4r+5|ξ2|−3 < 0.1. If Σr+1,0 6= 0, then by Lemma 2.10.1, c2(r + 1, 0)|ξ1|−4(r+1)−3|ξ2| > 0.9. Hence, |ξ2| > 0.9 c2(r + 1, 0) |ξ1|4(r+1)+3 > 4r+1 √ r + 1 27h2r+3 ( |A4|1/2 3 )1/2 (9 √ 3I)−2r−2|ξ1|4r+7. If, however, Σr+1,0 = 0, then by Lemma 2.9.2, both Σr+1,1 and Σr+2,1 are both non-zero and by Lemma 2.9.1, we have r > 5. Using the induction hypothesis, we get c1(r + 1, 1)|ξ1|4r+4|ξ2|−3 < 0.01 and thus by Lemma 2.10.1, (2.10) and (2.28), we conclude c2(r + 1, 1)|ξ1|−4r−4|ξ2| > 0.99. So, we obtain |ξ2| > 4 r+1 √ r + 1 27h2r+2 |A4|1/8 31/4 ( 9 √ 3I |A4| )−2r−1 |ξ1|4(r+1). Consequently, c1(r + 2, 1)|ξ1|4r+8|ξ2|−3 is less than 2 √ pi × 27 ( 3 |A4|1/2 )( 9 √ 3I |A4| )6r+3 h6r+7 42r+1(r + 1) √ (r + 1) (r + 2) |ξ1|−8r−4 < 0.1. A final application of Lemma 2.10.1 implies c2(r + 2, 1)|ξ1|−4r−8|ξ2| > 0.9 or |ξ2| > 0.9 c2(r + 2, 1) |ξ1|4r+8. 43 Chapter 2. The Method Of Thue-Siegel For Binary Quartic Forms It follows that |ξ2| > √ r + 24r+2 27 |A4|1/8 31/4h2r+4 ( 9 √ 3I |A4| )−2r−3 |ξ1|4(r+1)+4. Since |ξ1| > 0.1I9/8h11/4 |A4|1/8, we conclude that |ξ2| > 4 r+1 √ r + 1 27 √ 3 |A4|1/8 (9 √ 3I |A4|)−2r−2|ξ1|4r+7. 2.12 Representation Of Unity By F (x, y) 2.12.1 The Proof Of Theorem 2.1.1 for I > 1536 Let us now assume that there are 4 distinct solutions to |F (x, y)| = 1 related to ω, corresponding to ξ−1 , ξ0 , ξ1 and ξ2, where we have ordered these in nondecreasing modulus. From (2.28), we have |ξ1| > 0.06 I9/8. Similar to Lemma 2.11.1, one can show that for I > 1536, |ξ2| > 4 r√r 27 √ 3 |A4|1/8 (9 √ 3I |A4|)−2r|ξ1|4r+3. This, together with (2.28), provides a contradiction when I > 1536. So we conclude that related to each fourth root of unity, there are at most 3 solutions to equation (2.1). Since there are 4 such roots of unity, we conclude Theorem 2.1.1 when I > 1536. 2.12.2 Forms With Small Discriminant To finish the proof of Theorem 2.1.1, we need to study the quartic forms F (x, y) = a0x4 + a1x3y + a2x2y2 + a3xy3 + a4y4 with 0 < IF ≤ 1536 and A0 = 3(8a0a2 − 3a21) < 0. Quartic forms with I > 0 and A0 < 0 are called forms of type 2 (signature (4, 0) ) in [8]. We followed an algorithm of Cremona, in section 4.6 of [8], which gives all inequivalent integer quartics with given invariant I and J = 0. Using Magma, we counted the number of solutions to |F (x, y)| = 1 44 Chapter 2. The Method Of Thue-Siegel For Binary Quartic Forms for all reduced quartic forms F with IF ≤ 1536 and JF = 0. Regarding (x, y) and (−x,−y) as the same, we didn’t find any form F for which there are more than 4 solutions to F (x, y) = ±1. We have solved about 1200 equations |F (x, y)| = 1. Our algorithm is not efficient in the sense that it solves more than one equation from some equivalent classes. We finish this paper by giving some examples with IF ≤ 1536 and JF = 0, for which F (x, y) = ±1 has 4 solutions: If F (x, y) = x4 − 12x2y2 + 16xy3 + 4y4 then F (x, y) = 1 has 4 solutions (5, 2), (1, 3), (1, 1), (1, 0) and the equation F (x, y) = −1 has no solution. If F (x, y) = 4x4 − 12x2y2 − 8xy3 − y4 then F (x, y) = 1 has no solution and the equation F (x, y) = −1 has 4 solutions (−3, 5), (−1, 1), (0, 1), (2, 1). For the folowing forms the equation |F (x, y)| = 1 has 4 solutions: If F (x, y) = x4 − x3y − 6x2y2 + xy3 + y4 then the solutions are: (−1, 0), (0, 1), (−1, 2), (2, 1). 45 Chapter 2. The Method Of Thue-Siegel For Binary Quartic Forms If F (x, y) = 4x4 − 6x3y − 3x2y2 + 5xy3 − y4 then the solutions are (−1, 1), (2, 3), (0, 1), (1, 1). 46 Bibliography [1] S. Akhtari. The Diophantine equation aX4 − bY 2 = 1. (2007), to appear in J. Reine Angew. Math. [2] S. Akhtari and G. Walsh. The Diophantine equation aX4 − bY 2 = 2. (2007), to appear in Acta. Arith. [3] M.A. Bennett. On the representation of unity by binary cubic forms. Trans. Amer. Math. Soc. 353 (2001), 1507-1534. [4] E. Bombieri and W.M. Schmidt. On Thue’s equation. Invent. Math. 88 (1987), 69-81. [5] J.W. Cassels. An Introduction to the Geometry of Numbers. Springer- Verlag,1959. [6] G.V. Chudnovsky. On the method of Thue-Siegel. Ann. Math. II Ser. 117 (1983), 325-382. [7] H. Cohen, A Course in Computational Algebraic Number Theory, 3rd corrected printing, Graduate Texts in Mathematics 138 (Springer- Verlag, 1996). [8] J.E. Cremona. Reduction of binary cubic and quartic forms. London Math Soc. J. Comput. Math 2 (1999), 62-92. [9] J.H. Evertse. On the representation of integers by binary cubic forms of positive discriminant. Invent. Math.73(1983), 117-138. [10] V. Krechmar. On the superior bound of the number of representations of an integer by binary forms of the fourth degree (Russian). Bull. Acad. Sci. URSS, Ser. Math. (1939), 289-302 [11] W. Ljunggren. On the representation of integers by certain binary cubic and biquadratic forms. Acta. Arith. XVII (1971) 47 Bibliography [12] C.L. Siegel. Über einige Anwendungen diophantischer Approximatio- nen, Abh. Preuss. Akad. Wiss. 1 (1929). [13] C.L. Siegel. Einige Erläuterungen zu Thues Untersuchungen über Annäherungswerte algebraischer Zahlen und diophantische Gleichun- gen, Akad. D. Wiss. in Gottingen , Math.-Phys. Klasse 1970. no 8, (1970), 169-195. [14] C.L. Siegel. Die Gleichung axn − byn = c, Math. Ann. 114 (1937), 57-68. [15] K. Stromberg. An Introduction to Classical Real Analysis. Wadsworth international mathematical series, 1981. [16] A. Thue. Bemerkungen über gewisse Näherungsbrüche algebraischer Zahlen, Kra. Vidensk. Selsk. Skrifter. I. Mat. Nat. Kl. (1908), no 3, Kra, 1908. [17] A. Thue. Über Annäherungenswerte algebraischer Zahlen. J. reine angew. Math. 135 (1909), 284-305. [18] A. Thue. Über rationale Annäherungswerte der reellen Wuzel der ganzen Funktion dritten Grades x3 − ax − b. Kra. Vidensk. Selsk. Skrifter. I. Mat. Nat. Kl. (1908), no. 6, Kra, 1908. [19] A.Thue. Ein Fundamental theorem zur Bestimmunhg von Annäherungs-werten aller Wurzeln gewisser ganzer Funktionen. J. Reine Angew. Mat. 138 (1910), 96-108. 48 Chapter 3 The Diophantine equation aX4 − bY 2 = 1 2 3.1 Introduction In a series of papers over nearly forty years, Ljunggren (see e.g. [11], [12], [13], [14] and [15]) derived remarkable sharp bounds for the number of so- lutions to various quartic Diophantine equations, particularly those of the shape aX4 − bY 2 = ±1, (3.1) typically via sophisticated application of Skolem’s p-adic method. More recently, there has been a resurgence of interest in Ljunggren’s work; results along these lines are well surveyed in the paper of Walsh [26]. By way of example, using lower bounds for linear forms in logarithms, together with an assortment of elementary arguments, Bennett and Walsh [2] showed that the equation aX4 − bY 2 = 1 (3.2) has at most one solution in positive integers X and Y , when a is an integral square and b is a positive integer. For general a and b, however, there is no absolute upper bound for the number of integral solutions to (3.2) available in the literature, unless one makes strong additional assumptions (see e.g. [2], [3], [5], [7], [14], [15] and [22]). This lies in sharp contrast to the situation for the apparently similar equation aX4 − bY 2 = −1 (3.3) where Ljunggren [12] was able to bound the number of positive integral solutions by 2 for arbitrary fixed a and b. Moreover, it appears that the techniques employed to treat equation (3.3) and, in special cases, (3.2), do not lead to results for (3.2) in general. 2A version of this chapter has been accepted for publication. Akhtari. S. The Diophan- tine equation aX4 − bY 2 = 1. to appear in J. Reine Angew. Math. 49 Chapter 3. The Diophantine equation aX4 − bY 2 = 1 It is our goal in this paper to rectify this situation. To be precise, we will prove the following Theorem 3.1.1. Let a and b be positive integers. Then equation (3.2) has at most two solutions in positive integers (X,Y ). This resolves a conjecture of Walsh (see [2], [3], [7] and [26]), which had been suggested by computations and assorted heuristics. Since there are infinitely many pairs (a, b) for which two such solutions exist (see Section 3.2), this result is best possible. To prove this, we will appeal to classical results of Thue [21] from the theory of Diophantine approximation, together with modern refinements, particularly those of Evertse [8]. Such an approach, based on Padé approxi- mation to binomial functions, has been used in a number of previous works to explicitly solve Thue inequalities and equations (see e.g. [3], [5], [9], [22], [23], [24] ) or to bound the number of such solutions (see e.g. [1], [8], [10]). We will apply similar techniques to a certain family of quartic inequalities. 3.2 An Equivalent Problem Let a denote a non-square positive integer, and b a positive integer for which the quadratic equation aX2 − bY 2 = 1 (3.4) is solvable in positive integers X and Y . Let (v, w) be a pair of positive solutions to (3.4) so that τ = v √ a+ w √ b > 1, and τ is minimal with this property. All solutions in positive integers of (3.4) are given by (v2k+1, w2k+1), where τ2k+1 = v2k+1 √ a+ w2k+1 √ b (k ≥ 0) (see [25] for a proof). Solving the quartic equation (3.2) is thus equivalent to the problem of determining all squares in the sequence {v2k+1}. One can find a proof of the following result in [16]. Proposition 3.2.1. If v2k+1 is a square for some k ≥ 0, then v1 is also a square. 50 Chapter 3. The Diophantine equation aX4 − bY 2 = 1 Let us assume that equation (3.2) is solvable. Proposition 3.2.1 implies that τ = τ(a, b) is of the form τ = x2 √ a+ w √ b. We have τ = √ t+ 1 + √ t, where t = ax4 − 1. Thus, for k ≥ 0 τ2k+1 = V2k+1 √ t+ 1 +W2k+1 √ t, where V2k+1 = v2k+1 v1 . Hence, by Proposition 3.2.1, v2k+1 is a square if and only if V2k+1 is a square. In other words, in order to bound the number of positive integer solutions to an equation of the form aX4 − bY 2 = 1, it is sufficient to determine an upper bound for the number of integer solutions to Diophantine equations of the shape (t+ 1)X4 − tY 2 = 1. (3.5) The main result of [3] is the following. Proposition 3.2.2. Let m be a positive integer. Then the only positive integral solutions to the equation (m2 +m+ 1)X4 − (m2 +m)Y 2 = 1 are given by (X,Y ) = (1, 1) and (X,Y ) = (2m+ 1, 4m2 + 4m+ 3). In fact, these are the only values of t for which equation (3.5) is known to have as many as two positive solutions (suggesting a stronger version of Theorem 3.1.1). Note that if V3 = z2, where z is a positive integer, then since V3 = 1 + 4t, we have 4t = z2 − 1 = (z − 1)(z + 1) and therefore there exist positive integers m and n such that t = mn, 2m = z−1 and 2n = z+1. We conclude, therefore, that n = m+1 and t = m2+m. Proposition 3.2.2 thus implies the following. Corollary 3.2.3. If V3 is a square then for any k > 1, V2k+1 is not a square and there are only 2 solutions to equation (3.5) in positive integers X and Y . As it transpires, we will need to account for the possibility of V2k+1 being square, for odd values of k. The preceding result handles the case k = 1. For k = 3 and k = 5, we will appeal to 51 Chapter 3. The Diophantine equation aX4 − bY 2 = 1 Lemma 3.2.4. If t > 204, then neither V7 nor V11 is an integral square. Proof. The equation z2 = V7 = 64t3 + 80t2 + 24t + 1 was treated in- dependently using the function faintp on SIMATH and IntegralPoints on MAGMA, and found to have only the solutions corresponding to t = 0 and t = 1 . For the case z2 = V11, we first put x = 4t, and see that the de- sired result will follow by determining the set of rational points on the curve z2 = x5+9x4+28x3+35x2+15x+1. The proof now follows exactly as the proof for the case M2 = U11 on pages 8 − 10 of [4], but with x = P 2 and Q = −1, as the proof therein does not take into account the fact that P 2 is a square.. 3.3 Reduction To A Family Of Thue Equations We will begin by applying an argument of Togbe, Voutier and Walsh [22] to reduce (3.5) to a family of Thue equations. We subsequently apply the method of Thue-Siegel to find an upper bound for the number of solutions to this family. Let P (x, y) = x4 + 4tx3y − 6tx2y2 − 4t2xy3 + t2y4. The following is a modified version of Proposition 2.1 of [22]. We will include a proof primarily for completeness (and since we will have need of one of the inequalities derived therein). Proposition 3.3.1. Let t be a positive integer such that t 6= m2+m for all m ∈ Z. If (X,Y ) 6= (1, 1) is a positive integer solution to equation (3.5), then there is a solution in coprime positive integers (x, y) to the equation P (x, y) = t21, where t1 divides t, t1 ≤ √ t and xy > 64t3. Proof. For k ≥ 0, let us define τ , V2k+1 and W2k+1 as in Section 3.2, and choose Tk and Uk to satisfy τ2k = Tk + Uk √ t(t+ 1). Assume that V2k+1 = z2 for some integer z > 1. We will suppose that k is odd, k = 2n + 1 say, as the case that k is even is similar and discussed in [22]. When k = 2n+ 1, V4n+3 = z2 = V 22n+2 + V 2 2n+1 = tU 2 n+1 + V 2 2n+1, 52 Chapter 3. The Diophantine equation aX4 − bY 2 = 1 with, via Corollary 3.2.3, n > 0. Thus tU2n = z 2 − V 22n+1 = tU2n = z2 − (Tn + tUn)2. Since Un+1 = 2Tn + (2t+ 1)Un and gcd(Un, Tn) = 1, we have gcd(Un+1, Tn + tUn) = 1 and hence there exist positive integers G, H, t1, t2, with Un+1 = 2GH and t = t1t2, such that z − (Tn + tUn) = 2t1G2 and z + (Tn + tUn) = 2t2H2. Therefore, Tn + tUn = t2H2 − t1G2, and since 2GH = Un+1 = 2Tn + (2t+ 1)Un, we deduce that Un = 2GH − 2t2H2 + 2t1G2 and Tn = t2H2 − t1G2 − t(2GH − 2t2H2 + 2t1G2). Substituting for Tn and Un in the equation T 2n − t(t+ 1)U2n = 1, we obtain the equation t21G 4 − 4tt1G3H − 6tG2H2 + 4tt2GH3 + t22H4 = 1. Multiplying both sides by t21 and taking x = −t1G, y = H, we find that x and y are coprime positive integers satisfying P (x, y) = t21. To complete the proof, we observe that, since Lemma 3.2.4 and Corollary 3.2.3 imply that n ≥ 3, xy = t1GH = t1 2 Un+1 ≥ t12 U4 > 64t 3. (3.6) Our focus for the remainder of the paper will be to find, for fixed t, an upper bound upon the number of coprime positive integral solutions to the constrained inequality 0 < P (x, y) ≤ t2, xy > 64t3. (3.7) 53 Chapter 3. The Diophantine equation aX4 − bY 2 = 1 We should note that for t ≤ 204, Theorem 3.1.1 with (a, b) = (t + 1, t) has been verified in [22]. Here and henceforth, therefore, we will assume that t > 204. To proceed, let ξ = ξ(x, y) and η = η(x, y) be linear functions of (x, y) so that ξ4 = 4 ( √−t+ 1)(x−√−ty)4 and η4 = 4 (√−t− 1)(x+√−ty)4. We call (ξ, η), a pair of resolvent forms. Note that P (x, y) = 1 8 (ξ4 − η4) and if (ξ, η) is a pair of resolvent forms then there are precisely three others with distinct ratios, say (−ξ, η), (iξ, η) and (−iξ, η). Let ω be a fourth root of unity, (ξ, η) a fixed pair of resolvent forms and set z = 1− ( η(x, y) ξ(x, y) )4 . We say that the integer pair (x, y) is related to ω if∣∣∣∣ω − η(x, y)ξ(x, y) ∣∣∣∣ < pi12 |z|. It turns out that each nontrivial solution (x, y) to (3.7) is related to a fourth root of unity : Lemma 3.3.2. Suppose that (x, y) is a positive integral solution to inequality (3.7), with ∣∣∣∣ωj − η(x, y)ξ(x, y) ∣∣∣∣ = min0≤k≤3 ∣∣∣∣ekpii/2 − η(x, y)ξ(x, y) ∣∣∣∣ . Then |ωj − η(x, y) ξ(x, y) | < pi 12 |z(x, y)|. (3.8) Proof. We begin by noting that |z| = ∣∣∣∣ξ4 − η4ξ4 ∣∣∣∣ = 8P (x, y)|ξ4| , and, from xy 6= 0, |ξ4(x, y)| ≥ 4(√1 + t)5, whereby |z| ≤ 2t 2 ( √ t+ 1)5 < 1. 54 Chapter 3. The Diophantine equation aX4 − bY 2 = 1 Since η = −ξ̄, it follows that∣∣∣∣ηξ ∣∣∣∣ = 1, |1− z| = 1. Now let 4θ = arg ( η(x,y)4 ξ(x,y)4 ) . We have√ 2− 2 cos(4θ) = |z| < 1, and so |θ| < pi12 . Since ∣∣∣∣ωj − η(x, y)ξ(x, y) ∣∣∣∣ ≤ |θ|, it follows that∣∣∣∣ωj − η(x, y)ξ(x, y) ∣∣∣∣ ≤ 14 |4θ|√2− 2 cos(4θ) ∣∣∣∣1− η(x, y)4ξ(x, y)4 ∣∣∣∣ . From the fact that |4θ|√ 2−2 cos(4θ) < pi 3 whenever 0 < |θ| < pi12 , we obtain inequality (3.8), as desired. This lemma shows that each integer pair (x, y) is related to precisely one fourth root of unity. Let us fix such a fourth root, say ω, and suppose that we have distinct coprime positive solutions (x1, y1) and (x2, y2) to inequality (3.7), each related to ω. We will assume, as we may, that |ξ(x2, y2)| ≥ |ξ(x1, y1)|. For concision, we will write ηi = η(xi, yi) and ξi = ξ(xi, yi). Before we move into the heart of our proof, we will mention a pair of results that will be the starting point for our later proving that (x1, y1) and (x2, y2) are far apart in height. Since |z| = 8P (x, y)|ξ|4 ≤ 8t2 |ξ|4 , (3.9) it follows from (3.8) that |ξ1η2 − ξ2η1| = |ξ1(η2 − ωξ2)− ξ2(η1 − ωξ1)| (3.10) ≤ 2pi 3 t2 ( |ξ1| |ξ32 | + |ξ2| |ξ31 | ) ≤ 4pit 2 |ξ2| 3|ξ31 | . 55 Chapter 3. The Diophantine equation aX4 − bY 2 = 1 On the other hand, choosing our fourth root appropriately, we have( √ 2( √−t+ 1)1/4 −√2(√−t+ 1)1/4√−t√ 2( √−t− 1)1/4 √2(√−t− 1)1/4√−t )( x1 x2 y1 y2 ) = ( ξ1 ξ2 η1 η2 ) and so |ξ1η2 − ξ2η1| = ∣∣∣4(t+ 1)1/4√t (x1y2 − x2y1)∣∣∣ . Since x1y2 − x2y1 is a nonzero integer (recall that we assumed gcd(xi, yi) = 1), we have |ξ1η2 − ξ2η1| ≥ 4 √ t (t+ 1)1/4 (3.11) and thus, combining (3.10) and (3.11), we conclude that if (x1, y1) and (x2, y2) are distinct solutions to (3.7), related to ω, with |ξ(x2, y2)| ≥ |ξ(x1, y1)| then |ξ2| > 3 pi t−5/4 |ξ1|3. (3.12) As a final preliminary result, we have the following lemma, whose proof is an immediate consequence of the definition of resolvent forms : Lemma 3.3.3. If (x1, y1) and (x2, y2) are two pairs of rational integers then ξ(x1, y1)η(x2, y2) (−t− 1)1/4 , ξ(x1, y1) 3ξ(x2, y2) and η(x1, y1)3η(x2, y2) are integers in Q( √−t). 3.4 Padé Approximation The main focus of this section is to construct a family of dense approxima- tions to ξ/η from rational function approximations to the binomial function (1− z)1/4. Consider the system of linear forms Rr(z) = −Qr(z) + (1− z)1/4Pr(z), where Rr(z) = z2r+1R̄r(z), R̄r(z) is regular at z = 0, and Pr(z) and Qr(z) are polynomials of degree r. Thue [19], [20] explicitly found polynomials Pr(z) and Qr(z) that satisfy such a relationship, and Siegel [17] identi- fied them in terms of hypergeometric polynomials. Refining the work of Thue and Siegel, Evertse [8] used the theory of hypergeometric functions to 56 Chapter 3. The Diophantine equation aX4 − bY 2 = 1 sharpen Siegel’s upper bound for the number of solutions to the equation f(x, y) = 1, where f is a cubic binary form with positive discriminant. In this paper, we will apply similar arguments to certain quartic forms. We begin with some preliminaries on hypergeometric functions. A hy- pergeometric function is a power series of the shape F (α, β, γ, z) = 1 + ∞∑ n=1 α(α+ 1) · · · (α+ n− 1)β(β + 1) · · · (β + n− 1) γ(γ + 1) · · · (γ + n− 1)n! z n. Here z is a complex variable and α, β and γ are complex constants. If α or β is a non-positive integer and m is the smallest integer such that α(α+ 1) · · · (α+m)β(β + 1) · · · (β +m) = 0, then F (α, β, γ, z) is a polynomial in z of degree m. Furthermore, if γ is a non-positive integer, we will assume that at least one of α and β is also a non-positive integer, smaller than γ. We note that F (α, β, γ, z) converges for |z| < 1. By a result of Gauss, if α, β and γ are real with γ > α + β and γ, γ − α and γ − β are not non-positive integers, then F (α, β, γ, z) converges for z = 1 and we have F (α, β, γ, 1) = Γ(γ)Γ(γ − α− β) Γ(γ − α)Γ(γ − β) . For future use, it is worth noting that the hypergeometric function F (α, β, γ, z) satisfies the differential equation z(1− z)d 2F dz2 + (γ − (1 + α+ β)z)dF dz − αβF = 0. (3.13) Our family of dense approximations to ξ/η are as given in the following lemma; their connection to hypergeometric functions will be made apparent later. Lemma 3.4.1. Let r be a positive integer and g ∈ {0, 1}. Put Ar,g(z) = r∑ m=0 ( r − g + 14 m )( 2r − g −m r − g ) (−z)m, Br,g(z) = r−g∑ m=0 ( r − 14 m )( 2r − g −m r ) (−z)m. (3.14) 57 Chapter 3. The Diophantine equation aX4 − bY 2 = 1 (i) There exists a power series Fr,g(z) such that for all complex numbers z with |z| < 1 Ar,g(z)− (1− z)1/4Br,g(z) = z2r+1−gFr,g(z) (3.15) and |Fr,g(z)| ≤ ( r−g+1/4 r+1−g )( r−1/4 r )( 2r+1−g r ) (1− |z|)− 12 (2r+1−g). (3.16) (ii) For all complex numbers z with |1− z| ≤ 1 we have |Ar,g(z)| ≤ ( 2r − g r ) . (3.17) (iii) For all complex numbers z 6= 0 and for h ∈ {1, 0} we have Ar,0(z)Br+h,1,1(z) 6= Ar+h,1(z)Br,0(z). (3.18) Proof. Put Cr,g(z) = r∑ m=0 ( r − 1/4 r −m )( r − g + 1/4 m ) zm and Dr,g = r−g∑ m=0 ( r − 1/4 m )( r − g + 1/4 r − g +m ) zm. Note that, in terms of hypergeometric functions, Ar,g(z) = ( 2r − g r ) F (−1/4− r + g,−r,−2r + g, z), Br,g(z) = ( 2r − g r − g ) F (1/4− r,−r + g,−2r + g, z), Cr,g(z) = ( r − 1/4 r ) F (−1/4− r + g,−r, 3/4, z) and Dr,g(z) = ( r − g + 1/4 r − g ) F (1/4− r,−r + g, 5/4, z), We will begin by proving that Cr,g(z) = Ar,g(1− z), Dr,g(z) = Br,g(1− z). 58 Chapter 3. The Diophantine equation aX4 − bY 2 = 1 The power series F (z) = ∑∞ m=0 amz m is a solution to the differential equa- tion (3.13) precisely when (n+ 1)(γ + n)an+1 = (α+ n)(β + n)an for n = 0, 1, 2, . . . . (3.19) Both Ar,g(1− z) and Cr,g(z) satisfy (3.13) with α = −1/4− r + g , β = −r , γ = 3/4. Since γ is not a non-positive integer, all coefficient ai of power series y(z) are determined by a0. Hence the solution space of (3.13) is one- dimensional. Therefore, Ar,g(1− z) and Cr,g(z) are linearly dependent. On equating the coefficients of zr in (1 + z)2r+g = (1 + z)r−1/4(1 + z)r−g+1/4, we find that Cr,g(1) = r∑ m=0 ( r − 1/4 r −m )( r − g + 1/4 m ) = ( 2r − g r ) = Ar,g(0), and hence Cr,g(z) = Ar,g(1 − z). Similarly, Dr,g(z) = Br,g(1 − z). One can easily observe that Cr,g(z) has positive coefficients. Hence when |1− z| ≤ 1, |Ar,g(z)| = |Cr,g(1− z)| ≤ Cr,g(1) = Ar,g(0) = ( 2r − g r ) . This proves part (ii) of our lemma. To prove (3.15), we define Gr,g(z) = F (r + 1− g, r + 3/4, 2r + 2− g, z) and notice that, for |z| < 1, the functions Ar,g(z), (1 − z)1/4Br,g(z) and z2r+1−gGr,g(z) satisfy (3.13) with α = −1/4− r+ g , β = −r , γ = −2r+ g. Suppose Gr,g(z) = ∞∑ m=0 gmz m. We have g0 = 1 and, for m ≥ 0, gm+1 gm = (r + 1− g +m)(r + 3/4 +m) (m+ 1)(2r + 2− g +m) ≤ r + 1/2− g/2 +m m+ 1 = (−1)m+1(−r−1/2+g/2m+1 ) (−1)m(−r−1/2+g/2m ) . 59 Chapter 3. The Diophantine equation aX4 − bY 2 = 1 Therefore, |Gr,g(z)| ≤ r∑ m=0 (−r − 1/2 + g/2 m ) (−|z|)m = (1− |z|)− 12 (2r+1−g). Since r ≥ 1 and g ∈ {0, 1}, γ = −2r + g is a negative integer. By (3.19), If F (z) = ∑∞ m=0 amz m is a solution to (3.13), then since a0 and a2r−g+1 may vary independently, the solution space of (3.13) is two-dimensional. Therefore, there are constants c1, c2 and c3, not all zero, such that c1Ar,g(z) + c2(1− z)1/4Br,g(z) + c3z2r+1−gGr,g(z) = 0. Letting z = 0, since Ar,g(0) = Br,g(0) 6= 0, we find that c1 = −c2 6= 0. We may thus assume c1 = 1. Substituting z = 1 in above identity thus yields c3 = −Ar,g(z)Gr,g(z) , whence Fr,g(z) = Ar,g(1)Gr,g(1)−1Gr,g(z). In order to complete the proof of part (i), note that, by (3.13), we have Ar,g(1)Gr,g(1)−1 = ( r − 1/4 r ) Γ(r + 1)Γ(r + 5/4− g) Γ(2r + 2− g)Γ(1/4) = ( r−1/4 r )( r−g+1/4 r+1−g )( 2r+1−g r ) . It remains to prove part (iii). By (3.15), Ar,0(z)Br+h,1(z)−Ar+h,1(z)Br,0(z) = z2r+hPr,h(z), where Pr,h(z) is a power series. However, the left hand side of the above identity is a polynomial of degree at most 2r + h, and so Pr,h must be a constant. Letting z = 1, we obtain that Pr,h is not 0. Therefore, Ar,0(z)Br+h,1(z)−Ar+h,1(z)Br,0(z) = 0 if and only if z = 0. 60 Chapter 3. The Diophantine equation aX4 − bY 2 = 1 3.5 Some Algebraic Numbers Combining our polynomials of the previous section with the resolvent forms defined in Section 3.3, we will consider the complex sequences Σr,g given by Σr,g = η2 ξ2 Ar,g(z1)− (−1)r η1 ξ1 Br,g(z1) where z1 = 1− η41/ξ41 . Define Λr,g = ξ4r+1−g1 ξ2 (−t− 1)1/4Σr,g. Let OK be the ring of integers of the number field K = Q( √−t). Put O = {m+ n √−t 2 ∈ OK | m,n ∈ Z}. It is easy to check thatO is a subring ofOK. Let d be the largest square-free divisor of t. From the well-known characterization of algebraic integers in quadratic fields, we have OK = {a+ b 1 + √−d 2 | a, b ∈ Z} if d ≡ 1 mod 4 and OK = {a+ b √−d | a, b ∈ Z} if d ≡ 2, 3 mod 4. Therefore, θ ∈ O if and only if θ ∈ OK, θ − θ̄ ∈ Z. √−t, (3.20) where θ̄ is the complex conjugate of θ. This implies that |θ| ≥ | Im θ| ≥ 1 2 ( √ t). (3.21) We will show that Λr,g is either in O or a fourth root of such an algebraic integer. If Λr,g 6= 0, this provides a lower bound upon |Λr,g|. In conjunction with the inequalities derived in Lemma 3.4.1, this will induce a strong “gap principle”, guaranteeing that solutions to inequality (3.7) must, in a certain sense, increase rapidly in height. For a polynomial P (z) of degree n, we will denote by P ∗(x, y) = xnP (y/x) 61 Chapter 3. The Diophantine equation aX4 − bY 2 = 1 an associated binary form. Let Ar,g and Br,g be as in (3.14) and, as in the proof of Lemma 3.4.1, set Cr,g(z) = Ar,g(1− z) and Dr,g(z) = Br,g(1− z). For z 6= 0, we have Dr,0(z) = zrCr,0(z−1), hence A∗r,0 (z, z + z̄) = z rAr,0 ( 1 + z̄ z ) = zrCr,0 (−z̄ z ) (3.22) = (−1)rz̄rDr,0 (−z z̄ ) = (−1)rz̄rBr,0 ( 1 + z z̄ ) = (−1)rB∗r,0 (z̄, z̄ + z) = (−1)rB̄∗r,0 (z, z + z̄) . Lemma 3.5.1. For any pair of integers (x, y), both A∗r,g(ξ 4(x, y), ξ4(x, y)− η4(x, y)) and B∗r,g(ξ 4(x, y), ξ4(x, y)− η4(x, y)) are contained in the subring O. Proof. It is clear that A∗r,g(ξ 4(x, y), ξ4(x, y)− η4(x, y)) and B∗r,g(ξ 4(x, y), ξ4(x, y)− η4(x, y)) belong to Q( √−t); we need only show that they are of the shape m+n √−t 2 . From the definitions of A∗r,g(x, y), B∗r,g(x, y), ξ(x, y) and η(x, y) (in particu- lar, since ξ4(x, y) − η4(x, y) = 8P (x, y)), this is an immediate consequence of Lemma 4.1 of [6], which, in this case, implies that( a/4 n ) 8n is, for fixed nonnegative integers a and n, a rational integer. We now proceed to show that Λr,g has the desired property. We have Λr,g = ξ1−g1 η2 (−t− 1)1/4A ∗ r,g(ξ 4, ξ4 − η4)− (−1) rξ2g1 ξ2η1 (−t− 1)1/4 B ∗ r,g(ξ 4, ξ4 − η4). 62 Chapter 3. The Diophantine equation aX4 − bY 2 = 1 By Lemmas 3.3.3, 3.5.1 and (3.22), Λr,0 ∈ Z √−t. Similarly, Lemmas 3.3.3 and 3.5.1 imply that Λ4r,1 is an algebraic integer in Q( √−t). It can also be easily seen, from the definition of ξ and η, that Λ4r,1 belongs to O. We claim that it is not a rational integer. To see this, let us start by noting that Σr,g (−t− 1)1/4 = η2 ξ2 Ar,g(z1)− (−1)r η1 ξ1 Br,g(z1) = η ξ (η2/η ξ2/ξ Ar,g(z1)− (−1)r η1/η ξ1/ξ Br,g(z1) ) , where η = ( √−t− 1)1/4 and ξ = (√−t+ 1)1/4. By Lemma 3.5.1, η2/η ξ2/ξ Ar,g(z1)− (−1)r η1/η ξ1/ξ Br,g(z1) ∈ Q( √−t) and so f = Q( √−t,Σr,g) = Q( √−t, (−t− 1)1/4 η ξ ) (3.23) = Q( √−t, (−t+ 1− 2√−t)1/4). If we choose a complex number X so that ξ(X, 1) = η(X, 1) then X ∈ f and P (X, 1) = 1 8 (ξ4(X, 1)− η4(X, 1)) = 0. Since we have assumed that P is irreducible, X and Σr,g both have degree 4 over Q( √−t). Suppose that Λ4r,1 ∈ Z. Then we have for some ρ, ρ1 ∈ {±1,±i}, that Λr,1 = ρΛ̄r,1 and (−t− 1)1/4 = ρ1(−t− 1)1/4), whence, from Lemma 3.3.3, Σr,1 = (−t− 1)1/4ξ−4r1 ξ−12 ρΛ̄r,1 = ξ−4r1 ξ −1 2 η 4r 1 η2ρρ1 ( ξ2 η2 Ar,1 ( 1− ξ 4 η4 ) − (−1)r ξ1 η1 Br,1 ( 1− ξ 4 η4 )) = ρρ1 η4r1 ξ4r1 ( Ar,1 ( 1− ξ 4 1 η41 ) − (−1)r ξ1η2 ξ2η1 Br,1 ( 1− ξ 4 1 η41 )) . This together with Lemmas 3.3.3 and 3.5.1 imply that Σr,1 ∈ Q( √−t, ρρ1), which contradicts the fact that Σr,1 has degree 4 over Q( √−t). We conclude that Λr,1 can not be a rational integer. Therefore, we conclude that, if Λr,g 6= 0, g ∈ {0, 1}, then |Λr,g| ≥ 2 −g 4 t 1 2 − 3g 8 . (3.24) 63 Chapter 3. The Diophantine equation aX4 − bY 2 = 1 3.6 Three Auxiliary Lemmas We will now combine inequality (3.24) with upper bounds from Lemma 3.4.1 to show that solutions to (3.7) are widely spaced : Lemma 3.6.1. If Σr,g 6= 0, then c1(r, g) |ξ1|4r+1−g|ξ2|−3 + c2(r, g) |ξ1|−4r−3(1−g)|ξ2| > 1, where we may take c1(r, g) = 22r+1+g/4√ pir t5/4+3g/8 and c2(r, g) = 21/2+g/4−2r34r+2−2g pi √ r t4r+5/4−13g/8. Proof. By (3.15), we can write∣∣∣(t+ 1)1/4 Λr,g∣∣∣ = |ξ1|4r+1−g|ξ2| ∣∣∣∣(η2ξ2 − ω ) Ar,g(z1) + ωz 2r+1−g 1 Fr,g(z1) ∣∣∣∣ . Since |1 − z1| = 1 and |z1| ≤ 1, from (3.8), (3.9), (3.16), (3.17), and the inequality |ξ1|4 > 4 (1 + t)5/2, we have ∣∣∣(t+ 1)1/4 Λr,g∣∣∣ ≤ |ξ1|4r+1−g|ξ2|L, where L is equal to( 2r − g r ) 2t2 |ξ42 | ( r−g+1/4 r+1−g )( r−1/4 r )( 2r+1−g r ) ( 9t2|ξ41 | )2r+1−g . Comparing this with (3.24), we obtain c1(r, g) |ξ1|4r+1−g|ξ2|−3 + c2(r, g) |ξ1|−4r−3(1−g)|ξ2| > 1, where we may take c1 and c2 so that c1(r, g) ≥ 21+g/4 t5/4+3g/8 ( 2r r ) 64 Chapter 3. The Diophantine equation aX4 − bY 2 = 1 and c2(r, g) ≥ 2g/4 34r+2−2g t4r+5/4−13g/8 ( r−g+1/4 r+1−g )( r−1/4 r )( 2r+1−g r ) . Applying the following version of Stirling’s formula (see Theorem (5.44) of [18]) 1 2 √ k 4k ≤ ( 2k k ) < 1√ pik 4k, (valid for k ∈ N) leads immediately to the stated choice of c1. To evaluate c2(r, g), we begin by noting that( 2r + 1− g r ) ≥ ( 2r r ) ≥ 4 r 2 √ r . Next we will show that( r − g + 1/4 r + 1− g )( r − 1/4 r ) < 1√ 2pir , for r ∈ N and g ∈ {0, 1}, whence we may conclude that( r−g+1/4 r+1−g )( r−1/4 r )( 2r+1−g r ) < √2√ rpi4r . This gives the desired value for c2(r, g). To bound ( r−g+1/4 r+1−g )( r−1/4 r ) , first we note that ( r − 3/4 r ) > ( r + 1/4 r + 1 ) , for r ∈ N. Put Xr = ( r − 3/4 r )( r − 1/4 r ) = yr r , whereby Xr+1 = ( r + 1/4 r + 1 )( r + 3/4 r + 1 ) = ( r2 + r + 2/9 r2 + r ) yr r + 1 . Hence, y1 = 3/16 , yr = 3/16 r−1∏ k=1 k2 + k + 3/16 k2 + k . Since ∞∏ k=1 k2 + k + 3/16 k2 + k = 16 3Γ(1/4)Γ(3/4) = 16 3 √ 2pi , 65 Chapter 3. The Diophantine equation aX4 − bY 2 = 1 we obtain Xr < 1√ 2pir , which completes the proof. We will also have need of the following : Lemma 3.6.2. If r ∈ N and h ∈ {0, 1}, then at most one of {Σr,0,Σr+h,1} can vanish. Proof. Let r be a positive integer and h ∈ {0, 1} . Following an argument of Bennett [1], we define the matrix M: M =  Ar,0(z1) Ar+h,1(z1) (−1) r η1 ξ1 Ar,0(z1) Ar+h,1(z1) (−1)r η1ξ1 Br,0(z1) Br+h,1(z1) η2 ξ2  . The determinant of M is zero because it has two identical rows. Expanding along the first row, we find that Ar,0(z1)Σr+h,1 −Ar+h,1(z1)Σr,0 + (−1)r η1 ξ1 (Ar,0(z1)Br+h,1(z1)−Ar+h,1(z1)Br,0(z1)) vanishes and hence if Σr,0 = Σr+h,1 = 0, then Ar,0(z1)Br+h,1(z1)−Ar+h,1(z1)Br,0(z1) = 0, contradicting part (iii) of Lemma 3.4.1. Our final result of this section follows similar lines to an argument of Evertse [8]. We show : Lemma 3.6.3. Suppose that t > 204. For r ∈ {1, 2, 3, 4, 5}, we have Σr,0 6= 0. Proof. Let r ∈ {1, 2, 3, 4, 5} and suppose that Σr,0 = 0. From (3.15), for each r, the polynomial Ar,0(z)4 − (1− z)B4r,0 has a zero at 0 of order at least 2r + 1. We can thus find polynomials Ar(z), Br(z) and Fr(z) ∈ Z[z], satisfying Ar(z)4 − (1− z)B4r = z2r+1Fr(z). 66 Chapter 3. The Diophantine equation aX4 − bY 2 = 1 In fact, we have A1(z) = 4A1,0(z) = 8− 5z, B1(z) = 4B1,0(z) = 8− 3z, F1(z) = 320− 320z + 81z2, A2(z) = 32 3 A2,0(z) = 64− 72z + 15z2, B2(z) = 32 3 B2,0(z) = 64− 56z + 7z2, F2(z) = 86016− 172032z + 114624z2 − 28608z3 + 2401z4, A3(z) = 128A3,0(z) = 2560− 4160z + 1872z2 − 195z3, B3(z) = 128B3,0(z) = 2560− 3520z + 1232z2 − 77z3, F3(z) = 14057472000− 42172416000z + 48483635200z2 − 26679910400z3 +7150266240z4 − 839047040z5 + 35153041z6, A4(z) = 2048 5 A4,0(z) = 28672− 60928z + 42432z2 − 10608z3 + 663z4, B4(z) = 2048 5 B4,0(z) = 28672− 53760z + 31680z2 − 6160z3 + 231z4, F4(z) = 13989396348928− 55957585395712z + 91916125077504z2 − 79896826347520z3 + 39463764078592z4 − 11050000539648z5 + 1648475542656z6 − 113348764800z7 + 2847396321z8, A5(z) = 8192 21 A5,0(z) = 98304− 258048z + 243712z2 − 99008z3 + 15912z4 − 663z5, B5(z) = 8192 21 B5,0(z) = 98304− 233472z + 194560z2 − 66880z3 + 8360z4 − 209z5. 67 Chapter 3. The Diophantine equation aX4 − bY 2 = 1 and F5(z) = 121733331812352− 608666659061760z + 1301756554248192z2 − 1555026262622208z3 + 1136607561252864z4 − 523630732640256z5 + 151029162176512z6 − 26204424888320z7 + 2515441608384z8 − 113971885760z9 + 1908029761z10. We also define A∗r and B∗r via A∗r(x, y) = x rAr(y/x) and B∗r (x, y) = x rBr(y/x). Since Σr,0 is assumed to be zero, η42 ξ42 = η41(B ∗ r (ξ 4 1 , ξ 4 1 − η41))4 ξ41(A∗r(ξ41 , ξ41 − η41))4 . Let Ir be the integral ideal in Q( √−t) generated by ξ41(A∗r(ξ41 , ξ41−η41))4 and η41(B ∗ r (ξ 4 1 , ξ 4 1 − η41))4, and N(Ir) be the absolute norm of Ir. Since the ideal generated by ξ41(A ∗ r(ξ 4 1 , ξ 4 1 − η41))4− η41(B∗r (ξ41 , ξ41 − η41))4 divides (ξ42 − η42)Ir, we obtain |ξ1|4(4r+1) ∣∣A4r(z1)− (1− z1)B4r (z1)∣∣ =∣∣ξ41(A∗r(ξ41 , ξ41 − η41))4 − η41(B∗r (ξ41 , ξ41 − η41))4∣∣ . From the fact that Ir is contained in an imaginary quadratic field, |ξ1|4(4r+1)|A4r(z1)− (1− z1)B4r (z1)| ≤ N(Ir)1/2|ξ42 − η42|. By (3.15), A4r(z1)− (1− z1)B4r (z1) = z2r+11 Fr(z1), so we may conclude |z1|2r+1|Fr(z1)| ≤ N(Ir)1/2|ξ42 − η42||ξ1|−4(4r+1); i.e. 1 ≤ N(Ir) 1/2|ξ42 − η42||ξ1|−4(4r+1) |z1|2r+1|Fr(z1)| . Since |z1| = |ξ−41 ||ξ41 − η41| and |ξ4i − η4i | = 8P (x, y), it follows after a little work that |ξ1|8r ≤ N(Ir)1/2 ∣∣ξ41 − η41∣∣−4r−1 (8P (x, y))2r+1 |Fr(z1)|−1 . (3.25) 68 Chapter 3. The Diophantine equation aX4 − bY 2 = 1 To estimate N(Ir)1/2, we choose a finite extensionM of Q( √−t) so that the ideal generated by ξ41 and ξ 4 1−η41 inM is a principal ideal, with generator p, say. We denote the extension of Ir to M, by I′r. Let rr be the ideal in M generated by A∗r(u, v) and B∗r (u, v), where u = ξ41 p and v = ξ41−η41 p . Since A∗r(x, x− y) = B∗r (y, y − x), p4r+1r4rB ∗ r (0, 1) 4 ⊂ p4r+1r4r(u,B∗r (0, v)4)(u− v,B∗r (0, v)4) (3.26) ⊂ p4r+1r4r(u,B∗r (0, v)4)(u− v,A∗r(v, v)4) ⊂ p4r+1r4r(u, u− v)(u,B∗r (u, v)4)(u− v,A∗r(u, v)4) ⊂ p4r+1(uA∗(u, v)4, (u− v)B∗r (u, v)4) = I′r, where (m1, . . . ,mn) denote the ideal in M generated by m1, . . . ,mn. We have A∗1(x, y)−B∗1(x, y) = −2y. Therefore, 2(v) ⊂ (A∗1(u, v), B∗1(u, v)) ⊂ r1, where (v) is the ideal generated by v in M. Since B∗1(0, 1) = −3, it follows from (3.26) that 1296(ξ41 − η41)5 ⊂ 1296p(ξ41 − η41)4 = p516v4B∗1(0, 1)4 ⊂ I′1. For r = 2, we first observe that B∗1(x, y)A ∗ 2(x, y)−A∗1(x, y)B∗2(x, y) = −10y3 and (−32x+ 7y)A∗2(x, y)− (−32x+ 15y)B∗2(x, y) = 80xy2. Therefore, by (3.26) we have 80(v)2 ⊂ (−10v3, 80uv2) ⊂ (A∗2(u, v), B∗2(u, v)) ⊂ r2. Since B∗2(0, 1) = 7, we have 804 × 74(ξ41 − η41)9 ⊂ 804 × 74p(ξ41 − η41)8 = 804p9v8B∗2(0, 1)4 ⊂ I′2. When r = 3, we have B∗2(x, y)A ∗ 3(x, y)−A∗2(x, y)B∗3(x, y) = −210y5 and (1616x2 − 1078xy + 77y2)A∗3(x, y)− (1616x2 − 1482xy + 195y2)B∗3(x, y) = −16800x2y3. 69 Chapter 3. The Diophantine equation aX4 − bY 2 = 1 Substituting 77 for B∗3(0, 1), we conclude 168004×774(ξ41−η41)13 ⊂ 168004×774p(ξ41−η41)12 = 168004p13v12B∗3(0, 1)4 ⊂ I′3. For r = 4, setting G4(x, y) = 14178304x3 − 15889280x2y + 4071760xy2 − 162393y3, H4(x, y) = 14178304x3 − 19433856x2y + 6714864xy2 − 466089y3, we may verify that B∗3(x, y)A ∗ 4(x, y)−A∗3(x, y)B∗4(x, y) = −6006y7 and G4(x, y)A∗4(x, y)−H4(x, y)B∗4(x, y) = −150678528y4x3. This implies that 1506785284 × 2314(ξ41 − η41)17 ⊂ 1506785284 × 2314p(ξ41 − η41)16. Since this latter quantity is equal to 1506785284p17v16B∗4(0, 1)4, it follows that 1506785284 × 2314(ξ41 − η41)17 ⊂ I′4. Finally, for r = 5, we have B∗4(x, y)A ∗ 5(x, y)−A∗4(x, y)B∗5(x, y) = −14586y7 and G5(x, y)A∗5(x, y)−H5(x, y)B∗5(x, y) = −134424576y5x4, where G5(x, y) = 43706368x4 − 69346048x3y + 32767856x2y2 − 4764782xy3 + 123519y4, H5(x, y) = 43706368x4 − 80272640x3y + 46006896x2y2 − 8845746xy3 + 391833y4. This implies that 1344245764 × 2094(ξ41 − η41)21 ⊂ 1344245764 × 2094p(ξ41 − η41)20 whereby 1344245764 × 2094(ξ41 − η41)21 ⊂ 1344245764p21v20B∗5(0, 1)4 ⊂ I′5. 70 Chapter 3. The Diophantine equation aX4 − bY 2 = 1 From the preceding arguments, we are thus able to deduce the following series of inequalities : N(I1)1/2|ξ41 − η41|−5 ≤ 1296, N(I2)1/2|ξ41 − η41|−9 ≤ 5604, N(I3)1/2|ξ41 − η41|−13 ≤ (77× 16800)4, N(I4)1/2|ξ41 − η41|−17 ≤ (231× 150678528)4 and N(I5)1/2|ξ41 − η41|−21 ≤ (209× 134424576)4. These will enable us to contradict inequality (3.25) for r ≤ 5, provided we can find a suitably strong lower bound for |ξ1|. Since ξ4i = 4( √−t+ 1)(xi − √−tyi)4 and x1y1 > 64t3, via calculus we have that |ξ1|4 > 216 t15/2, (3.27) whence (3.25) and the assumption that P (x, y) ≤ t2 imply 226r−3t11r−2 < N(Ir)1/2 ∣∣ξ41 − η41∣∣−4r−1 |Fr(z1)|−1 . (3.28) From (3.27), we have |z1| = ∣∣∣∣8P (x, y)ξ41 ∣∣∣∣ < (213 t11/2)−1 < 0.001, and consequently, F1(z1) > 102, F2(z1) > 104, F3(z1) > 1010, F4(z1) > 1013 and F5(z1) > 1014. In case r = 1, inequality (3.28) thus implies that 223t9 < 6635.52× t6, a contradiction for all t. Arguing similarly for r = 2, 3, 4 and 5, and noting that t > 204, completes the proof of Lemma 3.6.3. 71 Chapter 3. The Diophantine equation aX4 − bY 2 = 1 3.7 The Proof Of Theorem 3.1.1 Assume that there are two distinct coprime solutions (x1, y1) and (x2, y2) to inequality (3.7) with |ξ2| > |ξ1|. We will show that |ξ2| is arbitrary large in relation to |ξ1|. In particular, we will demonstrate via induction that |ξ2| > √ r 5 t4r+7/4 ( 4 81 )r |ξ1|4r+3, (3.29) for each positive integer r. Since inequality (3.27) thus implies that |ξ2| > t7r/2+31/8, for arbitrary r, we deduce an immediate contradiction. We first prove inequality (3.29) for r = 1. By (3.12) and (3.27), c1(1, 0) |ξ1|5|ξ2|−3 < 2−13pi−1/2t−5/2 < 0.1, and hence, since Σ1,0 6= 0, Lemma 3.6.1 yields c2(1, 0)|ξ1|−7|ξ2| > 0.9, which, after a little work, implies (3.29). We now proceed by induction. Suppose that (3.29) holds for some r ≥ 1. Then c1(r + 1, 0)|ξ1|4r+5|ξ2|−3 < 2000√ pir2 t12r+13/2 ( 312 24 )r |ξ1|−8r−4, and hence, from (3.27), c1(r + 1, 0)|ξ1|4r+5|ξ2|−3 < 125212√pir2 t −3r−1 ( 312 236 )r < 0.1. If Σr+1,0 6= 0, then by Lemma 3.6.1, c2(r + 1, 0)|ξ1|−4(r+1)−3|ξ2| > 0.9, which leads to inequality (3.29) with r replaced by r + 1. If, however, Σr+1,0 = 0 then by Lemmas 3.6.2 and 3.6.3, both Σr+1,1 and Σr+2,1 are nonzero, and r ≥ 5. Using the induction hypothesis, we find as previously that c1(r + 1, 1)|ξ1|4r+4|ξ2|−3 < 0.1 72 Chapter 3. The Diophantine equation aX4 − bY 2 = 1 and thus by Lemma 3.6.1 conclude that c2(r + 1, 1)|ξ1|−4r−4|ξ2| > 0.9. It follows that |ξ2| > 0.08× √ r + 1 t4r+29/8 ( 4 81 )r |ξ1|4r+4. Consequently, c1(r + 2, 1)|ξ1|4r+8|ξ2|−3 < 24000(r + 1)2 ( 312 24 )r t12r+25/2 |ξ1|−8r−4, whereby, from (3.27) and the fact that r ≥ 5, c1(r + 2, 1)|ξ1|4r+8|ξ2|−3 < 12(r + 1)2 ( 312 236 )r t5−3r < 0.1. Lemma 3.6.1 thus implies the inequality c2(r + 2, 1)|ξ1|−4r−8|ξ2| > 0.9 and so |ξ2| > 0.08 √ r + 1 ( 4 81 )r+1 t−4r−61/8|ξ1|4r+8. From (3.27), it follows that |ξ2| > √ r + 1 5t4r+4+7/4 ( 4 81 )r+1 |ξ1|4r+7, as desired. This completes the proof of inequality (3.29) and hence we conclude that there is at most one solution to (3.7) related to each fourth root of unity. To finish the proof of Theorem 3.1.1, it is enough to show that three of the roots of unity under consideration do not have solutions to (3.7) associated to them. We recall that the polynomial P (x, 1) = x4 + 4tx3 − 6tx2 − 4t2x+ t2 has 4 real roots β1, β2, β3, β4, say, where √ t+ 1 2 + 1 8 √ t − 2 8t < β1 < √ t+ 1 2 + 1 8 √ t − 1 8t −√t+ 1 2 − 1 8 √ t − 1 8t < β2 < − √ t+ 1 2 − 1 8 √ t 1 4 − 5 64t + 22 512t2 < β3 < 1 4 − 5 64t + 23 512t2 −4t− 5 4 + 21 64t − 87 512t2 < β4 < −4t− 54 + 21 64t − 84 512t2 . 73 Chapter 3. The Diophantine equation aX4 − bY 2 = 1 (since t ≥ 18, the polynomial P (x, 1) changes sign between the given bounds). Since P (βi, 1) = 1 8 (ξ4(βi, 1)− η4(βi, 1)) = 0, it follows that, for each 1 ≤ i ≤ 4, η(βi,1)ξ(βi,1) is a fourth root of unity. Noting that η(βi, 1) ξ(βi, 1) − η(βj , 1) ξ(βj , 1) = (√−t− 1√−t+ 1 )1/4 2√−t(βj − βi) (βi − √−t)(βj − √−t) , they are in fact distinct. We now proceed to show that solutions to (3.7) necessarily correspond to fourth roots of unity related to β2. In [22], it is shown that for {V2n+1} defined in Section 3.2, the equation z2 = V4n+1 has no solution. Supposing that z2 = V4n+3, as in the proof of Proposition 3.3.1, there exist integers t1, t2, G and H, so that the integers x, y arising from Proposition 3.3.1 satisfy x = −t1G and y = H. We have x y = −t1G H = −2t1G2 2GH = − √ V4n+3 − V2n+1 Un+1 = − √ V 22n+2 + V 2 2n+1 − V2n+1 V2n+2√ t = −√t (√ 1 + V 22n+1 V 22n+2 − V2n+1 V2n+2 ) , using the fact that V2n+2 = √ t Un+1. Thus∣∣∣∣xy +√t ∣∣∣∣ = √t ∣∣∣∣∣ √ 1 + V 22n+1 V 22n+2 − V2n+1 V2n+2 − 1 ∣∣∣∣∣ . A crude application of the Mean Value Theorem therefore implies that∣∣∣∣xy +√t ∣∣∣∣ < √t and consequently, x/y ∈ (−2√t, 0), whereby the inequalities for βi yield∣∣∣∣xy − β1 ∣∣∣∣ ≥ ∣∣∣√t+ β1∣∣∣− ∣∣∣∣xy +√t ∣∣∣∣ > 2√t−√t = √t, 74 Chapter 3. The Diophantine equation aX4 − bY 2 = 1 ∣∣∣∣xy − β3 ∣∣∣∣ ≥ ∣∣∣√t+ β3∣∣∣− ∣∣∣∣xy +√t ∣∣∣∣ > √t+ 15 −√t = 15 and ∣∣∣∣xy − β4 ∣∣∣∣ ≥ ∣∣∣√t+ β4∣∣∣− ∣∣∣∣xy +√t ∣∣∣∣ > 3t−√t > 2t. Let β ∈ {β1, β3, β4}. We have just shown that if (x, y) is a solution to inequality (3.7), then ∣∣∣∣xy − β ∣∣∣∣ > 15 . If we suppose that ω = η(β,1)ξ(β,1) , then∣∣∣∣ω − ξ(x, y)η(x, y) ∣∣∣∣ = ∣∣∣∣∣η(β, 1)ξ(β, 1) − η( x y , 1) ξ(xy , 1) ∣∣∣∣∣ = ∣∣∣∣∣ 2 √−t(xy − β) (β −√−t)(xy − √−t) ∣∣∣∣∣ , whence the inequalities |β −√−t| < √ 16t2 + 17t and ∣∣∣∣xy −√−t ∣∣∣∣ < √5t (recall that |x/y| < 2√t) imply∣∣∣∣ω − ξ(x, y)η(x, y) ∣∣∣∣ > 25√80t2 + 85t . Since |z| = 8P (x, y)|ξ4(x, y)| ≤ 8t2 |ξ4(x, y)| , this, together with (3.27), contradicts Lemma 3.3.2. This shows that there is no solution related to three of the fourth roots of unity (those corresponding to β1, β3 and β4). Therefore, there is at most a single solution to inequality (3.7). Together with Propositions 3.2.2 and 3.3.1, this completes the proof of Theorem 3.1.1. 75 Bibliography [1] M.A. Bennett. On the representation of unity by binary cubic forms. Trans. Amer. Math. Soc. 353 (2001), 1507–1534. [2] M.A. Bennett and P.G. Walsh, The Diophantine equation b2X4−dY 2 = 1, Proc. A.M.S. 127 (1999) 3481–3491. [3] M.A. Bennett, A. Togbe, P.G. Walsh, A generalization of a theorem of Bumby on quartic Diophantine equations, International J. Number Theory 2 (2006), 195–206. [4] A. Bremner and N. Tzanakis, On squares in Lucas sequences, to appear in J. Number Theory. [5] J.H. Chen, P.M. Voutier, A complete solution of the Diophantine equa- tion x2 + 1 = dy4 and a related family of quartic Thue equations, J. Number Theory 62 (1997), 71–99. [6] G.V. Chudnovsky, On the method of Thue-Siegel, Ann. of Math. II Ser. 117 (1983), 325–382. [7] J.H.E. Cohn, The Diophantine equation x4 −Dy2 = 1 II, Acta Arith. 78 (1997) 401–403. [8] J.H. Evertse, On the representation of integers by binary cubic forms of positive discriminant. Invent. Math. 73 (1983), 117–138. [9] C. Heuberger, On general families of parametrized Thue equations. Al- gebraic number theory and Diophantine analysis (Graz, 1998), 215–238, de Gruyter, Berlin, 2000. [10] V. Krechmar, On the superior bound of the number of representation of an integer by binary forms of the fourth degree (Russian). Bull. Acad. Sci. URSS, Ser. Math. (1939), 289–302 [11] W. Ljunggren, Über die Gleichung x4−Dy2 = 1. (German) Arch. Math. Naturvid. 45, (1942). no. 5, 61–70. 76 Bibliography [12] W. Ljunggren, Ein Satz Über die diophantische Gleichung Ax2−By4 = C (C = 1, 2, 4). (German) Tolfte Skandinaviska Matematikerkon- gressen, Lund, 1953, pp. 188–194. Lunds Universitets Matematiska Inst., Lund, (1954). [13] W. Ljunggren, Some remarks on the diophantine equations x2−Dy4 = 1 and x4 −Dy2 = 1. J. London Math. Soc. 41 1966 542–544. [14] W. Ljunggren, On the diophantine equation Ax4−By2 = C (C = 1, 4). Math. Scand. 21 1967 149–158 (1969) [15] W. Ljunggren, A remark concerning the paper ”On the diophantine equation Ax4−By2 = C(C = 1, 4)”. Math. Scand. 22 1968 282 (1969). [16] A. Rotkiewicz, Application of Jacobi’s symbol to Lehmer’s numbers, Acta Arith. 42 (1983), 163–187. [17] C.L. Siegel, Über einige Anwendungen diophantischer Approximatio- nen, Abh. Preuss. Akad. Wiss. 1 (1929). [18] K. Stromberg, An Introduction to Classical Real Analysis. Wadsworth international mathematical series, 1981. [19] A. Thue, Bemerkungen Über gewisse Näherungsbrüche algebraischer Zahlen, Kra. Vidensk. Selsk. Skrifter. I. mat. nat. Kl. (1908), no 3, Kra, 1908. [20] A. Thue, Über rationale Annaḧerungswerte der reellen Wüzel der ganzen Funktion dritten Grades x3 − ax − b. Kra. Vidensk. Selsk. Skrifter. I. Mat. Nat. Kl. (1908), no. 6, Kra, 1908. [21] A. Thue, Ein Fundamental Theorem zur Bestimmung vou Annäherungs-Werten aller Würzeln gewisser ganzer Funktionen. J. Reine Angew. Mat. 138 (1910), 96–108. [22] A. Togbe, P.M. Voutier, P.G. Walsh, Solving a family of Thue equations with an application to the equation x2 − Dy4 = 1. Acta Arith. 120.1 (2005), 39–58. [23] I. Wakabayashi, On a family of quartic Thue inequalities. I. J. Number Theory 66 (1997), no. 1, 70–84. [24] I. Wakabayashi, Cubic Thue inequalities with negative discriminant. J. Number Theory 97 (2002), no. 2, 222–251. 77 Bibliography [25] D.T. Walker, On the Diophantine equation mX2 − nY 2 = ±1, Amer. Math. Monthly 74 (1967), 504–513. [26] P.G. Walsh, Diophantine equations of the form aX4 − bY 2 = ±1, in Algebraic Number Theory and Diophantine Analysis, Proceeding of In- ternational Conference in Graz 1998 (Walter de Gruyter, Berlin 2000), pp. 531–554. 78 Chapter 4 The Diophantine Equation aX4 − bY 2 = 2 3 4.1 Introduction The problem of determining upper bounds for the number of integer points on elliptic curves has received considerable attention, and is notoriously dif- ficult. In a series of papers [11],[12],[13],[14],[15], Ljunggren proved absolute upper bounds for the number of positive integer solutions to equations of the form aX4 − bY 2 = c c ∈ {±1,−2,±4}. In the case c = 1, Ljunggren [12] proved that the equation X4−bY 2 = 1 has at most two solutions in positive integers X,Y . This result was extended by Bennett and the third author in [3], wherein it was proved that all equations of the form a2X4−bY 2 = 1, with a > 1, have at most one solution in positive integers. In [5], Chen and Voutier proved that the equation aX4 − Y 2 = 1 (a > 2) has at most one solution in positive integers. These results have re- cently been extended by the first author [1], wherein it was proved that any equation of the form aX4 − bY 2 = 1 has at most two solutions in positive integers. A key fact in proving this result is that one needs only to prove the result for the subfamily of equations (t+ 1)X4 − tY 2 = 1. Noticeably absent from the above list of values for c is the particular value c = 2. The third author [19] has recently proved, under stringent conditions on a, b, that the equation aX4−bY 2 = 2 has at most one solution in positive integers. Proving a similar result for the general equation aX4 − bY 2 = 2 remains elusive. However, the methods of [1] can be employed for the particular subfamily of equations (t + 2)X4 − tY 2 = 2. The purpose of the present paper is to prove an upper bound in the case that c = 2. In particular, we prove 3 A version of this chapter has been accepted for publication. Akhtari. S, Togbe. A, and Walsh. P.G. On the equation aX4 − bY 2 = 2. Acta Arith. 131 (2008), 145-169. 79 Chapter 4. The Diophantine Equation aX4 − bY 2 = 2 Theorem 4.1.1. For all odd positive integers t > 40, 000, the equation (t+ 2)X4 − tY 2 = 2 (4.1) has at most two solutions in positive integers X,Y . For the remaining odd positive integers t, equation (4.1) has at most three solutions in positive integers X,Y . Theorem 4.1.1 is likely not best possible. We conjecture that the only posi- tive integer solution to equation (4.1) is (X,Y ) = (1, 1), and more generally, that any equation of the form aX4−bY 2 = 2, with a and b odd, has at most one solution in positive integers, and that such a solution must arise from the fundamental solution to the quadratic equation aX2 − bY 2 = 2. This conjecture was verified for (4.1) in the range 1 ≤ t < 1200. The organization of the paper is as follows. In Section 4.2 we reduce the solution of (4.1) to the problem of determining all squares in certain lin- ear recurrences, yielding equations of the form X2 = V2k+1(t). In Section 4.3, the problem is further reduced to a family of Thue equations with the property that the roots of the associated family of polynomials can be given explicitly in terms of the parameter t. We finish off the first part of the paper by proving some lower bounds in Section 4.4, which will be needed in the sequel. In Section 4.5 the family of Thue equations are shown to be written in terms of resolvent forms, and the concept of a solution being associated to a fourth root of unity is introduced. In Section 4.6 the Main Lemma is proved, which shows that for each fourth root of unity, there can be at most one associated solution to the Thue equation. The proof of this fact uses the hypergeometric method, and in particular, proves that for fixed t, there is at most one solution (k, x) to the equation X2 = V4k+3(t). In Section 4.7, we use Thue’s method to solve the equation X2 = V4k+1(t) completely, completing the proof of Theorem 4.1.1. 4.2 Linear Recurrences For t ≥ 1 and odd, let α = √ t+2+ √ t√ 2 , and for k ≥ 0, define sequences {Vi}, {Ui} by α2k+1 = V2k+1 √ t+ 2 + U2k+1 √ t√ 2 , α2k = V2k + U2k √ t(t+ 2). 80 Chapter 4. The Diophantine Equation aX4 − bY 2 = 2 All positive integer solutions (X,Y ) to the quadratic equation (t+ 2)X2 − tY 2 = 2 are given by (X,Y ) = (V2k+1, U2k+1). Thus, a positive integer solution (X,Y ), with X > 1, to equation (4.1) is equivalent to the existence of positive integers (t, k) satisfying X2 = V2k+1(t). Our strategy to prove Theorem 4.1.1 is to first prove that for fixed t, the equations X2 = V4k+1(t) (4.2) and X2 = V4k+3(t) (4.3) are solvable for at most one integer k > 0. This results in an upper bound of three solutions for equation (4.1). We then show that for t large enough, equation (4.2) has no solution with k > 0. 4.3 Reduction To Thue Equations We show here that a solution to equation (4.1) gives rise to a solution to a Thue equation. It is easily proved by induction that the following relation holds for {Vn} V4k+1 = V 22k+1 + 2tU 2 2k. (4.4) Therefore, if V4k+1 = X2, then X2 = V 22k+1 + 2tU 2 2k, and hence, (X−V2k+1)(X+V2k+1) = 2tU22k. It follows that there are positive integers r, s, A,B, with rs = 2t and U2k = 2AB, for which X − V2k+1 = 2rA2, X + V2k+1 = 2sB2. Consequently, V2k+1 = sB2−rA2, and from the easily seen identity V2k+1 = V2k + tU2k, one has that V2k = sB2 − rA2 − 2tAB. Substituting these ex- pressions for V2k and U2k into the equation V 22k − t(t+ 2)U22k = 1 results in the equation s2B4 − 4tsAB3 − 12tA2B2 + 4rtBA3 + r2A4 = 1. 81 Chapter 4. The Diophantine Equation aX4 − bY 2 = 2 Multiplying this equation through by s2, and setting x = −sB, y = A (4.5) shows that x and y satisfy the Thue equation x4 + 4tx3y − 12tx2y2 − 8t2xy3 + 4t2y4 = s2. (4.6) Similar to (4.4), one has the relation V4k+3 = V 22k+1 + 2tU 2 2k+2, and so if V4k+3 = X2, then X2 = V 22k+1 + 2tU 2 2k+2. Therefore, (X − V2k+1)(X + V2k+1) = 2tU22k+2, and so there are positive integers r, s, A,B, with rs = 2t and U2k+2 = 2AB, for which X − V2k+1 = 2rA2, X + V2k+1 = 2sB2. It follows that V2k+1 = sB2−rA2, and from identity V2k+2 = V2k+1+tU2k+2, one has that V2k+2 = sB2−rA2+2tAB. Substituting these expressions into the equation V 22k+2 − t(t+ 2)U22k+2 = 1 gives s2B4 + 4tsAB3 − 12tA2B2 − 4rtBA3 + r2A4 = 1. Multiplying through by s2 and letting x = sB, y = A (4.7) shows that x and y satisfy equation (4.6). Asymptotically, the roots of the polynomial pt(x) = x4 + 4tx3 − 12tx2 − 8t2x+ 4t2 (4.8) are given as follows. We adopt the L-notation defined in [8], pages 1151- 1152, that we recall here. Let c be a real number, assume f(x), g(x), and h(x) are real functions and h(x) > 0 for x > c. We will write f(x) = g(x) + Lc(h(x)) if g(x)− h(x) ≤ f(x) ≤ g(x) + h(x), for x > c. 82 Chapter 4. The Diophantine Equation aX4 − bY 2 = 2 Therefore we obtain β(1) = √ 2t+ 1 + 1 2 √ 2t − 12t − 916t√2t + L6 ( 0.59 t2 ) β(2) = −√2t+ 1− 1 2 √ 2t − 12t + 916t√2t + L792 ( 0.48 t2 ) β(3) = 12 − 516t + 2364t2 + L105 ( 0.49 t3 ) β(4) = −4t− 52 + 2116t − 8464t2 + L5 ( 1.349 t3 ) . Carefully analyzing the construction of the Thue equation (4.6), it is not difficult to verify that if X2 = V4k+1, with k > 0, then the corresponding positive integer solution (x, y), given by (4.5), to equation (4.6), satisfies the property that the closest root to x/y is β(4), whereas if X2 = V4k+3, then the corresponding positive integer solution (x, y), given by (3.4), to equation (4.6), satisfies the property that the closest root to xy is β (1). We make this comment more concrete in the following. Lemma 4.3.1. If X2 = V4k+1 is solvable with X an integer and k > 0, and if (x, y), given by (4.5), is the corresponding solution to the Thue equation (4.6), then |x/y − β(4)| < 1 16t|y|4 . If X2 = V4k+3 is solvable with X an integer and k > 0, and if (x, y), given by (4.7), is the corresponding solution to the Thue equation (4.6), then |x/y − β(1)| < 1 4t|y|4 . Proof. We will prove the result for the equation X2 = V4k+1, as the proof for the second case is essentially identical. We will use the fact that V2k/U2k is a close approximation to √ t(t+ 2). The definition of (x, y) gives x y = −sB A = −2sB2 2AB = −X + V2k+1 U2k = − √ V4k+1 + V2k+1 U2k . Using the relations V4k+1 = V 22k+1 + 2tU 2 2k and V2k+1 = V2k + tU2k, we find that x/y = −( √ (V2k/U2k)2 + 2t(V2k/U2k) + (t2 + t) + (V2k/U2k) + t). Using the fact that V2k/U2k is a close approximation to √ t(t+ 2), we see that the above expression is closely approximated by −4t−5/2, which shows 83 Chapter 4. The Diophantine Equation aX4 − bY 2 = 2 that the closest root to x/y is β(4). As |pt(x, y)| = |y|4 4∏ i=1 |x/y − β(i)| ≤ 4t2, we see that |x/y − β(4)| ≤ 4t 2 |y|4∏i6=4 |x/y − β(i)| . Using the crude estimate 4t for each factor |x/y − β(i)|, we see that |x/y − β(4)| ≤ 1 16t|y|4 . 4.4 Lower Bounds For k, t And |y| In order to prove Theorem 4.1.1, we first need to verify that equation (4.1) has only the positive integer solution (X,Y ) = (1, 1) for all t up to a certain bound. Two independent computations, using PARI and MAGMA, were run in order to verify that equation (4.1) has no positive integer solutions other than (X,Y ) = (1, 1) for all 1 ≤ t ≤ 1200. This was achieved by com- puting all integer solutions to each Thue equation of the form pt(x, y) = s2, where pt(x, y) is given in (4.8), and s is a divisor of 2t. We will also need a lower bound for k. The polynomial V4k+1(t) is monic and of even degree. Therefore, Runge’s method can be applied directly to equa- tion (4.2). Fortunately, Runge’s method has been implemented in MAGMA by Beukers and Tengely (see [4]). This rather short MAGMA computation verified that no positive integer solutions (X, t) to equation (4.2) exist for each 1 ≤ k ≤ 24. We now describe how to obtain the lower bound k > 6 in the case of solving X2 = V4k+3. Firstly, it is trivial to see that V4k+3 ≡ 3 (mod 4) for k even. To see this, first note that from the definition of Vk, V2k+1 = (t+ 1)V2k−1 + tU2k+1, and as (t+ 1 + √ t(t+ 2))−1 = t+ 1−√t(t+ 2), we also have the relation V2k−3 = (t+ 1)V2k−1 − tU2k+1. 84 Chapter 4. The Diophantine Equation aX4 − bY 2 = 2 Combining these two equations gives the second order linear recurrence V2k+1 = (2t+ 2)V2k−1 − V2k−3. Since t is odd, 4 divides 2t+ 2, and so for each k, V2k+1 ≡ −V2k−3 (mod 4). As V1 = 1 and V3 = 2t+ 1 ≡ 3 (mod 4), it follows that V2k+1 ≡ 1 (mod 4) for 2k + 1 ≡ 1, 7 (mod 8), and V2k+1 ≡ 3 (mod 4) for 2k + 1 ≡ 3, 5 (mod 8). If k = 1, then the equation is simply X2 = 8t3 + 20t2 + 12t + 1, which is easily seen to have no solutions in positive integers X, t using MAGMA. For the case k = 3, the equation V15 = X2 implies that V5 = 4t2 + 6t+ 1 is either a square or three times a square by elementary divisibility properties of terms in the sequence {Vn} (see [9] for details). Evidently, neither of these possibilities is possible. Finally, if k = 5, we use the fact that for each i ≥ 0, the Jacobi symbol (V16i+7/V16i+5) = −1, which is easily proved by induction, however we provide the details for the reader. The proof uses the above linear recurrence equation for {V2k+1}, the above congruences (mod 4) for V2k+1, along with basic manipulation of Jacobi symbols. First, (V7/V5) = (((2t+ 2)V5 − V3)/V3) = (−V3/V5) = −(V3/V5) = (V5/V3) = (−V1/V3) = (−1/V3) = −1. Thus, the result holds for i = 1. We next show that (V8j+7/V8j+5) = −(V8j−1/V8j−3), for all j ≥ 0, which upon putting j = 2i and j = 2i − 1 gives the desired 85 Chapter 4. The Diophantine Equation aX4 − bY 2 = 2 result. (V8j+7/V8j+5) = (−V8j+3/V8j+5) = −(V8j+3/V8j+5) = (V8j+5/V8j+3) = (−V8j+1/V8j+3) = −(V8j+1/V8j+3) = −(V8j+3/V8j+1) = −(V8j−1/V8j+1) = −(V8j+1/V8j−1) = −(V8j−3/V8j−1) = −(V8j−1/V8j−3). We use the above lower bounds for k to obtain lower bounds for |y|, where (x, y) is a solution to the Thue equation (4.6) arising from a solution to equation (4.1). In the case of a solution to (4.1) with X2 = V4k+3, we see from the above construction that 2ry2 = √ V4k+3 − V2k+1 = √ V 22k+1 + 2tU 2 2k+2 − V2k+1. It follows that 2ry2 = 2tU22k+2√ V 22k+1 + 2tU 2 2k+2 + V2k+1 . Dividing the numerator and denominator of this equation by √ tU2k+2 gives 2ry2 = 2 √ tU2k+2√ (V 22k+1/tU 2 2k+2) + 2 + (V2k+1/ √ tU2k+2) . Using the fact that V2k+1 < U2k+2, it follows that 2ry2 > √ tU2k+2, and from the lower bound U2k+2 > (2t)k > (2t)6, it follows that |y| > 2 √ 2t9/4. In the case of a solution to X2 = V4k+1, with k > 0, we obtain a much larger lower bound since the equation X2 = V4k+1 was solved using Runge’s method for 1 ≤ k ≤ 24. In this case, an analysis similar to the one given above shows that |y| > 210t11. 86 Chapter 4. The Diophantine Equation aX4 − bY 2 = 2 4.5 Associated Fourth Roots Of Unity Let pt(x, y) = x4 + 4tx3y − 12tx2y2 − 8t2xy3 + 4t2y4 and t be a positive integer. Our goal is to find, for fixed t, an upper bound upon the number of coprime nonzero integer solutions to the inequality 0 < pt(x, y) ≤ 4t2. (4.9) To proceed, let ξ = ξ(x, y) and η = η(x, y) be linear functions of (x, y) so that ξ4 = 4 (− √ − t 2 + 1) ( x+ √−2ty)4 and η4 = 4 (− √ − t 2 − 1) (x−√−2ty)4 . We refer to (ξ, η) as a pair of resolvent forms. Note that ξ4 = −η̄4 and that pt(x, y) = 1 8 (ξ4 − η4) and if (ξ, η) is a pair of resolvent forms then there are precisely three others with distinct ratios, say (−ξ, η), (iξ, η) and (−iξ, η). Let ω be a fourth root of unity, (ξ, η) a fixed pair of resolvent forms and set z = 1− ( η(x, y) ξ(x, y) )4 . We say that the integer pair (x, y) is related to ω if∣∣∣∣ω − η(x, y)ξ(x, y) ∣∣∣∣ < pi12 |z|. It turns out that each nontrivial solution (x, y) to (3.3) is related to a fourth root of unity : Lemma 4.5.1. Suppose that (x, y) is a positive integral solution to (4.9), with ∣∣∣∣ωj − η(x, y)ξ(x, y) ∣∣∣∣ = min0≤k≤3 ∣∣∣∣ekpii/2 − η(x, y)ξ(x, y) ∣∣∣∣ . Then |ωj − η(x, y) ξ(x, y) | < pi 12 |z(x, y)|. (4.10) 87 Chapter 4. The Diophantine Equation aX4 − bY 2 = 2 Proof. We begin by noting that |z| = ∣∣∣∣ξ4 − η4ξ4 ∣∣∣∣ = 8 pt(x, y)|ξ4| , and, from xy 6= 0, |ξ4(x, y)| ≥ 8 √ 2( √ t)5, whereby |z| ≤ 4t 2 √ 2( √ t)5 < 1. Since η = −ξ̄, it follows that∣∣∣∣ηξ ∣∣∣∣ = 1, |1− z| = 1. Now let 4θ = arg ( η(x,y)4 ξ(x,y)4 ) . We have√ 2− 2 cos(4θ) = |z| < 1, and so |θ| < pi12 . Since ∣∣∣∣ωj − η(x, y)ξ(x, y) ∣∣∣∣ ≤ |θ|, it follows that∣∣∣∣ωj − η(x, y)ξ(x, y) ∣∣∣∣ ≤ 14 |4θ|√2− 2 cos(4θ) ∣∣∣∣1− η(x, y)4ξ(x, y)4 ∣∣∣∣ . From the fact that |4θ|√ 2−2 cos(4θ) < pi/3 whenever 0 < |θ| < pi 12 , we obtain the desired inequality. We now put ωi = η(β(i),1) ξ(β(i),1) for 1 ≤ i ≤ 4. The ωi are the distinct fourth roots of unity. The following lemma represents a key step towards the proof of Theorem 4.1.1. Lemma 4.5.2. If X2 = V4k+1 is solvable, with X an integer and k > 0, and if (x, y), given by (4.5), is the corresponding solution to the Thue equation (4.6), then (x, y) is associated to ω4. If X2 = V4k+3 is solvable, with X an integer and k > 0, and if (x, y), given by (4.7), is the corresponding solution to the Thue equation (4.6), then (x, y) is associated to ω1. 88 Chapter 4. The Diophantine Equation aX4 − bY 2 = 2 Proof. Assume that X2 = V4k+3, as the other case is proved in the same way. The goal is to prove that∣∣∣∣∣η(β(1), 1)ξ(β(1), 1) − η(x, y)ξ(x, y) ∣∣∣∣∣ < pi12 ∣∣∣∣∣1− ( η(x, y) ξ(x, y) )4∣∣∣∣∣ . Obtaining a common divisor for the left side, expanding and simplifying, shows that the above inequality is the same as |2√−2t||x− β(1)y| |β(1) +√−2t||x+ y√−2t| < |2Pt(x, y)| |ξ(x, y)4| . Cross multiplying the above and dividing through by |y|4, we reduce the problem to proving the inequality |2√−2t||2−√−2t||x/y +√−2t|3 |β(1) +√−2t| (4.11) < |x/y − β(2)||x/y − β(3)||x/y − β(4)|. (4.12) Lemma 4.3.1 shows that x/y is very close to β(1). Indeed, Lemma 4.3.1, together with the lower bound for |y| determined in section 4.4, show that |x/y− β(1)| < 1/(28t10). This difference is sufficiently small that we will re- place x/y in (4.11) by β(1), which entails that we need to prove the inequality |2√−2t||2−√−2t||β(1)+√−2t|2 < |β(1)−β(2)||β(1)−β(3)||β(1)−β(4)|. (4.13) Expanding the left hand side of (4.13) gives an estimate with leading terms 16t2+20t, while that for the right hand side has leading terms 16t2+24t. 4.6 The Main Lemma The following represents the most crucial lemma of this paper, as it provides for an absolute bound for the number of integer solutions to equation (4.1). Lemma 4.6.1. There is at most one solution of (4.9) related to each fourth root of unity. Because of the lower bound for k obtained in section 4.4, we may assume that k > 6. Furthermore, since ξ4i = 4(− √ − t2 + 1)(xi + √−2tyi)4, via calculus one may conclude that |ξ1|4 > t23/2. (4.14) 89 Chapter 4. The Diophantine Equation aX4 − bY 2 = 2 4.6.1 Approximating Polynomials The following Lemma gives a family of dense approximations to ξ/η from rational function approximations to the binomial function (1− z)1/4. Lemma 4.6.2. Let r be a positive integer and g ∈ {0, 1}. Put Ar,g(z) = r∑ m=0 ( r − g + 14 m )( 2r − g −m r − g ) (−z)m, Br,g(z) = r−g∑ m=0 ( r − 14 m )( 2r − g −m r ) (−z)m. (i) There exists a power series Fr,g(z) such that for all complex numbers z with |z| < 1 Ar,g(z)− (1− z)1/4Br,g(z) = z2r+1−gFr,g(z) (4.15) and |Fr,g(z)| ≤ ( r−g+1/4 r+1−g )( r−1/4 r )( 2r+1−g r ) (1− |z|)− 12 (2r+1−g). (4.16) (ii) For all complex numbers z with |1− z| ≤ 1 we have |Ar,g(z)| ≤ ( 2r − g r ) . (4.17) (iii) For all complex numbers z 6= 0 and for h ∈ {1, 0} we have Ar,0(z)Br+h,1,1(z) 6= Ar+h,1(z)Br,0(z). (4.18) Proof. See the proof of Lemma 4.1 of [1]. Combining the polynomials of Lemma 4.6.2 with the resolvent forms ξ(x, y) and η(x, y), we will consider the complex sequences Σr,g given by Σr,g = η2 ξ2 Ar,g(z1)− (−1)r η1 ξ1 Br,g(z1) where z1 = 1− η41/ξ41 . Define Λr,g = ξ4r+1−g1 ξ2 (− t2 − 1)1/4 Σr,g. (4.19) 90 Chapter 4. The Diophantine Equation aX4 − bY 2 = 2 Let OK be the ring of integers of the number field K = Q( √−2t). Put O = {m+ n √−2t 2 ∈ OK | m,n ∈ Z}. It is easy to check thatO is a subring ofOK. Let d be the largest square-free divisor of 2t. From the well-known characterization of algebraic integers in quadratic fields, we have OK = {a+ b 1 + √−d 2 | a, b ∈ Z} if d ≡ 1 mod 4 and OK = {a+ b √−d | a, b ∈ Z} if d ≡ 2, 3 mod 4. Therefore, θ ∈ O if and only if θ ∈ OK, θ − θ̄ ∈ Z( √−2t), (4.20) where θ̄ is the complex conjugate of θ. This implies that |θ| ≥ | Im θ| ≥ 1 2 ( √−2t). (4.21) We will show that Λr,g is either in O or a fourth root of such an algebraic integer. If Λr,g 6= 0, this provides a lower bound upon |Λr,g|. In conjunction with the inequalities derived in Lemma 4.6.2, this will induce a strong “gap principle”. Lemma 4.6.3. If (x1, y1) and (x2, y2) are two pairs of rational integers then ξ(x1, y1)η(x2, y2) (− t2 − 1)1/4 , ξ(x1, y1)3ξ(x2, y2) and η(x1, y1)3η(x2, y2) are contained in O. Proof. This is an immediate consequence of the definition of resolvent forms. For a polynomial P (z) of degree n, we will denote by P ∗(x, y) = xnP (y/x) an associated binary form. Let Ar,g and Br,g be as in Lemma 6.2, and set Cr,g(z) = Ar,g(1− z) and Dr,g(z) = Br,g(1− z). 91 Chapter 4. The Diophantine Equation aX4 − bY 2 = 2 For z 6= 0, we have Dr,0(z) = zrCr,0(z−1), hence A∗r,0 (z, z + z̄) = z rAr,0 ( 1 + z̄ z ) = zrCr,0 (−z̄ z ) = (−1)rz̄rDr,0 (−z z̄ ) = (−1)rz̄rBr,0 ( 1 + z z̄ ) = (−1)rB∗r,0 (z̄, z̄ + z) = (−1)rB̄∗r,0 (z, z + z̄) . Lemma 4.6.4. For any pair of integers (x, y), both A∗r,g(ξ 4(x, y), ξ4(x, y)− η4(x, y)) and B∗r,g(ξ 4(x, y), ξ4(x, y)− η4(x, y)) belong to O. Proof. It is clear thatA∗r,g(ξ4(x, y), ξ4(x, y)−η4(x, y)) andB∗r,g(ξ4(x, y), ξ4(x, y)− η4(x, y)) belong to Q( √−2t); we need only show that they belong to O. From the definitions of A∗r,g(x, y), B∗r,g(x, y), ξ(x, y) and η(x, y) (in particu- lar, since ξ4(x, y) − η4(x, y) = 8pt(x, y)), this is an immediate consequence of Lemma 4.1 of [6], which, in this case, implies that( a/4 n ) 8n is, for fixed nonnegative integers a and n, a rational integer. Proposition 4.6.5. Let Λr,g be the complex number defined in (4.19). Then Λr,0 belongs to Z. √−2t and Λr,1 is a fourth root of an algebraic integer in O. Proof. We have Λr,g = ξ1−g1 η2 (− t2 − 1)1/4 A∗r,g(ξ 4 1 , ξ 4 1 − η41)− (−1)rξ2g1 ξ2η1 (− t2 − 1)1/4 B∗r,g(ξ 4 1 , ξ 4 1 − η41). By Lemmas 4.6.3 and (4.6.4), Λr,0 ∈ Z √−2t. Similarly, Lemmas 4.6.3 and 4.6.4 imply that Λ4r,1 is in O. We claim that it is not a rational integer. To see this, let us start by noting that Σr,g (− t2 − 1)1/4 = η2 ξ2 Ar,g(z1)− (−1)r η1 ξ1 Br,g(z1) = η ξ (η2/η ξ2/ξ Ar,g(z1)− (−1)r η1/η ξ1/ξ Br,g(z1) ) , 92 Chapter 4. The Diophantine Equation aX4 − bY 2 = 2 where η = ( √ − t2 − 1)1/4 and ξ = ( √ − t2 + 1)1/4. By Lemma 4.6.4, η2/η ξ2/ξ Ar,g(z1)− (−1)r η1/η ξ1/ξ Br,g(z1) ∈ Q( √−2t) and so f = Q( √−2t,Σr,g) = Q( √−2t, (− t 2 − 1)1/4 η ξ ) = Q( √−2t, (− t 2 + 1−√−2t)1/4). If we choose a complex number X so that ξ(X, 1) = η(X, 1) then X ∈ f and pt(X, 1) = 1 8 (ξ4(X, 1)− η4(X, 1)) = 0. Since pt is irreducible, X and Σr,g both have degree 4 over Q( √−2t). Suppose that Λ4r,1 ∈ Z. Then we have for some ρ, ρ1 ∈ {±1,±i}, that Λr,1 = ρΛ̄r,1 and (− t2 − 1)1/4 = ρ1(− t2 − 1)1/4), whence, from Lemma 4.6.3, Σr,1 = (− t2 − 1) 1/4ξ−4r1 ξ −1 2 ρΛ̄r,1 = ξ−4r1 ξ −1 2 η 4r 1 η2ρρ1 ( ξ2 η2 Ar,1 ( 1− ξ 4 η4 ) − (−1)r ξ1 η1 Br,1 ( 1− ξ 4 η4 )) = ρρ1 η4r1 ξ4r1 ( Ar,1 ( 1− ξ 4 1 η41 ) − (−1)r ξ1η2 ξ2η1 Br,1 ( 1− ξ 4 1 η41 )) . This together with Lemmas 4.6.3 and 4.6.4 imply that Σr,1 ∈ Q( √−2t, ρρ1), which contradicts the fact that Σr,1 has degree 4 overQ( √−2t). We conclude that Λr,1 can not be a rational integer. By (4.21), we may conclude that, if Λr,g 6= 0, g ∈ {0, 1}, then |Λr,g| ≥ 2 −g 4 (2t) 1 2 − 3g 8 . (4.22) 4.6.2 Gap Principles Lemma (4.9) shows that each integer pair (x, y) is related to precisely one fourth root of unity. Let us fix such a fourth root, say ω, and suppose that we have distinct coprime positive solutions (x1, y1) and (x2, y2) to inequality (4.9), each related to ω. We will assume that |ξ(x2, y2)| ≥ |ξ(x1, y1)|. Let 93 Chapter 4. The Diophantine Equation aX4 − bY 2 = 2 us write ηi = η(xi, yi) and ξi = ξ(xi, yi). We will use the following results to prove that (x1, y1) and (x2, y2) are far apart in height. Since |z| = 8pt(x, y)|ξ|4 ≤ 32t2 |ξ|4 , (4.23) it follows from (4.10) that |ξ1η2 − ξ2η1| = |ξ1(η2 − ωξ2)− ξ2(η1 − ωξ1)| (4.24) ≤ 8pit 2 3 ( |ξ1| |ξ32 | + |ξ2| |ξ31 | ) ≤ 16pit 2 |ξ2| 3|ξ31 | . On the other hand, choosing our fourth root appropriately, we have √2(−√− t2 + 1)1/4 √2(−√− t2 + 1)1/4√−2t√ 2(− √ − t2 − 1)1/4 − √ 2(− √ − t2 − 1)1/4 √−2t ( x1 x2 y1 y2 ) = ( ξ1 ξ2 η1 η2 ) and so |ξ1η2 − ξ2η1| = ∣∣∣∣4( t2 + 1)1/4√2t (x1y2 − x2y1) ∣∣∣∣ . Since x1y2 − x2y1 is a nonzero integer (recall that we assumed gcd(xi, yi) = 1), we have |ξ1η2 − ξ2η1| ≥ 4 √ 2t( t 2 + 1)1/4, (4.25) and thus, combining (4.24) and (4.25), we conclude that if (x1, y1) and (x2, y2) are distinct solutions to (4.9), related to ω, with |ξ(x2, y2)| ≥ |ξ(x1, y1)| then |ξ2| > 3 t −5/4 4pi |ξ1|3. (4.26) We will now combine inequality (4.22) with upper bounds from Lemma 4.6.2 to show that solutions to (4.9) are widely spaced: Lemma 4.6.6. If Σr,g 6= 0, then c1(r, g) |ξ1|4r+1−g|ξ2|−3 + c2(r, g) |ξ1|−4r−3(1−g)|ξ2| > 1, 94 Chapter 4. The Diophantine Equation aX4 − bY 2 = 2 where we may take c1(r, g) = √ pi22r+11/4+5g/8 3 √ r t5/4+3g/8 and c2(r, g) = 21/4+5g/8−2r332r+1−g pi √ r t4r+5/4−13g/8. Proof. By (4.15), we can write∣∣∣∣( t2 + 1)1/4 Λr,g ∣∣∣∣ = |ξ1|4r+1−g|ξ2| ∣∣∣∣(η2ξ2 − ω ) Ar,g(z1) + ωz 2r+1−g 1 Fr,g(z1) ∣∣∣∣ . Since |1 − z1| = 1 and |z1| ≤ 1, from (4.10), (4.23), (4.16), (4.17), and the inequality |ξ1|4 > 8 √ 2 (t)5/2, we have∣∣∣∣( t2 + 1)1/4 Λr,g ∣∣∣∣ ≤ |ξ1|4r+1−g|ξ2| (( 2r − g r ) 2pit2 3|ξ42 | + ( r−g+1/4 r+1−g )( r−1/4 r )( 2r+1−g r ) ( 9t2|ξ41 | )2r+1−g) . Comparing this with (4.22), we obtain c1(r, g) |ξ1|4r+1−g|ξ2|−3 + c2(r, g) |ξ1|−4r−3(1−g)|ξ2| > 1, where we may take c1 and c2 so that c1(r, g) ≥ 2 11/4+5g/8 t5/4+3g/8pi 3 ( 2r r ) and c2(r, g) ≥ 25g/8−1/4 332r+1−g t4r+5/4−13g/8 ( r−g+1/4 r+1−g )( r−1/4 r )( 2r+1−g r ) . Applying the following version of Stirling’s formula (see Theorem (5.44) of [16]) 1 2 √ k 4k ≤ ( 2k k ) < 1√ pik 4k, 95 Chapter 4. The Diophantine Equation aX4 − bY 2 = 2 (valid for k ∈ N) leads immediately to the stated choice of c1. One can also show via Stirling’s formula that for r ∈ N and g ∈ {0, 1}, we have( r−g+1/4 r+1−g )( r−1/4 r )( 2r+1−g r ) < √2√ rpi4r (see the proof of Lemma 6.1 of [1] for more details). This gives the desired value for c2(r, g). 4.6.3 The Proof of Lemma 4.6.1 We will start this section with the statements of two Lemmas from [1]. These Lemmas allow us to apply the strong gap principle provided by Lemma 4.6.5. We note here that although ξ(x, y) and η(x, y) are defined differently in [1], those properties of ξ and η used in the proofs of Lemmas 6.2 and 6.3 of [1], hold for our choice of ξ and η in the present paper. Lemma 4.6.7. If r ∈ N and h ∈ {0, 1}, then at most one of {Σr,0,Σr+h,1} can vanish. Lemma 4.6.8. Suppose that t > 1200. For r ∈ {1, 2, 3, 4, 5}, we have Σr,0 6= 0. Assume that there are two distinct coprime solutions (x1, y1) and (x2, y2) to inequality (4.9) with |ξ2| > |ξ1|. We will show that |ξ2| is arbitrary large in relation to |ξ1|. In particular, we will demonstrate via induction that |ξ2| > √ r t4r+7/4 ( 4 332 )r |ξ1|4r+3, (4.27) for each positive integer r. So by (4.14), |ξ2| > t7r for arbitrary r, a contradiction. To prove inequality (4.27) for r = 1, we use (4.26) and (4.14) to get c1(1, 0)|ξ1|5|ξ2|−3 < 2 43/4√pi 3 t5 |ξ1|−4 < 0.01, and hence, since Σ1,0 6= 0, Lemma 4.6.6 yields c2(1, 0)|ξ1|−7|ξ2| > 0.09, 96 Chapter 4. The Diophantine Equation aX4 − bY 2 = 2 which, immediately, implies (4.27). We now proceed by induction. Suppose that (4.27) holds for some r ≥ 1. Then c1(r + 1, 0)|ξ1|4r+5|ξ2|−3 < √ pi√ 3r2 t12r+26/4 ( 336 26 )r |ξ1|−8r−4, and hence, from (6.1), and the fact that t ≥ 1200, c1(r + 1, 0)|ξ1|4r+5|ξ2|−3 < 0.1. If Σr+1,0 6= 0, then by Lemma 4.6.6, c2(r + 1, 0)|ξ1|−4(r+1)−3|ξ2| > 0.9, which again leads to (4.27). If, however, Σr+1,0 = 0, then by Lemmas 4.6.7 and 4.6.8, both Σr+1,1 and Σr+2,1 are nonzero, and r ≥ 2. Using the induction hypothesis, we find as previously that c1(r + 1, 1)|ξ1|4r+4|ξ2|−3 < 0.001, and thus by Lemma 4.6.6 conclude that c2(r + 1, 1)|ξ1|−4r−4|ξ2| > 0.999. It follows that |ξ2| > √ r + 1 27/8 t4r+29/8 ( 4 332 )r+1 |ξ1|4r+4. Consequently, from (4.14), r ≥ 6 and t > 1200, c1(r + 2, 1)|ξ1|4r+8|ξ2|−3 < 0.001. Therefore Lemma 6.5 implies that c2(r + 2, 1)|ξ1|−4r−8|ξ2| > 0.999, and so |ξ2| > pi27/8 √ r + 2 ( 4 332 )r+2 t−4r−61/8|ξ1|4r+8. From (4.14), it follows that |ξ2| > √ r + 1 t4r+4+7/4 ( 4 332 )r+1 |ξ1|4r+7, 97 Chapter 4. The Diophantine Equation aX4 − bY 2 = 2 as desired. This completes the proof of inequality (4.27), and hence we con- clude that there is at most one solution to (4.9) related to each fourth root of unity. We conclude from the above, in conjunction with Lemma 4.5.2, that there are at most three solutions in positive integers to equation (4.1). In partic- ular, there is the solution (X,Y ) = (1, 1), and for both i = 1 and i = 3, at most one integer k for which X2 = V4k+i is solvable. We now proceed to the proof that for t large enough, the equation X2 = V4k+1 is not solvable for all k > 0. 4.7 An Effective Measure Of Approximation In this section we will apply the hypergeometric method to obtain effective measures of approximation to the two roots β(3) and β(4). Because of the relation β(3)β(4) = −2t, we will only need to deal with one of the roots, say β(3). Our first lemma is Thue’s “Fundamentaltheorem” [17] together with its relation to the hypergeometric function, as discovered by Siegel. The reader is also referred to Proposition 1 in [10], or Lemma 3.1 in [18]. Lemma 4.7.1. Let α1, α2, c1 and c2 be complex numbers with α1 6= α2. For n ≥ 2, we define the following polynomials a(X) = n2 − 1 6 (α1 − α2) (X − α2) , c(X) = n2 − 1 6 α1 (α1 − α2) (X − α2) , b(X) = n2 − 1 6 (α2 − α1) (X − α1) , d(X) = n2 − 1 6 α2 (α2 − α1) (X − α1) , u(X) = −c2 (X − α2)n , z(X) = c1 (X − α1)n . Putting λ = (α1 − α2)2 /4, 98 Chapter 4. The Diophantine Equation aX4 − bY 2 = 2 for any positive integer r, we define ( √ λ)rAr(X) = a(X)X∗n,r(z, u) + b(X)X ∗ n,r(u, z) and ( √ λ)rBr(X) = c(X)X∗n,r(z, u) + d(X)X ∗ n,r(u, z). Then, for any root β of P (X) = z(X)− u(X), the polynomial Cr(X) = βAr(X)−Br(X) is divisible by (X − β)2r+1. Proof. This is a simplified version Lemma 2.1 from [5], obtained by noting that if P (X) satisfies the differential equation given there, with U(X) = (X − α1)(x− α2), then P (X) must be of the form given here, which allows us to determine the above expressions. Lemma 4.7.2. With the above notation, put w(x) = z(x)/u(x) and write w(x) = µeiϕ with µ ≥ 0 and −pi < ϕ ≤ pi. Put w(x)1/n = µ1/neiϕ/n. (i) For any x ∈ C such that w = w(x) is not a negative real number or zero,(√ λ )r Cr(x) ={ β ( a(x)w(x)1/n + b(x) ) − ( c(x)w(x)1/n + d(x) )} Xn,r(u, z) − (βa(x)− c(x))u(x)rRn,r(w), with Rn,r(w) = Γ(r + 1 + 1/n) r!Γ(1/n) ∫ w 1 ((1− t)(t− w))r t1/n−r−1 dt, where the integration path is the straight line from 1 to w. (ii) Let w = eiϕ, 0 < ϕ < pi and put √ w = eiϕ/2. Then |Rn,r(w)| ≤ nΓ(r + 1 + 1/n) r!Γ(1/n) ϕ ∣∣1−√w∣∣2r . Proof. This is Lemma 2.5 of [5]. Lemma 4.7.3. Let u,w and z be as above. Then∣∣X∗n,r(u, z)∣∣ ≤ 4|u|r Γ(1− 1/n)r!Γ(r + 1− 1/n) ∣∣1 +√w∣∣2r−2 . Proof. This is Lemma 2.6 of [5]. 99 Chapter 4. The Diophantine Equation aX4 − bY 2 = 2 Lemma 4.7.4. Let N4,r be the greatest common divisor of the numerators of the coefficients of X4,r(1− 2x) and let D4,r be the least common multiple of the denominators of the coefficients of X4,r(x). Then the polynomial (D4,r/N4,r)X4,r(1− 2x) has integral coefficients. Moreover, N4,r = 2r and D4,r Γ(3/4)r! Γ(r + 3/4) < 0.8397·5.342r and D4,rΓ(r + 5/4)Γ(1/4)r! < 0.1924·5.342 r. Proof. Using the so-called Kummer transformation, we can write X4,r(1− 2x) = 2F1 (−r,−r − 1/4;−2r; 2x) . Expanding the right-hand side, we find that X4,r(1−2x) = r∑ i=0 (−1)i (r + 1) · · · (2r − i) 3 · 7 · · · (4r − 1) ( r i ) (4r−4i+1) · · · (4r+1)22r−ixi. Therefore, 2r divides N4,r and by examining the coefficient of xr, we see that N4,r = 2r. We know turn to the inequalities. From the arguments in the proof of Proposition 2(c) from [10], we obtain D4,r < exp ( 1.6708r + 3.43 3 √ r ) < 5.341227r for r ≥ 20 000. Since exp(0.000073r) > exp(1.46) > 2 for such values of r, the upper bound for D4,r holds for r ≥ 20 000. For r ≥ 2, Γ(r + 5/4) Γ(1/4)r! = 5 16 r∏ i=2 i+ 1/4 i < 5 16 exp (∫ r 1 log ( x+ 1/4 x ) dx ) < 5 16 exp (∫ r 1 dx 4x ) ≤ 5 16 r1/4. As a consequence, the inequalities in the statement of the lemma hold for r ≥ 20 000. A computation, similar to those described in the proof of Proposition 2 from [10], shows that the same inequalities holds for all smaller values of r. 100 Chapter 4. The Diophantine Equation aX4 − bY 2 = 2 Lemma 4.7.5. Let α1, α2, Ar(X), Br(X) and P (X) be defined as in Lemma 3.1 and let a, b, c and d be complex numbers satisfying ad− bc 6= 0. Define Kr(X) = aAr(X) + bBr(X) and Lr(X) = cAr(X) + dBr(X). If (x− α1) (x− α2)P (x) 6= 0, then Kr+1(x)Lr(x) 6= Kr(x)Lr+1(x), for all r ≥ 0. Proof. This is Lemma 2.7 of [5]. Lemma 4.7.6. Let θ ∈ R. Suppose that there exist k0, l0 > 0 and E,Q > 1 such that for all r ∈ N, there are rational integers pr and qr with |qr| < k0Qr and |qrθ − pr| ≤ l0E−r satisfying prqr+1 6= pr+1qr. Then for any rational integers p and q with |q| ≥ 1/(2l0), we have∣∣∣∣θ − pq ∣∣∣∣ > 1c|q|κ+1 , where c = 2k0Q(2l0E)κ and κ = logQlogE . Proof. This is Lemma 2.8 from [5]. For the remainder of this section, we shall assume that t is a fixed integer greater than 37. We shall also simplify our notation here to reflect the fact that we have n = 4. We shall use Rr and Xr instead of R4,r and X4,r. We now determine the quantities defined in the Lemma 4.7.1. Put α1 = √−2t, α2 = − √−2t, c1 = ( 1 + √ −t/2 ) /2, c2 = ( 1− √ −t/2 ) /2, then pt(X) = X4 + 4tX3 − 12tX2 − 8t2X + 4t2. We define τ = √ t+ √ t+ 2√ 2 and ρ = √ τ2 + 1 = √ t+ 2 + √ t2 + 2t. for any positive integer t. The preliminary results above will now be used in order to obtain an effective measure of approximation to β(3). By Lemma 4.7.2, we want to choose x so that β(3) = c(x)w(x)1/4 + d(x) a(x)w(x)1/4 + b(x) , 101 Chapter 4. The Diophantine Equation aX4 − bY 2 = 2 and for this purpose we will select x = 0. We have w = w (0) = 2 + √−2t −2 +√−2t ,( τ − i τ + i )2 = w, ( τ − i ρ )2 = τ − i τ + i , and so w1/4 = τ − i ρ . Using the fact that ρ2 = τ2 + 1, one can check that −iτ − 1 + iρ −τ + i− ρ = −τ + ρ, and since a(0) = −10t, b(0) = −10t, c(0) = −10t√−2t and d(0) = 10t√−2t, it follows that β(3) = c(0)w1/4 + d(0) a(0)w1/4 + b(0) . Therefore, the first term in the expression for (−2t)r/2Cr(0) in Lemma 4.7.2 disappears. We now construct our sequence of rational approximations to β(3). By Lemma 4.7.1, Lemma 4.7.2, we have that λ = −2t, and moreover (−2t)r/2Ar(0) = a(0)X∗r (z(0), u(0)) + b(0)X∗r (u(0), z(0)), (−2t)r/2Br(0) = c(0)X∗r (z(0), u(0)) + d(0)X∗r (u(0), z(0)), (−2t)r/2Cr(0) = − ( β(3)a(0)− c(0))u(0)rRr(w). Therefore, we have (−2t)r/2Ar(0) = −10t [X∗r (z(0), u(0)) +X∗r (u(0), z(0))] , (−2t)r/2Br(0) = 5(−2t)3/2 [X∗r (z(0), u(0))−X∗r (u(0), z(0))] , (−2t)r/2Cr(0) = 10t2r+1 [ β(3) −√−2t] (−2 +√−2t)r Rr(w). These quantities will form the basis for our approximations. We first elim- inate some common factors. We can write u(0) = t2 (−2 +√−2t) and 102 Chapter 4. The Diophantine Equation aX4 − bY 2 = 2 z(0) = t2 ( 2 + √−2t). Using Lemma 4.7.2, Lemma 4.7.3, and triangular inequality, we obtain |Ar(0)| ≤ 80t(3r+2)/2 (2 + t)r/2 Γ(3/4)r!Γ(r + 3/4) ∣∣1 +√w∣∣2r−2 ; |Br(0)| ≤ 80 √ 2 t(3r+3)/2 (2 + t)r/2 Γ(3/4)r! Γ(r + 3/4) ∣∣1 +√w∣∣2r−2 ; and |Cr(0)| ≤ 40 t(3r+2)/2 ϕ ∣∣∣β(3) −√−2t∣∣∣ (2 + t)r/2Γ(r + 5/4) r!Γ(1/4) ∣∣1−√w∣∣2r . In the other hand, after some routine manipulations, we find that (−2t)r/2Ar(0) = −10t2r+1 N4,r D4,r { D4,r N4,r [( 2 + √−2t)rXr (1− 2η) + (−2 +√−2t)rXr (1− 2η̄)]} and (−2t)r/2Br(0) = 5(−2)3/2 t2r+3/2N4,r D4,r { D4,r N4,r [(−2 +√−2t)rXr (1− 2η̄)− (2 +√−2t)rXr (1− 2η)]} ; where η = 2 2+ √−2t . By Lemma 4.7.4, the quantities inside the braces can be expressed as (−1)r(e− f√−2t)± (e− f√−2t), where e and f are rational integers, and recalling from Lemma 4.7.4 that N4,r = 2r, considering the cases of r being even or odd separately, we find that Pr = D4,rBr(0) 20 · t[(3r+3)/2] and Qr = D4,rAr(0) 20 · t[(3r+3)/2] (4.28) are rational integers. We note for future reference that if r is even, then Pr will be divisible by t. The numbers Pr/Qr are those that will be used as the rational approximations to β(3). We have Qrβ (3) − Pr = Sr, where Sr = D4,rCr(0) 20 · t[(3r+3)/2] . 103 Chapter 4. The Diophantine Equation aX4 − bY 2 = 2 We wish to show that these are good approximations. This will be done by estimating |Pr|, |Qr| and |Sr| from above. It is readily verified that |1 + √ w(0)| = 2τ√ τ2 + 1 = 2− 1 2t +O ( 11 16t2 ) ; and in particular we have that |1 + √ w(0)| < 2. Therefore, we have, for t ≥ 37, that |Qr| ≤ 3.36 (22 √ t)r. Similarly, one obtains for t ≥ 37 that |Pr| ≤ 4.75 √ t (22 √ t)r. Also we have ∣∣∣1−√w(0)∣∣∣2 = 4 τ2 + 1 ≤ 2 t . With φ as in Lemma 4.7.2, it can be shown that 2ϕ/pi ≤ sinϕ and sinϕ = Im w(0) = −2 √ 2t/(t+ 2) ≥ −2 √ 2/t. From the estimates for the roots of pt(X) given in section 3, we know that 0 < β(3) < 0.5, and so ϕ ∣∣∣β(3) −√−2t∣∣∣ ≤ pi(2 + 1√ 2t ) ≤ pi ( 2 + 1√ 2 ) . Combining these inequalities with Lemma 4.7.4, we obtain |Sr| < 3.3 ( 11√ t )r ; for t ≥ 37. Note also that since β(3)β(4) = −2t, we have 2tQr + β(4)Pr = −β(4)Sr. With these estimates, Lemma 4.7.6 gives the following. 104 Chapter 4. The Diophantine Equation aX4 − bY 2 = 2 Lemma 4.7.7. Suppose that t ≥ 37. Define κ = log ( 22 √ t ) log (√ t/11 ) . For j = 3, 4 and any rational integers p and q, we have∣∣∣p− β(j)q∣∣∣ > 1 cj |q|κ for |q| ≥ 1, where c3 = 147.84 √ t ( .6 √ t )κ and c4 = 4598 √ t (.44t)κ . Proof. In each case we will apply Lemma 4.7.5 and Lemma 4.7.6. First notice that PrQr+1 − Pr+1Qr is a non-zero multiple of Ar+1(0)Br(0) − Ar(0)Br+1(0). Applying Lemma 4.7.5, with a = d = 1, b = c = 0 and x = 0, we see that PrQr+1 6= Pr+1Qr. For β(3), we put pr = Pr and qr = Qr, and apply Lemma 7.6 with k0 = 3.36, l0 = 3.3, E = √ t/11 and Q = 22 √ t. For β(4), we take advantage of the fact that P2r is divisible by t. In this case, we set pr = −2Q2r, qr = P2r/t, sr = S2rβ(4)/t, and apply Lemma 7.6 accordingly. Since −4t − 2 < β(4) < −4t, we can put k0 = 4.75/ √ t, l0 = 13.2, E = t/112 = ( √ t/11)2, and Q = 484t = (22 √ t)2. We see therefore that the value of κ in this case is the same as in the case of β(3). 4.8 Completion Of The Proof Of Theorem 4.1.1 The goal now is to solve equation (4.2) for all t > 40, 000. We remark that the analysis here will use the fact that, by not restricting that a solution (x, y) have the property that x/y is close to β(4), as asserted in Lemma 4.3.1, we can restrict to the case that s in equation (4.6) satisfies the inequality s < √ 2t. The can be seen by considering once again the equation s2B4 − 4tsAB3 − 12tA2B2 + 4rtBA3 + r2A4 = 1 appearing in section 3. Since rs = 2t, we define s0 = min(r, s), and multiply the above equation through by s20. By defining (x, y) = (−sB,A) if s0 = s and (x, y) = (rA,B) if s0 = r, one obtains the Thue equation x4 + 4tx3y − 12tx2y2 − 8t2xy3 + 4t2y4 = s20, 105 Chapter 4. The Diophantine Equation aX4 − bY 2 = 2 where, as discussed above, s0 divides 2t, and also, s0 ≤ √ 2t. It is not difficult to verify that if s0 = s, then the closest root of pt(X) to x/y is β(4), while if s0 = r, then the closest root of pt(X) to x/y is β(3). The consequence of this remark is that one then needs only to solve the Thue inequality |pt(x, y)| ≤ 2t, as opposed to having 4t2 on the right hand side. We first obtain a lower bound for |y| for any solution to the Thue inequality. We will assume, in the construction of pt, given in section 3, that r < √ 2t, as the case s < √ 2t actually gives a larger lower bound for |y|. Recall that y = A, where X − V2k+1 = √ V4k+1 − V2k+1 = 2rA2. Recall also that r < 2t. We make use of the inequality U2k > (2t)k−1, which is easily proved by induction. We note that because of the relation V4k+1 = V 22k+1 + 2tU 2 2k, we can deduce the following expression √ V4k+1 − V2k+1 = √ 2tU2k√ V2k+1/ √ 2tU2k + 1 + V2k+1/ √ 2tU2k . Therefore, y2 = A2 > √ 2tU2k 4t > ( √ 2t)2k−1. By the fact that k > 24, we deduce that |y| > (2t)47/4. As remarked earlier, the assumption s < √ 2t implies that x/y is closest to β(3). In other words, |x − β(3)y| = mini=1,2,3,4 |x − β(i)y|, and since |Pt(x, y)| ≤ 2t, it follows that |x − β(3)y| < (2t)1/4. Therefore, as y > 4, x/y > β(3) − (2t)1/4/4, and so |x/y − β(4)| > β(3) − (2t)1/4/4− β(4) > 4t− (2t)1/4/4 + 3− 21/(16t) . . . . Similarly for i = 1, 2, |x/y − β(i)| > √ 2t− (2t)1/4/4− 1/2 . . . . 106 Chapter 4. The Diophantine Equation aX4 − bY 2 = 2 Therefore, because t > 40, 000, it is readily deduced that |x/y − β(3)| < 1 15.9ty4 . A similar argument for β(4) gives |x/y − β(4)| < 1 31.9t2y4 . Combining the above upper bound for |x/y − β(3)| with the lower bound proved in Lemma 4.7.7, it follows that |y|3−κ < c3(t) 15.9t . For t > 40, 000, κ < 2.9, and we conclude that 3t2 > 147.84 √ t(.6 √ t)3 15.9t > 147.84 √ t(.6 √ t)κ 15.9t > |y|.1 > (221t23).1 > 4t2, which is not possible. Similarly for β(4), combining the upper and lower bounds for |x/y − β(4)| gives |y|3−κ < c3(t) 31.9t2 . Again since t > 40000, we have that κ < 2.9, and we therefore conclude that 13t1.5 > 4598 √ t(.44t)3 31.9t2 > 4598 √ t(.44t)κ 31.9t2 > |y|.1 > (221t23).1 > 4t2, which is not possible for t > 10. 107 Bibliography [1] S. Akhtari, The equation aX4−bY 2 = 1. To Appear in J. Reine Angew. Math. [2] M.A. Bennett. On the representation of unity by binary cubic forms, Trans. Amer. Math. Soc. 353 (2001), 1507-1534. [3] M.A. Bennett and P.G. Walsh, The Diophantine equation b2X4 − dY 2 = 1. Proc. A.M.S. 127 (1999), 3481-3491. [4] F. Beukers and S. Tengely, An implementation of Runge’s method for Diophantine equations. preprint, 2005. [5] J.H. Chen and P.M. Voutier. A complete solution of the Diophantine equation x2 + 1 = dy4 and a related family of quartic Thue equations. J. Number Theory 62 (1997), 71-99. [6] G.V. Chudnovsky, On the method of Thue-Siegel, Ann. of Math. II Ser. 117 (1983), 325-382. [7] J.H. Evertse. On the representation of positive integers by binary cubic forms of positive discriminant, Invent. Math. 73 (1983), 117-138. [8] C. Heuberger, A. Togbé, and V. Ziegler. Automatic solutions of families of Thue equations and an example of degree 8, J. Symbolic Comput. 38 (2004), 1145-1163. [9] D.H. Lehmer. An extended theory of Lucas functions. Ann. Math. 31 (1930), 419-448. [10] G. Lettl, A. Pethö and P.M. Voutier. Simple families of Thue inequali- ties. Trans. Amer. Math. Soc. 351 (1999), 1871-1894. [11] W. Ljunggren. Zur Theorie der Gleichung x2 + 1 = Dy4. Avh. Norsk. Vid. Akad. Oslo (1942), 1-27. 108 Bibliography [12] W. Ljunggren. Über die Gleichung x4−Dy2 = 1. Arch. Math. Naturv. 45 (1942) no.5. [13] W. Ljunggren. Einige Eigenschaften der Einheiten reeller quadratis- cher und reinbiquadratischer Zahl-Körper usw. Oslo Vid.-Akad. Skrifter (1936), nr. 12. [14] W. Ljunggren. Ein Satz über die Diophantische Gleichung Ax2−By4 = C (C = 1, 2, 4) Tolfte Skand. Matemheikerkongressen, Lund, 1953, 188- 194, (1954). [15] W. Ljunggren. On the Diophantine equation Ax4−By2 = C(C = 1, 4). Math. Scand. 21 (1967), 149-158. [16] K. Stromberg. An Introduction to Classical Real Analysis, Wadsworth International Mathematical Series, 1981. [17] A. Thue. Ein Fundamentaltheorem zur Bestimmung von Annäherungswerten aller Wurzeln gewisser ganzer Funktionen, J. Reine Angew. Math. 138 (1910), 96-108. [18] A. Togbe, P.M. Voutier, and P.G. Walsh. Solving a family of Thue equations with an application to the equation x2 − Dy4 = 1, Acta Arith. 120 (2005), 39-58. [19] P.G. Walsh. On a family of quartic equations and a Diophantine prob- lem of Martin Gardner. Glasnik Matematiki 41 (2006), 217-221. 109 Chapter 5 Addendum on the equation aX4 − bY 2 = 2 4 In a recent paper [2], the authors proved that for t > 40, 000, the Diophan- tine equation (t + 2)X4 − tY 2 = 2 has at most two solutions in positive integers X,Y . In this addendum, we provide a simple argument due to Ljunggren [3] which shows that the above result implies that any equation of the form aX4 − bY 2 = 2 has at most two solutions in positive integers X,Y . We necessarily restrict our attention to pairs of odd integers a, b for which the quadratic equation ax2 − by2 = 2 is solvable in odd integers x, y. Given such a pair of integers a, b, let (x, y) = (u1, v1) denote the smallest solution in positive integers to ax2 − by2 = 2, and define τ = τa,b = u1 √ a+ v1 √ b√ 2 . For i ≥ 1 odd, define sequences {ui}, {vi} by τ i = ui √ a+ vi √ b√ 2 . Then all positive integer solutions (x, y) to the quadratic equation ax2 − by2 = 2 are given by (x, y) = (ui, vi). Theorem 5.0.1. If a, b are odd positive integers with τ = τa,b > 280, then the equation (1.1) aX4 − bY 2 = 2 4 A version of this chapter has been submitted for publication. Akhtari. S, Togbe. A, and Walsh. P.G. Addendum On the equation aX4 − bY 2 = 2. 110 Chapter 5. Addendum on the equation aX4 − bY 2 = 2 has at most two solutions in positive integers X,Y . For the remaining finite set of pairs of positive integers a, b, equation (1.1) has at most three solutions in positive integers X,Y . Theorem 1.1 is likely not best possible. We conjecture that any equation of the form aX4 − bY 2 = 2, with a and b odd, has at most one solution in positive integers, and that such a solution must arise from the fundamental solution to the quadratic equation aX2 − bY 2 = 2. This conjectured was verified for (a, b) = (t+ 2, t), with t in the range 1 ≤ t < 1200. Proof We will assume that a, b are odd positive integers for which there is at least one solution in odd integers (X,Y ) to the equation aX4 − bY 2 = 2. Thus, there is at least one odd positive integer k with the property that uk is a square, and we assume that k represents the smallest such integer. We there- fore define the positive integer X specifically by uk = X2. Write u1 = l1s21 with l1 squarefree. Then by elementary divisibility properties of terms in the sequence {ui}, it follows that l1 divides k. If l1 > 1, write ul1 = l2s22 with l2 squarefree. By the same reasoning as before, l1l2 must divide k. If l2 > 1, then we write ul1l2 = l3s 2 3 with l3 squarefree, and again it follows that l1l2l3 must divide k. Since k is finite, this process must stop, and so there are squarefree integers l1, l2, . . . , lj such that k = l1l2 · · · lj . Furthermore, arguing as above, if k1 is any odd positive integer for which uk1 is a square, then k = l1l2 · · · lj must be a divisor of k1. Thus, we define t = au2k − 2 = aX4 − 2, and put γ = √ t+ 2 + √ t√ 2 . For i ≥ 1 odd, we define new sequences {Ui}, {Vi} by γi = Ui √ t+ 2 + Vi √ t√ 2 . It follows that for each odd i ≥ 1, UiX 2 = uki, and so uki is a square precisely when Ui is a square. Since the set of squares in the sequence {ui} are contained in the subsequence {uki}, we see that 111 Chapter 5. Addendum on the equation aX4 − bY 2 = 2 there is a one to one correspondence between the set of squares in {ui}, and the set of squares in {Ui}. Finally, if τa,b ≥ 280, then it is easily verified that t = aX4 − 2 > 40, 000, and so we may deduce that there are at most two squares in the sequence {ui}. 112 Bibliography [1] S. Akhtari, The Diophantine equation aX4− bY 2 = 1. To Appear in J. Reine Angew. Math. [2] S. Akhtari, A. Togbe, P.G. Walsh, On the equation aX4 − bY 2 = 2. Acta Arith. 131 (2008), 145-169. [3] W. Ljunggren, Über die Gleichung x4−Dy2 = 1. Arch. Math. Naturv. 45 (1942) no.5. 113 Chapter 6 Geometry Of Quartic Thue Equations 5 6.1 Introduction In this paper, we will consider irreducible binary quartic forms with integer coefficients; i.e. polynomials of the shape F (x, y) = a0x4 + a1x3y + a2x2y2 + a3xy3 + a4y4. The discriminant D of F is given by D = a60(α1 − α2)2(α1 − α3)2(α1 − α4)2(α2 − α3)2(α2 − α4)2(α3 − α4)2, where α1 , α2, α3 and α4 are the roots of F (x, 1) = a0x4 + a1x3 + a2x2 + a3x+ a4. The invariants of F form a ring, generated by two invariants of weights 4 and 6, namely I = IF = a22 − 3a1a3 + 12a0a4 and J = JF = 2a32 − 9a1a2a3 + 27a21a4 − 72a0a2a4 + 27a0a23. These are algebraically independent and every invariant is a polynomial in I and J . For the invariant D, we have 27D = 4I3 − J2. In [1], we show that if JF = 0 and F splits in R then the equation |F (x, y)| = 1 (6.1) 5A version of this chapter will be submitted for publication. Akhtari. S. Geometry of quartic Thue equations. 114 Chapter 6. Geometry Of Quartic Thue Equations has at most 12 solutions in integers x, y. In this paper we will give upper bounds for the number of integral solutions to (6.1) with large discriminant and no restriction on J . We will use some ideas of Stewart [16] to prove Theorem 6.1.1. Let F (x, y) be an irreducible binary form with integral coefficients and degree 4. The Diophantine equation (6.1) has at most 61 solutions in integers x and y (with (x, y) and (−x,−y) regarded as the same), provided that the discriminant of F is greater than 10500. We also combine some analytic methods from [16] with some geometric methods from [13] to show that Theorem 6.1.2. Let F (x, y) be an irreducible binary form with integral coefficients and degree 4 that splits in R. Then the Diophantine equation (6.1) has at most 37 solutions in integers x and y (with (x, y) and (−x,−y) regarded as the same), provided that the discriminant of F is greater than 10500. Note that If (x, y) is a solution to (6.1) then (−x,−y) is also a solution to (6.1). So here we will only count the solutions with y ≥ 0. Let F (x, y) = a0(x− α1y)(x− α2y)(x− α3y)(x− α4y). We call forms F1 and F2 equivalent if they are equivalent under GL2(Z)- action; i.e. if there exist integers a1 , a2 , a3 and a4 such that F1(a1x+ a2y, a3x+ a4y) = F2(x, y) for all x, y, where a1a4 − a2a3 = ±1. We denote by NF the number of solutions in integers x and y of the Diophantine equation (6.1). If F1 and F2 are equivalent then NF1 = NF2 and DF1 = DF2 . Suppose there is a solution (x0, y0) to the equation (6.1). Since gcd(x0, y0) = 1, there exist integers x1, y1 ∈ Z with x0y1 − x1y0 = 1. Then F ∗(1, 0) = 1, 115 Chapter 6. Geometry Of Quartic Thue Equations where, F ∗(x, y) = F (x0x+ x1y, y0x+ y1y). Therefore, F ∗ is a monic form equivalent to F . From now on we will assume F is monic. In this paper we give an upper bound for the number of integral solutions to F (x, y) = ±1. For the equation F (x, y) = h of degree 4, one may use an argument of Bombieri and Schmidt [2] to prove that if N is a given bound in the special case h = 1, then N4ν is a corre- sponding bound in the general case, where ν is the number of distinct prime factors of h. 6.2 Quartic Forms And Elliptic Curves Let us define, for a quartic form F , a quartic covariant, the Hessian G, by G(x, y) = d2F dx2 d2F dy2 − ( d2F dxdy )2 . Then G(x, y) = A0x4 +A1x3y +A2x2y2 +A3xy3 +A4y4, where A0 = 3(8a0a2 − 3a21), A1 = 12(6a0a3 − a1a2), A2 = 6(3a1a3 + 24a0a4 − 2a22), (6.2) A3 = 12(6a1a4 − a2a3), A4 = 3(8a2a4 − 3a23). For the binary form F (x, y) with HessianG(x, y), the sextic covariantQ(x, y) is defined by Q(x, y) = δF δx . δG δy − δF δy . δG δx . We have the following syzygy between the covariants: (see [4] for a proof) 27Q2 = −G3 + 48IF 2G+ 64JF 3. (6.3) 116 Chapter 6. Geometry Of Quartic Thue Equations Theorem 6.2.1. [Hermite] The binary forms with integer coefficients, with given invariants, can be arranged in a finite number of equivalent classes. The proof to this theorem can be found in chapter 18 of [12]. Suppose that F1(x, y) = ax4 + 4bx3y + 6cx2y2 + 4dxy3 + ey4, (6.4) where a, b, c, d, e are integers. Let us define, for the quartic form F1 in (6.4), the invariants g2 = I 12 , g3 = J 432 , where I and J are the invariants of F1. Observe that g2, g3 ∈ Z and for covariants G and Q of F1 we have G 144 , Q 8 ∈ Z[x, y]. By (6.3), ( Q 8 )2 = −4 ( G 144 )3 + g2F 21G+ g3F 3 1 . (6.5) We can now find the integer solutions of the equation Z2 = X3 −G2XY 2 −G3Y 3, (6.6) where G2, G3 are given integers and the right-hand side has no square linear factor. Let F1(x, y) = ax4 + 4bx3y + 6cx2y2 + 4dxy3 + ey4 be a binary quartic form with invariants g2 = 4G2, g3 = 4G3. Then from (6.5), a solution to (6.6) is given by X = G 144 (p, q), Y = F (p, q), 2Z = Q 8 , (6.7) and, further, the X, Y and Z are all integers, provided p, q are integers. We may assume that gcd(p, q) = 1, because the 4th power of gcd(p, q) will divide X,Y and its 6th power will divide Z. In chapter 25 of [12], an elementary proof is given to show that all the integer solutions (X,Y, Z) to (6.6) with gcd(X,Y ) = 1 are included in the formulae (6.7). 117 Chapter 6. Geometry Of Quartic Thue Equations 6.3 Heights For any algebraic number α, we define the (naive) height of α, denoted by H(α), by H(α) = max (|an|, |an−1|, . . . , |a0|) where f(x) = anxn+ . . .+a1x+a0 is the minimal polynomial of α. Suppose that over C, f(x) = an(x− α1) . . . (x− αn). We put M(α) = |an| n∏ i=1 max(1, |αi|). M(α) is known as the M ahler measure of α. We have the following result of Landau: Lemma 6.3.1. Let α be an algebraic number of degree n. then M(α) ≤ (n+ 1)1/2H(α). For any polynomial G in C[z1, . . . , zn] that is not identically zero the Mahler measure M(G) is defined by M(G) = exp ∫ 1 0 dt1 . . . ∫ 1 0 dtn log ∣∣G(e2piit1 , . . . , e2piitn)∣∣ . Thus if n = 1 and G(z) = an(z − α1) . . . (z − αn) with an 6= 0, by Jensen’s theorem, M(G) = |an| n∏ i=1 max(1, |ai|). In [9], Mahler showed that M(F ) ≥ ( DF 44 ) 1 6 . (6.8) Following Matveev [10, 11], we will define the absolute logarithmic height of an algebraic number. Let Q(α1)σ be the embeddings of the real number field Q(α1) in R, 1 ≤ σ ≤ n, where {α1, α2, . . . , αn} are roots of F (x, 1) = 0. We respectively have n Archimedean valuations of Q(α1): |ρ|σ = ∣∣∣ρ(σ)∣∣∣ , 1 ≤ σ ≤ n. 118 Chapter 6. Geometry Of Quartic Thue Equations We enumerate simple ideals of Q(α) by indices σ > n and define non- Archimedean valuation of Q(α) by the formulas |ρ|σ = (Norm p)−k, Where k = ordp(α), p = pσ, σ > n, for any ρ ∈ Q∗(α). Then we have the product formula : ∞∏ 1 |ρ|σ = 1, ρ ∈ Q(α). Note that |ρ|σ 6= 1 for only finitely many ρ. We should also remark that if σ2 = σ̄1, i.e., σ2(x) = ¯σ1(x) for x ∈ Q(α), then the valuations | . |σ1 and | . |σ2 are equal. We define the absolute loga- rithmic height of ρ as h(ρ) = 1 2n ∞∑ σ=1 |log |ρ|σ| . Lemma 6.3.2. Suppose α is an algebraic number of degree n over Q. Then h(α) = 1 n logM(α). Proof. It is well-known that∏ σ max(1, |α|σ) =M(α). Since h(ρ) = 1 2n ∞∑ σ=1 |log |ρ|σ| , by the product formula, h(α) = 2 2n log ∏ σ max(1, |α|σ). Therefore, h(α) = 1 n logM(α). 119 Chapter 6. Geometry Of Quartic Thue Equations Let α and β be two algebraic numbers. Then the following inequalities hold (see [3]): h(α+ β) ≤ log 2 + h(α) + h(β) (6.9) and h(αβ) ≤ h(α) + h(β). (6.10) Two algebraic integers α and α′ are called equivalent if their minimal polynomials are equivalent. Proposition 6.3.3. (Győry [8]) Suppose that α is an algebraic integer with discriminant D(α) and |D| ≥ 3. Then there is an algebraic number α′ equivalent to α for which we have H(α′) ≤ exp exp [2(logD)13] . This allows us to assume H(α) ≤ exp exp [2(logD)13], where need be. Lemma 6.3.4. (Mahler [9]) If a and b are distinct zeros of polynomial P (x) with degree n, then we have |a− b| ≥ √ 3(n+ 1)−nM(P )−n+1, where M(P ) is the Mahler measure of P . Since M(P ) ≤ (n+ 1)1/2H(P ), we have |a− b| ≥ √ 3(n+ 1)−(2n+1)/2H(P )−n+1. 6.4 The Thue-Siegel Principle Let α be an algebraic number of degree n and f be its minimal polynomial over the integers. Let t and τ be positive numbers such that t < √ 2/n and√ 2− nt2 < τ < t, and put λ = 2t−τ and A1 = t2 2− nt2 ( logM(α) + n 2 ) . Suppose that λ < n. A rational number xy is said to be a very good approx- imation to α if |α− x/y| < (4 eA1 max(|x|, |y|))−λ . The following result of Bombieri and Schmidt is based on a classical work of Thue and Siegel. 120 Chapter 6. Geometry Of Quartic Thue Equations Proposition 6.4.1. (Thue-Siegel principle) If α is of degree n ≥ 3 and x/y and x′/y′ are two very good approximations to α then log ( 4eA1 ) + log ( max(|x′|, |y′|)) ≤ γ−1 (log (4eA1)+ log (max(|x|, |y|))) , where γ = nt 2+τ2−2 n−1 . We also need the following refinement of an inequality of Lewis and Mahler : Lemma 6.4.2. Let F be a binary form of degree n ≥ 3 with integer coef- ficients and nonzero discriminant D. For every pair of integers (x, y) with y 6= 0 min α ∣∣∣∣α− xy ∣∣∣∣ ≤ 2n−1nn−1/2 (M(F ))n−2 |F (x, y)||D(F )|1/2|y|n , where the minimum is taken over the zeros α of F (z, 1). Proof. This is Lemma 3 of [16]. 6.5 Large Solutions We will now estimate the number of solutions (x, y) of (6.1) with y > M(F )2. Suppose that (x, y) is an integral solution to (6.1). Then we have (x− α1y)(x− α2y)(x− α3y)(x− α4y) = ±1. Therefore, for some 1 ≤ i ≤ 4, |x− αiy| < 1. Definition. We say the pair of solution (x, y) is related to αi if |x− αiy| = min 1≤j≤4 |x− αjy| . Suppose (x1, y1), (x2, y2), . . . are the solutions to (6.1) which are related to αi with yj > M(F )2, for j = 1, 2, . . ., ordered so that y1 ≤ y2 ≤ . . .. By Lemma 6.4.2, ∣∣∣∣αi − xjyj ∣∣∣∣ ≤ 210M(F )2|D(F )|1/2y4j (6.11) for j = 1, 2, . . .. Therefore,∣∣∣∣xj+1yj+1 − xjyj ∣∣∣∣ ≤ 211M(F )2|D(F )|1/2y4j 121 Chapter 6. Geometry Of Quartic Thue Equations Since |xj+1yj − xjyj+1| ≥ 1 and D > 222, we have y3j M(F )2 ≤ yj+1. (6.12) To each solution (xj , yj), we associate a real number δj > 1 by yj =M(F )1+δj . (6.13) From (6.12), we have 3δj ≤ δj+1. Therefore, 3j−1 ≤ δj . (6.14) Moreover, if the pairs of solutions (xk, yk) and (xk+l, yk+l) are both related to αi then 3lδk ≤ δk+l. (6.15) Let us now apply the Thue-Siegel principle (Lemma 6.4.1) with t = √ 2 4.01 and τ = 1.2 √ 2− 4t2 = 0.12 t. Then λ = 2 t− τ = 2 0.88t < 3.22, A1 = 100 (log(M(F )) + 2) and γ−1 < 1368, (6.16) where, γ = 4t 2+τ2−2 3 . Since we have assumed ∣∣∣αi − xjyj ∣∣∣ < 1, |xj | < |yj |(|αi|+ 1) ≤ 2M(F )yj . Whereby H(xj , yj) < 2M(F )yj . By (6.8) and since D > 10500, we have 8eA1 = 8e200M(F )100 < M(F )102. (6.17) 122 Chapter 6. Geometry Of Quartic Thue Equations So by (6.13), (4eA1H(xj , yj))λ < M(F )(103+δj)λ. (6.18) From (6.11), ∣∣∣∣αi − xjyj ∣∣∣∣ < M(F )−4δj . Hence, xjyj is a very good approximation to αi whenever 4δj ≥ (103 + δj)λ. Since λ ≤ 3.22, if δj > 414 then xjyj is a very good approximation to αi. So by (6.14), whenever k > 1 + log 415 log 3 , then xkyk is a very good approximation to αi. This means there are at most 6 large solutions (x1,y1),. . . , (x6, y6) to (6.1) which are related to αi for which x1 y1 ,. . . , x6y6 are not good approximations to αi. Suppose that there are l pairs of solutions (x7, y7), . . . , (x6+l, y6+l) (l > 1) which are both related to αi, and for which xjyj are good approximations to αi. Then by the Thue-Siegel principle (Lemma (6.4.1)) and (6.16), log ( 4eA1 ) + log y7+l ≤ 1368 ( log ( 4eA1 ) + log(2M(F )y8) ) , and so, by (6.17), log y7+l ≤ 1368 (103 logM(F ) + log(y8))− 102 logM(F ) + log(2). Since δ8 > 414, by (6.13) and (6.15), 3l−1δ8 ≤ δ7+l < 1368 δ8 + 139435 < 336 δ8. Thus, l ≤ log 336 log 3 + 1 ≤ 6.30. This means there are at most 12 large solution related to each root of F (x, 1). 6.6 Small Solutions Here we will count the number of solutions to (6.1) with 1 ≤ y ≤M(F )2. We will follow Stewart’s[16] results for Thue inequalities with arbitrary degree and sharpen them for quartic Thue equations. Suppose that Y0 is a fixed 123 Chapter 6. Geometry Of Quartic Thue Equations positive number. For each root αi of F (x, 1), let (x(i), y(i)) be the solution to (6.1) related to αi with the largest value of y among those with 1 ≤ y ≤ Y0 . Let X be the set of solutions of (6.1) with 1 ≤ y ≤ Y0 minus the elements (x(1), y(1)), (x(2), y(2)), (x(3), y(3)), (x(4), y(4)). From inequality (60) of [16], we have (( 2 7 )4 M(F ) )|X| ≤ Y 40 , (6.19) where |X| denotes the cardinality of X. By (6.8), when D > 10500, we have( 2 7 )4 M(F ) ≥M(F )64/65. By (6.19), |X| < 4 65 log Y0 64 logM(F ) . (6.20) So when Y0 = M(F )2, we have |X| ≤ 8. Therefore the number of small solutions does not exceed 12. We have seen that there are at most 48 large solutions and 12 small ones to (6.1), when the discriminant is large. Since we assumed the quartic form F (x, y) is monic, (1, 0) is also a solution to (6.1). Thus, the proof of Theorem 6.1.1 is complete. In the next section, we will consider quartic forms F (x, y) for which all roots of F (x, 1) are real. There we will call a solution (x, y) a large solution if y > M(F )6. Lemma 6.6.1. There are at most 14 solutions to (6.1) with y ≤M(F )6. Proof. Choose θ > 0 such that 65 16 ( 8 3 + θ ) < 11. From (6.19), we conclude that (6.1) has at most 10 solutions with 1 ≤ y < M(f) 8 3 +θ. Further, by (6.12), equation (6.1) has at most 4 solutions with M(f) 8 3 +θ ≤ y < M(f)6. So altogether (6.1) has at most 14 solutions with 1 ≤ y < M(f)6. 124 Chapter 6. Geometry Of Quartic Thue Equations 6.7 Forms With Real Roots In this section, we will assume αi, the roots of F (x, 1), are real. Define φm(x, y) = log ∣∣∣∣∣D 1 12 (x− yαm) |f ′(αm)| 1 3 ∣∣∣∣∣ (6.21) and φ(x, y) = (φ1(x, y), φ2(x, y), φ3(x, y), φ4(x, y)) . Let ‖φ(x, y)‖ be the L2 norm of the vector φ(x, y). Lemma 6.7.1. Suppose that (x, y) is a solution to the equation F (x, y) = 1 for the binary form F in Theorem 6.1.2. If |x− αiy| = min 1≤j≤4 |x− αjy| Then ‖φ(x, y)‖ ≤ 6 log 1|x− αiy| + 4 log ( D 1 12 (5)4M(F )3√ 3 ) . Proof. Let us assume that∣∣x− αsjy∣∣ < 1, for 1 ≤ j ≤ p and |x− αbky| ≥ 1, for 1 ≤ k ≤ 4− p, where 1 ≤ p, sj , bk ≤ 4. We have∏ k |x− αbky| = 1∏ j ∣∣x− αsjy∣∣ . Therefore, for any 1 ≤ k ≤ 4− p, we have log |x− αbky| ≤ p log 1 |x− αiy| . Since |x− αiy| = min 1≤j≤4 |x− αjy| , 125 Chapter 6. Geometry Of Quartic Thue Equations we also have ∣∣log ∣∣x− αsjy∣∣∣∣ ≤ |log |x− αiy|| . From here, we conclude that ‖φ(x, y)‖ ≤ 4∑ m=1 log ∣∣∣∣∣ D 1 12 |f ′(αm)| 1 3 ∣∣∣∣∣+ (4− p)p |φi(x, y)|+ p |φi(x, y)| = 4∑ m=1 log ∣∣∣∣∣ D 1 12 |f ′(αm)| 1 3 ∣∣∣∣∣+ (5p− p2) |φi(x, y)| . The function f(p) = 5p− p2 gets its maximum value 6 over p ∈ {1, 2, 3, 4}. Our proof is complete by recalling the fact that if a and b are distinct zeros of f(x) = F (x, 1) , then by Lemma 6.3.4, we have |a− b| ≥ √ 3 54 M(f)−3. (6.22) 6.7.1 Exponential Gap Principle Here, our goal is to show Theorem 6.7.2. Suppose that (x1, y1), (x2, y2) and (x3, y3) are three pairs of non-trivial solutions to (6.1) with |xj − α4yj | < 1, for j ∈ {1, 2, 3}. If r1 ≤ r2 ≤ r3 then r3 > exp (r1 6 ) 2 √ 3 log4 1 + √ 5 2 , where rj = ‖φ(xj , yj)‖. We note that for three pairs of solutions in Theorem 6.7.2, the three points φ1 = φ(x1, y1), φ2 = φ(x2, y2) and φ3 = φ(x3, y3) form a triangle ∆. To establish Theorem 6.7.2, we will find a lower bound and an upper bound for the area of ∆. Then comparing these bounds, Theorem 6.7.2 will be proven. The length of each side of ∆ is less than 2r3. Lemma 6.7.3 gives an upper bound for the height of ∆. Suppose that (x, y) 6= (1, 0) is a solution to (6.1) and let t = xy . We have φ(x, y) = φ(t) = 4∑ i=1 log |t− αi| |f ′(αi)| 1 3 bi, 126 Chapter 6. Geometry Of Quartic Thue Equations where, b1 = 1 4 (3,−1,−1,−1), b2 = 14(−1, 3,−1,−1), b3 = 1 4 (−1,−1, 3,−1), b4 = 14(−1,−1,−1, 3), Without loss of generality, we will suppose that for the pair of solution (x, y) we have |x− α4y| < 1. We may write φ(x, y) = φ(t) = 3∑ i=1 log |t− αi| |f ′(αi)| 1 3 ci + E4b4, (6.23) where, for 1 ≤ i ≤ 3, ci = bi + 1 3 b4, E4 = log |t− α4| |f ′(α4)| 1 3 − 1 3 3∑ i=1 log |t− αi| |f ′(αi)| 1 3 One can easily observe that ci ⊥ b4, for 1 ≤ i ≤ 4. Lemma 6.7.3. Let L4 = 3∑ i=1 log |α4 − αi| |f ′(αi)| 1 3 ci + zb4, z ∈ R. Suppose that (x, y) 6= (1, 0) is a pair of solution to (6.1) with |x− α4y| = min 1≤j≤4 |x− αjy| and y ≥ M(F )6. Then the distance between φ(x, y) and the line L4 is less than exp (−r 6 ) , where r = ‖φ(x, y)‖. Proof. The distance between φ(x, y) and L4 is equal to∥∥∥∥∥ 3∑ i=1 log |t− αi| |α4 − αi|ci ∥∥∥∥∥ , 127 Chapter 6. Geometry Of Quartic Thue Equations where t = xy . If |t− αi| > |α4 − αi|, then∣∣∣∣log |t− αi||α4 − αi| ∣∣∣∣ = log |t− αi||α4 − αi| ≤ log ( |t− α4| |α4 − αi| + 1 ) < |t− α4| |αi − α4| . If |t− αi| < |α4 − αi|, then∣∣∣∣log |t− αi||α4 − αi| ∣∣∣∣ = log |α4 − αi||t− αi| ≤ log ( |t− α4| |t− αi| + 1 ) < |t− α4| |αi − t| . Note that when i 6= 3, either |t− αi| > |α4 − αi| or |t− αi| > |α3 − αi|. Therefore, for i 6= 3, ∣∣∣∣log |t− αi||α4 − αi| ∣∣∣∣ < |t− α4|m , where m = mini6=j{|αj − αi|}. Moreover, since we assumed t is closer to α4, |t− α3| ≥ |α4 − α3|2 . Consequently, ∣∣∣∣log |t− α3||α4 − α3| ∣∣∣∣ < 2|t− α4|m . Therefore ∥∥∥∥∥ 3∑ i=1 log |t− αi| |α4 − αi|ci ∥∥∥∥∥ < 4 √ 2 3 |u| m , (6.24) where u = t− α4. On the other hand, by Lemma 6.7.1 r − 4 log ( D 1 12 54M(F )3√ 3 ) ≤ 6 log 1|x− α4y| , which implies log |yu| < −r 6 + 16 25 log ( D 1 12 54M(F )3√ 3 ) . 128 Chapter 6. Geometry Of Quartic Thue Equations Therefore, |u| < exp (−r 6 ) exp(1625 log(D 112 54M(F )3√3 )) |y| Comparing this with(6.24), since |y| > M(F )6 and (by (6.8)) we have D1/12 < 41/3M(F )1/12, our proof is complete (note that by (6.3.4), m ≥ √ 3 54M(f)3 ). Lemma 6.7.3 shows that the height of ∆ is at most 2 exp (−r1 6 ) . Therefore, the area of ∆ is less than 2r3 exp (−r1 6 ) . (6.25) To estimate the area of ∆ from below, we appeal to Pohst’s lower bound for units. Since F (x, y) = (x− α1y)(x− α2y)(x− α3y)(x− α4y) = ±1, we conclude that x−αiy is a unit in Q(αi) when (x, y) is a pair of solution to (6.1). Suppose that (x1, y1) and (x2, y2) are two pairs of non-trivial solutions to (6.1). Then φ(x1, y1)− φ(x2, y2) = ( log x1 − α1y x2 − α1y2 , . . . , log x1 − α4y1 x2 − α4y2 ) = ~e. Since x1−αiyx2−αiy2 is a unit in Q(αi), we have ‖~e‖ ≥ 4 log2 1 + √ 5 2 (see exercise 2 on page 367 of [14]). Now we can estimate each side of ∆ from below to conclude that the area of the triangle ∆ is greater than 16 √ 3 4 log4 1 + √ 5 2 . Comparing this with (6.25) we conclude that 2 r3 exp (−r1 6 ) > 16 √ 3 4 log4 1 + √ 5 2 . Theorem 6.7.2 is immediate from here. 129 Chapter 6. Geometry Of Quartic Thue Equations 6.7.2 Geometry Of The Curve φ(t) In order to study the curve φ(t), we will consider some well-known geometric properties of the unit group U of Q(α), where α is a root of F (x, 1) = 0. Theorem 6.7.4 (Dirichlet’s Unit Theorem). Let K be an algebraic number field of degree n. Let r be the number of real conjugate fields of K and 2s the number of complex conjugate fields of K. Then the ring of integers OK contains r + s − 1 fundamental units 1, . . . , r+s−1 such that each unit of OK can be expressed uniquely in the form un11 . . .  nr+s−1 r+s−1 , where u is a root of unity in OK and n1, . . . , nr+s−1 are integers. For a real algebraic number field Q(α) of degree 4, in Dirichlet’s Unit Theorem we have r = 4 and s = 0. By Dirichlet’s unit theorem, we have a sequence of mappings τ : U 7−→ V ⊂ R4 (6.26) and log : V 7−→ Λ, (6.27) where V is the image of the map τ , Λ is a 3-dimensional lattice, τ is the obvious restriction of the embedding of Q(α) in R4, and the mapping log is defined as follows: For (x1, x2, x3, x4) ∈ V , log(x1, x2, x3, x4) = (log |x1|, log |x2|, log |x3|, log |x4|). If (x, y) is a pair of solutions to (6.1) then (x− αjy) is a unit in Q(αi). Suppose that λ2, λ3, λ4 are fundamental units of Q(αi) and are chosen so that log (τ(λ2)) , log (τ(λ3)) , log (τ(λ4)) form a reduced basis for the lattice Λ. Let us assume that ‖log (τ(λ2))‖ ≤ ‖log (τ(λ3))‖ ≤ ‖log (τ(λ4))‖ . φ(x, y) = φ(1, 0) + 4∑ k=2 mk log (τ(λk)) mk ∈ Z (6.28) 130 Chapter 6. Geometry Of Quartic Thue Equations Lemma 6.7.5. For every fixed integer m, there are at most 6 solutions (x, y) to (6.1) for which in (6.28), m4 = m. Proof. Let S be the 3-dimensional space of all points φ(1, 0)+ ∑4 i=2 µi log (τ(λi)) (µ4 ∈ R). Let µ4 = m. Then the points φ(1, 0) + 3∑ i=2 µi log (τ(λi)) +m log (τ(λ4)) form a linear subvariety S1 of S. Let ~N = (N1, . . . , N4) ∈ S be the normal vector of S1. Then the number of times that the curve φ(t) intersects S1 equals the number of solutions in t to ~N.φ(t) = 0, (6.29) where ~N.φ(t) is the inner product of two vectors ~N and φ(t). We have d dt ( ~N.φ(t) ) = P (t) F (t) , where F (t) = (t− α1) . . . (t− α4) and P (t) is a polynomial of degree 3. Therefore, since lim t→α+i log |t− αi| = −∞ and lim t→α−i log |t− αi| −∞, the derivative has at most 3 zeros and consequently, the equation (6.29) can not have more than 6 solutions. Definition of the set A. Assume that equation (6.1) has more than 6 solutions. Then we can list 6 solutions (xi, yi) (1 ≤ i ≤ 6), so that ri = ‖φ(xi, yi)‖ are the smallest among all ‖φ(x, y)‖, where (x, y) varies over all non-trivial pairs of solutions. We call the set of all these 6 solutions A. 131 Chapter 6. Geometry Of Quartic Thue Equations Corollary 6.7.6. Let (x, y) 6∈ A be a solution to (6.1). Then ‖log (τ(λ2))‖ ≤ ‖log (τ(λ3))‖ ≤ ‖log (τ(λ4))‖ ≤ 2 ‖φ(x, y)‖ . Proof. Since we have assumed that ‖log (τ(λ2))‖ ≤ ‖log (τ(λ3))‖ ≤ ‖log (τ(λ4))‖, it is enough to show that ‖log (τ(λ4))‖ ≤ ‖φ(x, y)‖. By Lemma 6.7.5, there is at least one solution (x0, y0) ∈ A so that φ(x, y)− φ(x0, y0) = 4∑ i=2 ki log (τ(λi)) , with k4 6= 0. Since {log (τ(λi))} is a reduced basis for the lattice Λ in (6.27), we conclude that ‖log (τ(λ4))‖ < ‖φ(x, y)− φ(x0, y0)‖ ≤ 2 ‖φ(x, y)‖ . Lemma 6.7.7. Suppose (x, y) 6∈ A. Then for r(x, y) = ‖φ(x, y)‖, we have r(x, y) ≥ 1 2 log ( |D|1/12 2 ) . Proof. Let (x′, y′) ∈ A be a pair of solutions to equation (6.1) and αi and αj be two distict roots of quartic polynomial F (x, 1). We have∣∣∣eφi(x′,y′)−φi(x,y) − eφj(x′,y′)−φj(x,y)∣∣∣ = ∣∣∣∣x′ − y′αix− yαi − x ′ − y′αj x− yαj ∣∣∣∣ = |αi − αj | |xy′ − yx′| |x− yαi||x− yαj | ≥ |αi − αj ||x− yαi||x− yαj | . The last inequality follows from the fact that |xy′ − yx′| is a non-zero integer. Since |φi| < ‖φ‖ = r and r(x′, y′) < r(x, y), we may conclude( 2e2r(x,y) )6 ≥ ∏ 1≤i<j≤4 ∣∣∣∣x′ − y′αix− yαi − x ′ − y′αj x− yαj ∣∣∣∣ ≥ √D. 132 Chapter 6. Geometry Of Quartic Thue Equations Let us define Ti,j(t) := log ∣∣∣ (t−αi)(α4−αj)(t−αj)(α4−αi) ∣∣∣ , so that for a pair of solutions (x, y) 6= (1, 0), Ti,j(x, y) = Ti,j(t) = log ∣∣∣∣α4 − αiα4 − αj ∣∣∣∣+ log ∣∣∣∣ t− αjt− αi ∣∣∣∣ = log ∣∣∣∣α4 − αiα4 − αj ∣∣∣∣+ log ∣∣∣∣x− αjyx− αiy ∣∣∣∣ = log |λi,j |+ 4∑ k=2 mi log |λk| |λ′k| , (6.30) where t = xy , λi,j = log ∣∣∣∣α4 − αiα4 − αj ∣∣∣∣ and λk and λ′k are fundamental units in Q(αj) and Q(αi), respectively. Note that the mk ∈ Z in (6.28) and (6.30) are the same integers. We will end this section by giving an upper bound for |T | and will estimate |T | from below in the next section. Lemma 6.7.8. Let (x, y) be a pair of solution to (6.1) with |y| > M(F )6. Then there exists a pair (i, j) for which |Ti,j(x, y)| < exp (−r 6 ) , where r = ‖φ(t)‖. Proof. Let us define βi = { αi if i ≤ 3 βi−3 if i ≥ 4. 133 Chapter 6. Geometry Of Quartic Thue Equations Note that 2∑ k=1 3∑ i=1 log2 ∣∣∣∣(t− βi)(α4 − βi+k)(α4 − βi)(t− βi+k) ∣∣∣∣ = 4 3∑ i=1 log2 ∣∣∣∣ (t− αi)(α4 − αi) ∣∣∣∣− 4∑ i6=j log ∣∣∣∣ (t− αi)(α4 − αi) ∣∣∣∣ log ∣∣∣∣ (t− αj)(α4 − αj) ∣∣∣∣ = 4 3∑ i=1 log2 ∣∣∣∣ (t− αi)(α4 − αi) ∣∣∣∣− 2 3∑ i=1 log ∣∣∣∣ (t− αi)(α4 − αi) ∣∣∣∣∑ j 6=i log ∣∣∣∣ (t− αj)(α4 − αj) ∣∣∣∣ = 4 3∑ i=1 log2 ∣∣∣∣ (t− αi)(α4 − αi) ∣∣∣∣− 2 3∑ i=1 log ∣∣∣∣ (t− αi)(α4 − αi) ∣∣∣∣ log ∣∣∣∣ (α4 − αi)y4f ′(α4)(t− α4)(t− αi) ∣∣∣∣ = 6 3∑ i=1 log2 ∣∣∣∣ (t− αi)(α4 − αi) ∣∣∣∣− 2 log ∣∣∣∣ 1ynf ′(α4)(t− αn) ∣∣∣∣ 3∑ i=1 log ∣∣∣∣ (t− αi)(α4 − αi) ∣∣∣∣ = 6 3∑ i=1 log2 ∣∣∣∣ (t− αi)(α4 − αi) ∣∣∣∣− 2 log2 ∣∣∣∣ 1y4f ′(α4)(t− α4) ∣∣∣∣ On the other hand, from the proof of Lemma 6.7.3 the distance between φ(x, y) and the line L4 = ∑3 i=1 log |α4−αi| |f ′(αi)| 1 3 ci + zb4, z ∈ R. is equal to∥∥∥∑3i=1 log |t−αi||α4−αi|ci∥∥∥ and by the definition of ci in section 6.7.1, we have∥∥∥∥∥ 3∑ i=1 log |t− αi| |α4 − αi|ci ∥∥∥∥∥ 2 = ∥∥∥∥∥ 3∑ i=1 log ( |t− αi| |α4 − αi| − 1 3 ∣∣∣∣log 1y4f ′(α4)(t− α4) ∣∣∣∣) ei ∥∥∥∥∥ 2 = 3∑ i=1 log2 ( |t− αi| |α4 − αi| − 1 3 ∣∣∣∣log 1y4f ′(α4)(t− α4) ∣∣∣∣) = 3∑ i=1 log2 ∣∣∣∣ (t− αi)(α4 − αi) ∣∣∣∣− 13 log ∣∣∣∣ 1y4f ′(α4)(t− α4) ∣∣∣∣ 3∑ i=1 log ∣∣∣∣ (t− αi)(α4 − αi) ∣∣∣∣ where {ei} is the standard basis for R3. So, there must be a pair of (i, j), 134 Chapter 6. Geometry Of Quartic Thue Equations for which log2 ∣∣∣∣(t− αi)(α4 − αj)(t− αj)(α4 − αi) ∣∣∣∣ < 1 6 2∑ k=1 3∑ i=1 log2 ∣∣∣∣(t− βi)(α4 − βi+k)(α4 − βi)(t− βi+k) ∣∣∣∣ = = ∥∥∥∥∥ 3∑ i=1 log |t− αi| |α4 − αi|ci ∥∥∥∥∥ 2 . Therefore, by Lemma 6.7.3 |Ti,j(x, y)| = ∣∣∣∣log ∣∣∣∣(t− αi)(α4 − αj)(t− αj)(α4 − αi) ∣∣∣∣∣∣∣∣ < exp(−r6 ) . 6.7.3 Linear Forms In Logarithms Theorem 6.7.9 (Matveev). Suppose that K is a real algebraic number field of degree d. We are given numbers α1, . . . αn ∈ K∗ with absolute logarithm heights h(αj). Let logα1 , . . . , logαn be arbitrary fixed non-zero values of the logarithms. Suppose that Aj ≥ max{dh(αj), | logαj |}, 1 ≤ j ≤ n. Now consider the linear form L = b1 logα1 + . . .+ bn logαn, with b1, . . . , bn ∈ Z and with the parameter B = max{1,max{bjAj/An : 1 ≤ j ≤ n}} . Put Ω = A1 . . . An, C(n) = 16 n! en(2n+ 2)(n+ 2)(4n+ 4)n+1( 1 2 en), C0 = log(e4.4n+7n5.5d2 log(en)), W0 = log(1.5eBd log(ed)). If bn 6= 0, then log |L| > −C(n)C0W0d2Ω. Proof. See [11] for proof. 135 Chapter 6. Geometry Of Quartic Thue Equations Let index σ be the isomorphism from Q(αi) to Q(αj) such that σ(αi) = αj . We may assume that σ(λi) = λ′i for i = 2, 3, 4. Let (x1, y1) , (x2, y2) , (x3, y3),(x4, y4),(x5, y5) be five distinct ” large” solutions to (6.1) with (xk, yk) 6∈ A, yk > M(F )6 and |xk − α4yk| = min 1≤i≤4 |xk − αiyk| k ∈ {1, 2, 3, 4, 5} and r1 ≤ r2 ≤ r3 ≤ r4 ≤ r5 where rk = ‖φ(xk, yk)‖. We will apply Matveev’s lower bound to Ti,j(x5, y5) = log |λi,j |+ 4∑ k=2 mk log |λk| |λ′k| , where (i, j) is chosen so that Lemma 6.7.8 is satisfied and mk ∈ Z. In the above representation, λk are multiplicatively dependent if and only if λi,j is a unit. If λi,j is a unit then we can write Ti,j(x, y) as a linear form in 3 logarithms. Since theorem 6.7.9 gives a better lower bound for linear forms in 3 logarithms, we will assume that λi,j , λ2, λ3 and λ4 are multiplicatively independent and Ti,j(x, y) is a linear form in 4 logarithms. Suppose that λ is a unit in the number field and λ′ is its algebraic conjugate. We have h(λ′) = h(λ) = 1 8 |log (τ(λ))|1 , where h is the logarithmic height and | |1 is the L1 norm on R4 and the mappings τ and log are defined in (6.26) and (6.27) . So we have h(λ) = 1 8 |log (τ(λ))|1 ≤ √ 4 8 ‖log (τ(λ))‖ , where ‖‖ is the L2 norm on R4. Since α4, αi and αj have degree 4 over Q, the number field Q(α4, αi, αj) has degree d ≤ 24 over Q. So when λ is a unit max{dh( λ λ′ ), ∣∣∣∣log(∣∣∣∣ λλ′ ∣∣∣∣)∣∣∣∣} ≤ max{24h( λλ′ ), | log( ∣∣∣∣ λλ′ ∣∣∣∣)|} ≤ 12 ‖log (τ(λ))‖ . (6.31) Therefore, to apply Theorem 6.7.9 to Ti,j(x, y), by Lemma 6.7.6, we may take Ai = 24r1, for 2 ≤ i ≤ 4. 136 Chapter 6. Geometry Of Quartic Thue Equations By Lemma 6.3.2, Proposition 6.3.3, (6.9) and (6.10), we may take A1 24 = 2 log 2 + 4 exp [ 2(logD)13 ] , (note that α1, αi , αj are algebraic conjugates). To estimate B, we note that since λi (2 ≤ i ≤ 4) form a reduced basis for the lattice Λ, we have mi ‖log τ(λi)‖ ≤ ‖φ(x5, y5)‖+ ‖φ(1, 0)‖ ≤ r5 + 2 logD1/12 + 2 log 5 4M(F )3√ 3 ≤ r5 + 2 logD1/12 + 2 log 5 11/2H(F )3√ 3 , where the inequalities are from Lemmas 6.3.1 and (6.22). Therefore, by Proposition 6.3.3, B = max{1,max{bjAj/A1 : 1 ≤ j ≤ n}} < r5. Theorem 6.7.9 implies that for a constant number K, log Ti,j(x5, y5) > −K exp [ 2(logD)13 ] r31 log r5. Comparing this with Lemma 6.7.8, we have(−r5 6 ) > −K exp [2(logD)13] r31 log r5. Thus we may compute the constant number K1, so that r5 < K1 exp [ 2(logD)13 ] r31 log r1, This contradicts Lemma 6.7.2 when D > 10500, for by Lemma 6.7.7, r1 ≥ 12 log ( |D| 112 2 ) . Thus, there are at most 16 solutions (x, y) 6∈ A with y > M(F )6. By Lemma 6.6.1 and since |A| = 6, counting the solution (1, 0), Theorem 6.1.2 is proven. 137 Bibliography [1] S. Akhtari, The method of Thue-Siegel for binary quartic forms, sub- mitted (2007). [2] E. Bombieri, W. M. Schmidt, On Thue’s equation, Invent. Math. 88 (1987), 69-81. [3] Y. Bugeaud, Bornes effectives pour les solutions des équations en S- units et des équations de Thue-Mahler, J. Number Theory 71(1998), 227-244. [4] J.E. Cremona. Reduction of binary cubic and quartic forms. LMS JMC 2 (1999), 62-92. [5] B.N. Delone and D.K. Fadeev. The Theory of Irrationalities of the Third Degree. Translation of Math. Monographs, AMS 10 (1964). [6] J.H. Evertse and K. Gyory, Effective finiteness results for binary forms with given discriminant, Compositio Math. 79(1991), 169-204. [7] K. Győry, Polynomials and binary forms of given discriminant, Publ. Math. Debrecen 69 (2006), 473-499. [8] K. Győry, Sur les polynomial a coefficients entiers et the discriminant donne II, Pub. Math. Debrecen 21 (1974), 125-144. [9] K. Mahler, An inequality for the discriminant of a polynomial, Michigan Math. J. 11 (1964), 257-262. [10] E. M. Matveev, An explicit lower bound for a homogeneous rational linear form in logarithms of algebraic numbers, Izv. Ross. Akad. Nauk Ser. Mat. 62 (1998), 81-136, translation in Izv. Math. 62 (1998), 723- 772. [11] E. M. Matveev, An explicit lower bound for a homogeneous rational linear form in logarithms of algebraic numbers, Izv. Ross. Akad. Nauk 138 Bibliography Ser. Mat. 64 (2000), 125-180, translation in Izv. Math. 64 (2000), 1217- 1269. [12] L.J. Mordell, Diophantine Equations, Academic Press, London and New York (1969). [13] R. Okazaki. Geometry of a cubic Thue equation, Publ. Math. Debrecen. 61 (2002),267-314. [14] M. Pohst and H. Zassenhause. Algorithmic Algebraic Number Theory, Cambridge University Press (1989). [15] W.M. Schmidt, Diophantine approximations and Diophantine equa- tions, Lecture Notes in Mathematics, Springer Verlag (2000). [16] C. L. Stewart, On the number of solutions of polynomial congruences and Thue equations, Journal of American Math. Soc., 4 (1991), 793- 838. 139 Chapter 7 Geometry of Binary Cubic Forms 6 7.1 Introduction Let F (x, y) be an irreducible binary cubic form with integral coefficients and negative discriminant. More than 80 years ago, Delone and Nagell established independently that the equation F (x, y) = 1 (7.1) has at most five solutions in integers x, y. This result is proved by consid- ering units in the algebraic number field Q(ρ), where ρ is the real root of F (x, 1) = 0. In their proofs the fact that the group of units in the ring of integers of Q(ρ) is generated by one fundamental unit is essential. The situation where the discriminant of F (x, y) is positive is complicated by the fact that the number field Q(ρ) (where ρ is any real root of F (x, 1) = 0) has a ring of integers generated by a pair of fundamental units. However, it is possible to reduce (7.1) to a set of exponential equations to which a local method of Skolem can be applied. In this way, Ljunggren [14] and Baulin [4], solved (7.1) respectively for F (x, y) = x3− 3xy2+ y3 of discriminant 81 and F (x, y) = x3+x2y−2xy2−y3 of discriminant 49. In the first case there are 6 solutions and in the second case there are 9 solutions to (7.1). In 1929, Siegel [16] used the theory of Padé approximation to binomial functions (via the hypergeometric functions), to show for F cubic of positive discriminant, that equation (7.1) has at most 18 solutions in integers x and y. Refining these techniques, Evertse [5] reduced this upper bound to 12. Later, Bennett [2] showed that if F (x, 1) has at least two distinct complex roots, then the equation F (x, y) = 1 possesses at most 10 solutions in integers x and y. In 2003, by studying the geometry of numbers in the “logarithmic space”, Okazaki [15] proved that when discriminant of F is greater than 5.65×1065, 6A version of this chapter will be submitted for publication. Akhtari. S. Geometry of binary cubic forms 140 Chapter 7. Geometry of Binary Cubic Forms equation (7.1) has at most 7 solutions. Okazaki’s method is essentially different from Evertse’s. In this paper, we will relate some geometric ideas of Okazaki [15] to the method of Thue-Siegel as refined by Evertse [5], in conjunction with lower bounds for linear forms in logarithms of algebraic numbers. The following are the main results of this paper: Theorem 7.1.1. If F (x, y) is a binary cubic form with discriminant D > 1.4× 1057, then the equation F (x, y) = 1 possesses at most 7 solutions in integers x and y. We will define the reduced forms in Section 7.2. It is well-known that every cubic form of positive discriminant is equivalent to a reduced form F (x, y) . Theorem 7.1.2. Let F (x, y) be a cubic form with discriminant D > 9 × 1058. If F (x, y) is equivalent to a reduced form which is not monic, then the equation F (x, y) = 1 posses at most 6 solutions in integers x and y. In 1990, using the fact that the underlying number fields are the so-called “simplest cubics”, Thomas [17] showed that the equations G1,n(x, y) = x3 + nx2y − (n+ 3)xy2 + y3 = 1 have only the solutions (1, 0) , (0, 1) and (−1,−1) in integers, provided n ≥ 1.365× 107. This restriction was later removed by Mignotte [11] except for the equation with n ∈ {−1, 0, 2}. It is known that G1,n(x, y) = 1 has 9 solutions for n = −1 ([4]), 6 solutions for n = 0 ([14]) and 6 solutions for n = 2 ([6]) . Define Fm(x, y) by Fm(x, y) = x3 − (m+ 1)x2y +mxy2 + y3 for m ∈ Z. Provided m 6= −2 , −1 or 1 , the equation Fm(x, y) = 1 has the five distinct integral solutions (x, y) = (1, 0) , (1, 1) , (1,−m− 1) , (0, 1) and (m, 1). That this list is complete was proven, independently, by Lee [7] and Mignotte and Tzanakis [13], for m suitably large and later, by Mignotte [12], for m > 2. The cases m = 0 and m = 1 correspond to discriminant −23 and −31, respectively. 141 Chapter 7. Geometry of Binary Cubic Forms All known irreducible cubic forms F (x, y), for which the equation (7.1) has more than 5 solutions, have discriminant less than 362. The following conjecture is essentially due to Nagell and refined by Pethö and Lippok. Conjecture If F is a binary cubic form with positive discriminant DF > 361, then the number of solutions of equation (7.1) is less than 6. 7.2 The Covariants of Binary Cubic Forms Let us define, for a cubic form F with discriminantD, an associated quadratic form, the Hessian H = HF , and a cubic form G = GF , by H(x, y) = −1 4 (δ2F δx2 δ2F δy2 − ( δ2F δxδy )2) = Ax2 +Bxy + Cy2 and G(x, y) = δF δx δH δy − δF δy δH δx . These forms satisfy a covariance property; i.e. HF◦γ = HF ◦ γ and GF◦γ = GF ◦ γ for all γ ∈ GL2(Z). We call forms F1 and F2 equivalent if they are equivalent under GL2(Z)- action; i.e. if there exist integers a1 , a2 , a3 and a4 such that F1(a1x+ a2y, a3x+ a4y) = F2(x, y) for all x, y , where a1a4 − a2a3 = ±1. We denote by NF the number of solutions in integers x and y of the diophantine equation (7.1). If F1 and F2 are equivalent, then NF1 = NF2 and DF1 = DF2 . Therefore, we can assume that F is monic (the coefficient of x3 in F (x, y) is 1). The discriminant D of such a form is given by D = 18abcd+ b2c2 − 27a2d2 − 4ac3 − 4b3d = a4 ∏ i,j (αi − αj)2 where α1, α2 and α3 are the roots of polynomial F (x, 1). For F (x, y) = ax3 + bx2y + cxy2 + dy3, it follows by routine calculation that A = b2 − 3ac, B = bc− 9ad, C = c2 − 3bd 142 Chapter 7. Geometry of Binary Cubic Forms and B2 − 4AC = −3D. Further, these forms are related to F (x, y) via the identity 4H(x, y)3 = G(x, y)2 + 27DF (x, y)2 (7.2) Binary cubic form F is called reduced if the Hessian of F , H(x, y) = Ax2 +Bxy + Cy2, satisfies C ≥ A ≥ |B| It is a basic fact (see [1]) that every cubic form of positive discriminant is equivalent to a reduced form F (x, y) . The reader is directed to [1] (chapter III and supplement I) for more details on reduction. We will later use the following lemma to bound the discriminant D from above. Lemma 7.2.1. Let F be an irreducible binary cubic form with positive dis- criminant D and Hessian H. For all integer solutions (x1, y1) to equation F (x, y) = 1, except possibly one solution, we have H(x1, y1) ≥ 12 √ 3D. Proof. If F1 is an equivalent reduced form to F and F1(a1x+ a2y, a3x+ a4y) = F (x, y), then H1(a1x+ a2y, a3x+ a4y) = H(x, y), where H and H1 are the Hessians of F and F1 respectively. This means the set of values of the Hessian at solutions is fixed under GL2(Z)-action. So we may assume that F is reduced. Now following the proof of lemma 5.1. of [2], we suppose (x, y) is a solution to F (x, y) = 1 with y 6= 0. If |y| ≤ |x|, then, since A > |B| and B2 − 4AC = −3D, we have that H(x, y) ≥ Cy2 ≥ C ≥ 1 2 √ 3D. If, on the other hand, |y| ≥ |x|+ 1, then H(x, y) ≥ (C − |B|)y2 + |B||y|+Ax2. Since this is an increasing function of |y| and y 6= 0, we have H(x, y) ≥ C +Ax2 ≥ C ≥ 1 2 √ 3D. Therefore, if H(x, y) < 12 √ 3D, then y = 0 and so x = ±1 accordingly . 143 Chapter 7. Geometry of Binary Cubic Forms Remark. The above proof shows the only possibility for the Hessian H(x, y) to assume a value less than 12 √ 3D, at a pair of solutions (x, y), is when the equivalent reduced form is monic. This is because (1, 0) is a solution to (7.1) if and only if F is monic. 7.3 Some Functions In The Number Field Q( √−3D) Let √−3D be a fixed choice of the square-root of −3D. we will work in the number field M = Q( √−3D). It is well-known that if F has positive discriminant then H is positive definite. By (7.2), we can write H3(x, y) = U(x, y)V (x, y) where U(x, y) = G(x, y) + 3 √−3DF (x, y) 2 , V (x, y) = G(x, y)− 3√−3DF (x, y) 2 . Then U and V are cubic forms with coefficients belonging to M such that corresponding coefficients of U and V are complex conjugates. Since F must be also irreducible over M , U and V do not have factors in common. It follows that U(x, y) and V (x, y) are cubes of linear forms over M , say ξ(x, y) and η(x, y). Note that ξ(x, y)η(x, y) must be a quadratic form which is cube root of H(x, y)3 and for which the coefficient of x3 is a positive real number. Hence we have ξ(x, y)3 − η(x, y)3 = 3√−3DF (x, y), ξ(x, y)3 + η(x, y)3 = G(x, y), (7.3) ξ(x, y)η(x, y) = H(x, y). and for x, y ∈ Z, ξ(x, y) ξ(1, 0) and η(x, y) η(1, 0) ∈M. The reason for the last identity is that for any pair of rational integers x0 , y0, ξ(x0, y0) and η(x0, y0) are complex conjugates and the discriminant of H is −3D. 144 Chapter 7. Geometry of Binary Cubic Forms We call a pair of forms ξ and η satisfying the above properties a pair of resolvent forms. Note that there are exactly three pairs of resolvent form, given by (ξ, η), (ωξ, ω2η), (ω2ξ, ωξ), where ω is a primitive cube root of unity. We say that a pair of rational integers (x, y) is related to the pair of resolvent forms (ξ, η) if |1− η(x, y) ξ(x, y) | = min 0≤k≤2 |ωk − η(x, y) ξ(x, y) | (7.4) Following a discussion of Delone and Faddeev in [1], we call the roots ρ1, ρ ′ 1, ρ ′′ 1 of the equation F (x, a) = 0 the left roots of the form F , while the roots ρ2, ρ′2, ρ′′2 of F (d,−y) are called the right roots of the form F . If t1 is a left root, then it is easily seen that t2 = −ad/t1 is a right root of F . Two such roots of F will be called corresponding roots and we will assume that ρ1 and ρ2 , ρ′1 and ρ′2, ρ′′1 and ρ′′2 correspond in pairs. The following lemma is a statement of Lagrange’s method for solution of cubic equations by means of the resolvent adapted to the case of binary cubic forms. Lemma 7.3.1. For the cubic form F (x, y) the following identity holds F (x, y) = 1 3 √−3D (ξ 3 − η3), where ξ = ξ1x+ ξ2y, η = η1x+ η2y, ξ1 = ρ1 + ωρ′1 + ω 2ρ′′1, η1 = ρ1 + ω2ρ′1 + ωρ ′′ 1, ξ2 = ρ2 + ωρ′2 + ω 2ρ′′2, η2 = ρ2 + ω2ρ′2 + ωρ ′′ 2 and ω = e 2pii 3 . Proof. One can find the complete proof of Lemma 7.3.1 in [1]. 145 Chapter 7. Geometry of Binary Cubic Forms We continue with the following definitions of p, q and ui : p = η + ξ√ 2 , q = √−1(η − ξ)√ 2 , u1 = D−1/6( q√ 6 + p√ 2 ) , u2 = D−1/6( q√ 6 − p√ 2 ) , u3 = D−1/6 2√ 6 q. (7.5) Since η and ξ are functions of x and y, so are p, q and ui The reason of our interest in the new functions p(x, y), q(x, y) and ui(x, y), despite their apparent complication, is that they explain the re- lation between the method of Evertse [5] and the method of Okazaki [15] for finding an upper bound for the number of integral solutions of (7.1). In other words, these functions allow us to recast the resolvent forms ξ and η in a geometric setting. By Lemma 7.3.1, we have q√ 6 = √−1(η − ξ) 2 √ 3 = √−1 ω 2 − ω 2 √ 3 [(ρ′1 − ρ′′1)x− (ρ′2 − ρ′′2)y]. We also have ω2−ω = cos(4pi/3)+√−1 sin(4pi/3)− (cos(2pi/3)+√−1 sin(2pi/3)) = √−3. so we get q√ 6 = −(ρ ′ 1 − ρ′′1)x+ (ρ′2 − ρ′′2)y 2 . (7.6) Further p√ 2 = (2ρ1 + ω(ρ′1 + ρ′′1) + ω2(ρ′1 + ρ′′1))x+ (2ρ2 + ω(ρ′2 + ρ′′2) + ω2(ρ′2 + ρ′′2))y 2 . Since ω is a primitive third root of unity, ω + ω2 = −1. Hence p√ 2 = 2(ρ1x+ ρ2y)− (ρ′1 + ρ′′1)x− (ρ′2 + ρ′′2)y 2 (7.7) Substituting −ad/ρ1, −ad/ρ′1 and −ad/ρ′′1 for ρ2, ρ′2 and ρ′′2 respectively, and noting that ρ1ρ′1ρ′′1 = −a2d, we obtain the following identities: u1 = D−1/6(ρ1 − ρ′′1)(x− ρ′1y/a), (7.8) 146 Chapter 7. Geometry of Binary Cubic Forms u2 = D−1/6(ρ′1 − ρ1)(x− ρ′′1y/a), (7.9) u3 = D−1/6(ρ′1 − ρ′′1)(x− ρ1y/a), (7.10) where ρ1, ρ′1 and ρ′′1 are left roots of F . Here we note that if we start with another choice of resolvent forms, only the order of ui changes. In other words, all three resolvent forms can be indexed so that qi = √−1(ηi − ξi)√ 2 Let us assume that F is monic, as we may. Therefore (x− ρ1y)(x− ρ′1y)(x− ρ′′1y) = F (x, y). If the pair (x0, y0) is a solution to (7.1), we conclude that (x0 − ρ1y0), (x0 − ρ′1y0) and (x0 − ρ′′1y0) are units in Q(ρ1). Moreover, u1u2u3 = D−1/2(ρ1 − ρ′′1)(ρ′1 − ρ1)(ρ′1 − ρ′′1)F (x, y) = ±F (x, y). When (x, y) is a solution to F (x, y) = 1, since log |u1| − log |u2| = log |ρ1 − ρ ′′ 1 ρ′1 − ρ1 |+ log |x− ρ ′ 1y x− ρ′′1y | and x−ρ ′ 1y x−ρ′′1y is a unit, we can write log |u1| − log |u2| = log |λ1|+m log |λ2|+ n log |λ3| , (7.11) where λ1 = ρ1−ρ′′1 ρ′1−ρ1 and where λ2 and λ3 are fundamental units in the ring of integers of Q(ρ1) (when DF > 0, the number field Q(ρ1) is real and has a ring of integer generated by a pair of fundamental units). Let us fix a resolvent forms (ξi, ηi) and corresponding pi and qi. We get |1− ηi ξi | = |1− pi − √−1qi pi + √−1qi | = 2|qi|/|ξi|. Since |ηi| = |ξi|, by (7.3), |ξi| = √ H. Hence, |1− ηi ξi | = 2|qi|/ √ H. 147 Chapter 7. Geometry of Binary Cubic Forms Suppose that (x, y) is a solution to (7.1) and related to resolvent form (ξi, ηi). Since |1− ηi ξi | = min k=1,2,3 ∣∣∣∣1− ηkξk ∣∣∣∣ , we conclude that |qi| = min k=1,2,3 |qk|. On the other hand, 3∏ k=1 |qi| = |η − ξ||ωη − ω 2ξ||ω2η − ωξ| 2 √ 2 = |η3 − ξ3| 2 √ 2 = 3 √ 3 2 √ 2 √ D, where the last equality comes from the equation (7.3). If the solution (x, y) is related to (ξi, ηi), then |u3| = |D−1/6 2√ 6 qi| < 1. (7.12) So we have log |u3| < 0. The identity |u1(x, y)u2(x, y)u3(x, y)| = 1 holds when (x, y) is a pair of solution to |F (x, y)| = 1. Therefore, log |u1|+ log |u2|+ log |u3| = 0 and log |u1u2| > 0. 7.4 Geometric Gap Principles We will study the geometric properties of the functions ui defined in section 2, by considering the well-known geometric properties of the unit group U of Q(ρ1), where ρ1 is a root of F (x, 1) = 0. 148 Chapter 7. Geometry of Binary Cubic Forms Theorem 7.4.1 (Dirichlet’s unit theorem). Let K be an algebraic number field of degree n. Let r be the number of real conjugate fields of K and 2s the number of complex conjugate fields of K. Then the ring of integers OK contains r + s − 1 fundamental units 1, . . . , r+s−1 such that each unit of OK can be expressed uniquely in the form un11 . . .  nr+s−1 r+s−1 , where u is a root of unity in OK and n1, . . . , nr+s−1 are integers. Since F has positive discriminant, for the algebraic number field Q(ρ1), in the notation of Dirichlet’s unit theorem, we have r = 3 and s = 0. Suppose that λ2 and λ3 are fundamental units of Q(ρ1). By Dirichlet’s unit theorem, we have a sequence of mappings τ : U 7−→ V ⊂ R3 and log : V 7−→ Λ, where Λ is a 2-dimensional lattice, τ is the obvious restriction of the em- bedding of K in R3, V is the image of mapping τ on U and log is defined as follows: For (x1, x2, x3) ∈ V , log(x1, x2, x3) = (log |x1|, log |x2|, log |x3|). We define τ to be the embedding from the unit group U to the lattice Λ: τ : U 7−→ Λ− {0}. By identities (7.8), (7.9) and (7.10), the vector ~u = (log |u1|, log |u3|, log |u3|) can be considered as ~v + (log |x− ρ′1y|, log |x− ρ′′1y|, log |x− ρ1y|), (7.13) where ~v = (log |D−1/6(ρ1 − ρ′′1)|, log |D−1/6(ρ′1 − ρ1)|, log |D−1/6(ρ′′1 − ρ′1)|. We have assumed that F (x, y) is monic, so we can suppose that (1, 0) is a pair of integer solutions to F (x, y) = 1. Note that the vector ~v in (7.13) is a permutation of the vector ~u(1, 0). 149 Chapter 7. Geometry of Binary Cubic Forms If (x, y) is a pair of solutions to |F (x, y)| = 1, then ~u ∈ ~v + Λ = Λ1. Note that Vol(Λ) = Vol(Λ1), where Vol(Λ) is the volume of fundamental parallelepiped of lattice Λ. Since ~u belongs to a 2-dimensional lattice, we can find a 2-dimensional representation for ~u, say (t, s). Specifically, let (x, y) be a pair of solution to F (x, y) = 1 and define functions t and s of x and y as follows t = −√6 2 log |u3| , s = log |u1| − log |u2|√ 2 . Then we have log |u1| = s/ √ 2 + t/ √ 6 (7.14) log |u2| = −s/ √ 2 + t/ √ 6 log |u3| = −2t/ √ 6. Therefore, it can be easily verified that ~u = (log |u1|, log |u2|, log |u3|) = s~α+ t~β, where ~α = 1√ 2 (1,−1, 0) and ~β = 1√ 6 (1, 1,−2) are two orthonormal vectors in R3. Hence, we can write ~u = (t, s) and ‖~u‖ = √s2 + t2 , where ‖ ‖ is the L2 norm. By (7.14,) we get∥∥∥∥(log |u1u2 |, log |u2u3 |, log |u3u1 |) ∥∥∥∥ = √3s~α′ +√3t~β′, where ~α′ = 1√ 3 ( √ 2, −1√ 2 , −1√ 2 ) and ~β′ = 1√ 3 (0, 3√ 6 , −3√ 6 ). Since ~α′ and ~β′ are orthonormal vectors in R3, we get∥∥∥∥(log |u1u2 |, log |u2u3 |, log |u3u1 |) ∥∥∥∥ = √3√s2 + t2 = √3|~u|2. (7.15) Remark. Since log |u3| < 0, the function t is a positive-valued function. 150 Chapter 7. Geometry of Binary Cubic Forms Lemma 7.4.2. If s ≥ 0 then log |u1| ≥ log |u2| and 2 sinh(s/ √ 2) = exp(− √ 6t/2), and if s < 0 then log |u1| < log |u2| and 2 sinh(−s/ √ 2) = exp(− √ 6t/2). Proof. From (7.5), the definition of ui, we have u1+u2+u3 = 0. Assume that s > 0. From (7.14), since s and t are both nonnegative, we get |u1| ≥ |u2| and |u1| ≥ |u3|. Therefore, es/ √ 2+t/ √ 6 − e−s/ √ 2+t/ √ 6 − e−2t/ √ 6 = |u1| − |u2| − |u3| = 0, and et/ √ 6(es/ √ 2 − e−s/ √ 2) = e−2t/ √ 6. Noting that es/ √ 2 − e−s/ √ 2 = 2 sinh(s/ √ 2), will complete the proof. One can give a similar proof for negative s . Let us define g(t) := √ 2 sinh−1( exp(−√6t/2) 2 ). Then s = ±g(t). In the following theorem, we summarize the properties of function g, which will be used later. Theorem 7.4.3. Let g(t) = √ 2 sinh−1( exp(− √ 6t/2) 2 ). We have: (i) g is decreasing . (ii) For any t > 0, |s| = g(t) < e− √ 6t/2/ √ 2. (iii) The function g(t)eat is decreasing when a ≤ √ 6√ 5 . Proof. (i) Since sinh(g/ √ 2) = exp(− √ 6t/2)/2, we have the following implicit differentiation: dg dt cosh(g/ √ 2) = −√3 2 exp(− √ 6t/2). Since cosh(g/ √ 2) and exp(−√6t/2)/2) are both positive, dg dt < 0 151 Chapter 7. Geometry of Binary Cubic Forms (ii) Define the function f(x) = √ 2 sinh(x/ √ 2)− x. The first derivative test shows that f is an increasing function and for positive x, f(x) > f(0) = 0. So √ 2 sinh(x/ √ 2) > x, when x > 0. Put x = |s| to get |s| = g(t) < √ 2 sinh(|s|/ √ 2). (iii) Set A(t) = g(t)eat, then A′(t) = eat(g′(t) + ag(t). For a ≤ 0, A′ < 0 since g′ < 0. For positive a, by part (i) and (ii), we have A′(t) ≤ eate− √ 6t/2( −√3 2 cosh(g/ √ 2) + a√ 2 ). Since g is a decreasing and positive-valued function, cosh(g(t)/ √ 2) is a decreasing function of t. So we have cosh(g(t)/ √ 2) < cosh(g(0)/ √ 2). An easy way to evaluate cosh(g(0)/ √ 2) is to recall that sinh(g(0)/ √ 2) = exp(0)/2 = 1/2. Therefore, cosh(g(0)/ √ 2) = √ 1 + 1 4 = √ 5/2. We conclude that A′ is negative if − √ 3/ √ 5 + a/ √ 2 ≤ 0. This means a ≤ √ 6√ 5 . 152 Chapter 7. Geometry of Binary Cubic Forms Lemma 7.4.4. Let (x, y) and (x′, y′) be two distinct solutions to equation (7.1), related to (η, ξ). Put p = p(x, y), p′ = p(x′, y′), q = q(x, y), q′ = q(x′, y′). We have |pq′ − p′q| ≥ √ 3D. Proof. By definition |pq′ − p′q| = |η + ξ√ 2 √−1(η′ − ξ′)√ 2 − η ′ + ξ′√ 2 √−1(η − ξ)√ 2 | = |ηξ′ − η′ξ|. Since ξ(x, y)η(x, y) = H(x, y) is a quadratic form of discriminant −3D , it follows that ηξ′ − η′ξ = ±√−3D(xy′ − x′y). Since (x, y) and (x′, y′) are distinct solutions to F (x, y), xy′−x′y is a nonzero integer. Lemma 7.4.5. Let (x, y) and (x′, y′) be two distinct solutions to equation (7.1), related to (ξ, η). Assume that t(x′, y′) ≥ t(x, y). Then we have t(x′, y′) ≥ 2t(x, y) + √ 6 6 logD − √ 6 log(2 + 1√ 2 ). Proof. Put p = p(x, y) p′ = p(x′, y′), q = q(x, y) q′ = q(x′, y′), s = s(x, y) s′ = s(x′, y′), t = t(x, y) t′ = t(x′, y′) and ui = ui(x, y), u′i = ui(x ′, y′). First we show that |p| ≤ √ 2D1/6et/ √ 6 cosh(s/ √ 2). 153 Chapter 7. Geometry of Binary Cubic Forms By the triangle inequality we have: |p| = 1√ 2 | √ 2p| ≤ 1√ 2 (|(p/ √ 2) + (q/ √ 6)|) + |(q/ √ 6)− (p/ √ 2)|) = 1√ 2 |(q/√6) + (p/√2)|1/2 |(q/√6)− (p/√2)|1/2 (|(q/ √ 6) + (p/ √ 2)|1/2|(q/ √ 6)− (p/ √ 2)|1/2) + 1√ 2 |(q/√6)− (p/√2)|1/2 |(q/√6) + (p/√2)|1/2 (|(q/ √ 6) + (p/ √ 2)|1/2|(q/ √ 6)− (p/ √ 2)|1/2). By (7.14) and (7.5), we have |(q/ √ 6)+(p/ √ 2)|1/2|(q/ √ 6)−(p/ √ 2)|1/2 = D1/6u1/21 u1/22 = D1/6 exp(t/ √ 6). Equations (7.14) and (7.5) also give us the following identities: ( |(q/√6) + (p/√2)|1/2 |(q/√6)− (p/√2)|1/2 + |(q/√6)− (p/√2)|1/2 |(q/√6) + (p/√2)|1/2 ) = e (log |u1|−log |u2|) 2 + e− (log |u1|−log |u2|) 2 = 2 cosh( s√ 2 ) and |q| = ( √ 6/2)D1/6e−2t/ √ 6. Using Lemma 7.4.4, we get D1/6 ≤ e(t′−2t)/ √ 6 cosh(|s′|/ √ 2) + e(t−2t ′)/ √ 6 cosh(|s|/ √ 2). One can express the above equation in terms of sinh instead of cosh by substituting cosh(|s|/√2) with sinh ( |s|√ 2 ) + e−|s|/ √ 2. Now we use the assumption that t′ ≥ t and the fact that e−|s|/ √ 2 ≤ 1. By Lemma 7.4.2, we get D1/6 ≤ e(t′−2t)/ √ 6(1 + e−3(t ′−t)/√6)(1 + e− √ 6t/2 2 ) . 154 Chapter 7. Geometry of Binary Cubic Forms Note that by Theorem (7.4.3), t ≥ log(2)/√6, whereby taking the logarithm of both sides of the above equality yields t′ − 2t ≥ √ 6 6 log(D)− √ 6 log((1 + e−3(t ′−t)/√6)(1 + 1 2 √ 2 )) So we have t′ − 2t ≥ √ 6 6 log(D)− √ 6 log(2 + 1√ 2 ). Lemma 7.4.6. Suppose that (7.1) has three distinct solutions related to (ξ, η). Then three distinct corresponding points (t, s) , (t′, s′) and (t′′, s′′) form a triangle. Proof. Suppose (t, s) , (t′, s′) and (t′′, s′′) are collinear and t ≤ t′ ≤ t′′ . Then s′ − s t′ − t = s′′ − s′ t′′ − t′ . Assume, without loss of generality, s′ > 0. Since g(t) = |s| is a decreasing function of t, we have s′′ − s′ < 0 and consequently, s′ − s < 0. Therefore, s > 0. Since we assumed (t, s) , (t′, s′) and (t′′, s′′) to be collinear, s′′ − s t′′ − t = s′ − s t′ − t By Lemma 7.4.5, and since |s| ≥ |s′| ≥ |s′′| > 0, we get s′ − s t′ − t < −s′ t < −2s′ t′ < s′′ − s′ t′′ − t′ < 0. This contradiction shows that (t, s) , (t′, s′) and (t′′, s′′) are not collinear (note that any vertical or horizontal line intersects the graph of g and g′ at most in two points). 155 Chapter 7. Geometry of Binary Cubic Forms Suppose that (7.1) has three distinct solutions related to (ξ, η) and A is the area of the triangle formed by three distinct corresponding points (t, s), (t′, s′) and (t′′, s′′). Then vectors (t− t′, s− s′) and (t− t′′, s− s′′) generate a sub-lattice of Λ1 with the volume of fundamental parallelepiped equal to 2A. Therefore, 2A ≥ Vol(Λ1). Now let us estimate 2A, the area of rectangle which has (t, s) , (t′, s′) and (t′′, s′′) as three of its edges. Recall that s(x, y) = ±g(t(x, y)) and g is a decreasing function. Suppose that t ≤ t′ ≤ t′′. Then g(t′′) ≤ g(t′) ≤ g(t) and we have 2A ≤ (t′′ − t)(g(t) + g(t′)) = (t′′ − t)(|s|+ |s′|). Part (iii) of Theorem 7.4.3 shows that |s′| < |s|e −√6(t′−t)√ 5 , Therefore, Vol(Λ) = Vol(Λ1) ≤ (t′′ − t)|s|(1 + e −√6(t′−t)√ 5 ). Using Theorem 7.4.3 again, we get the following gap principle of this paper which is essentially Theorem 5.5 of [15]: Theorem 7.4.7. Suppose that F (x, y) has three distinct solutions (x, y), (x′, y′) and (x′′, y′′), all related to (ξ, η). Assume that t = t(x, y) ≤ t′ = t(x′, y′) ≤ t′′ = t(x′′, y′′), where t is the function defined in the begining of this section. We have t′′ ≥ √ 2Vol(Λ)exp( √ 6t/2) 1 + exp(−√6(t′ − t)/√5) , where Vol(Λ) is the volume of fundamental parallelepiped of lattice Λ. 7.5 Linear Forms In Logarithms We have seen that √ 2s = log |u1| − log |u2| = log λ1 +m log λ2 + n log λ3. Where s is a function of (x, y) defined in Section 3 and ui are also functions of (x, y) defined in Section 7.2. By Lemma 7.4.3, we have log( √ 2|s|) ≤ −( √ 6/2)t. Here, we will use a well-known lower bound for linear forms in logarithms of algebraic numbers, to find an upper bound for log( √ 2|s|). 156 Chapter 7. Geometry of Binary Cubic Forms Theorem 7.5.1 (Matveev). Suppose that K is a real algebraic number field of degree d. We are given numbers α1, . . . αn ∈ K∗ with absolute logarithmic heights h(αj). Let logα1, . . . , logαn be arbitrary fixed non-zero values of the logarithms. Suppose that Aj ≥ max{dh(αj), | logαj |}, 1 ≤ j ≤ n. Now consider the linear form L = b1 logα1 + . . .+ bn logαn, with b1, . . . , bn ∈ Z and with the parameter B = max{1,max{bjAj/An : 1 ≤ j ≤ n}} . Put Ω = A1 . . . An, C(n) = 16 n! en(2n+ 2)(n+ 2)(4n+ 4)n+1( 1 2 en), C0 = log(e4.4n+7n5.5d2 log(en)), W0 = log(1.5eBd log(ed)). If bn 6= 0, then log |L| > −C(n)C0W0d2Ω. Proof. See mat26 for the proof. Here, we recall the definition of absolute logarithmic height from [9, 10]. Let Q(ρ)σ be the embeddings of the real number field Q(ρ) in R, 1 ≤ σ ≤ 3, where ρ is a root of F (x, 1) = 0. We respectively have 3 Archimedean valuations of Q(ρ): |α|σ = |α(σ)|, 1 ≤ σ ≤ 3. We enumerate prime ideals of Q(ρ) by indices σ > 3 and define non- Archimedean valuation of Q(ρ) by the formulas |α|σ = (Norm p)−k, where k = ordp(α), p = pσ, σ > d, for any α ∈ Q(ρ)∗. Then we have the product formula : ∞∏ 1 |α|σ = 1, α ∈ Q(ρ)∗. 157 Chapter 7. Geometry of Binary Cubic Forms Note that |α|σ 6= 1 for only finitely many α . We define the absolute loga- rithmic height of α as h(α) = 1 6 ∞∑ σ=1 | log |α|σ|. We will apply Matveev’s lower bound to log |u1| − log |u2| = log λ1 +m1 log λ2 + n1 log λ3. Suppose that ‖~u(x0, y0)‖ = min (x,y)∈S ‖~u(x, y)‖ and log |u1(x0, y0)| − log |u2(x0, y0)| = log ∣∣∣∣ρ− ρ′′ρ′ − ρ ∣∣∣∣+ a log λ1 + b log λ2 then for any solution (x, y), we can write log |u1(x, y)| − log |u2(x, y)| = log λ+m log λ1 + n log λ2, where m = m1 − a, n = n1 − a and λ = ∣∣∣∣ρ− ρ′′ρ′ − ρ ∣∣∣∣λa1λb2. Since λ2 and λ3 are the fundamental units of the ring of integers of Q(ρ), λ1, λ2 and λ3 are multiplicatively dependent if and only if λ1 is a unit. If λ1 is a unit then we can write log |u1| − log |u2| as a linear form in two logarithms. Since Theorem 7.5.1 gives a better lower bound for linear forms in two logarithms, we can assume that λ1, λ2 and λ3 are multiplicatively independent and log |u1| − log |u2| is a linear form in three logarithms. First, suppose that λ is a unit in the number field. We have h(λ) = 1 6 (| log(|λ|)|+ | log(|λ′|)|+ | log(|λ′′|)|) = 1 6 |τ(λ)|1 , where λ′ and λ′′ are the conjugates of λ , τ is the embedding of units to the lattice Λ and | |1 is the L1 norm on R3 . So we have h(λ) = 1 6 |τ(λ)|1 ≤ √ 3 6 ‖τ(λ)‖ , 158 Chapter 7. Geometry of Binary Cubic Forms where ‖ ‖ is the L2 norm on R3. So when λ is a unit max{3h(λ), | log(|λ|)|} ≤ ‖τ(λ)‖, (7.16) since | log(λ)| ≤ √ log2(|λ|) + log2(|λ′|) + log2(|λ′′|) = |τ(λ)|2. In the identity log |u1| − log |u2| = log λ+m1 log λ2 + n1 log λ3, λ2 and λ3 are fundamental units of Q(ρ). Therefore, in Theorem 7.5.1, Ai can be taken equal to |τ(λi)|2, for i = 2, 3. Bases ~b1 and ~b2 of lattice Λ are called reduced if the following conditions are satisfied : (i) |~b1|2 ≤ |~v|2 for every vector ~v ∈ Λ− {~0}; (ii) |~b2|2 ≤ |~v|2 for every vector ~v ∈ Λ− Z~b1. Remark. Although the definitions of reduced basis for lattices and reduced forms are somehow related, one should note that we define them separately and they are not to be confused. It is a fact that we can always choose a reduced basis for a two dimensional lattice. So we choose the fundamental units λ2 and λ3 such that the basis τ(λ2) and τ(λ3) are reduced basis for Λ. When ~b1 and ~b2 are the reduced basis of Λ, since ~b1, ~b2 ≤ ~b1 ± ~b2, we conclude that the angle between vectors ~b1 and ~b2 must be between pi/3 and 2pi/3. Therefore, λ2 and λ3 can be chosen so that |τ(λ2)|2|τ(λ3)|2| ≤ 2√ 3 Vol(Λ). Hence, in our case, A2A3 ≤ 2√ 3 Vol(Λ). By (7.15), we have∥∥∥∥(log |u1||u2| , log |u2||u3| , log |u3||u1|) ∥∥∥∥ = √3‖~u‖. The well-known inequality a+b+c3 ≤ [a 2+b2+c2 3 ] 1/2 shows that |~v|1 ≤ √ 3‖~v‖ (7.17) 159 Chapter 7. Geometry of Binary Cubic Forms for every vector ~v ∈ R3. Therefore,∣∣log |(ρ− ρ′′)(x− ρ′y)|∣∣+ ∣∣log |(ρ′ − ρ)(x− ρ′′y)|∣∣+ ∣∣log |(ρ′ − ρ′′)(x− ρy)|∣∣ ≤ 3 ‖~u(x0, y0)‖ . Now, we note that∑ σ>3 ∣∣∣∣log ∣∣∣∣(ρ− ρ′′)(x− ρ′y)(ρ′ − ρ)(x− ρ′′y) ∣∣∣∣∣∣∣∣ σ ≤ ∑ σ>3 ∣∣log |(ρ− ρ′′)(x− ρ′y)|∣∣ σ + ∑ σ>3 ∣∣log |(ρ′ − ρ)(x− ρ′′y)|∣∣ σ . We know that the Archimedean valuations of λ := (ρ− ρ′′)(x− ρ′y) are λ , (ρ′ − ρ′′)(x− ρy) and (ρ′ − ρ)(x− ρ′y). So by the product formula, since (x, y) is a solution to (7.1), the product of all non-Archimedean valuations of λ equals D−1/2. Therefore, ∣∣∣∣∣∑ σ>3 log ∣∣(ρ− ρ′′)(x− ρ′y)∣∣ σ ∣∣∣∣∣ = 12 logD, and similarly ∣∣∣∣∣∑ σ>3 log ∣∣(ρ′ − ρ)(x− ρ′′y)∣∣ σ ∣∣∣∣∣ = 12 logD. Since (ρ′ − ρ)(x− ρ′′y) and (ρ− ρ′′)(x− ρ′y) are algebraic integers, we get h(λ1) ≤ 16 (3‖~u(x0, y0)‖+ logD) . (7.18) This gives an estimate for A1. Let B1 = BA3, where B is as in theorem 7.5.1. Then B1 = max{bjAj , : 1 ≤ j ≤ 3}. Since log ∣∣∣∣u1u2 ∣∣∣∣ = log ∣∣∣∣(ρ− ρ′′)(x− ρ′y)(ρ′ − ρ)(x− ρ′′y) ∣∣∣∣+mλ1 + nλ2, we can write( log ∣∣∣∣u1u2 ∣∣∣∣ , log ∣∣∣∣u2u3 ∣∣∣∣ , log ∣∣∣∣u3u1 ∣∣∣∣) =( log |(ρ− ρ′′)(x− ρ′y)| |(ρ′ − ρ)(x− ρ′′y)| , log |(ρ′ − ρ)(x− ρ′′y)| |(ρ′ − ρ′′)(x− ρy)| , log |(ρ′ − ρ′′)(x− ρy)| |(ρ− ρ′′)(x− ρ′y)| ) + m~λ1 + n ~λ2, 160 Chapter 7. Geometry of Binary Cubic Forms where ~λi = τ(λi) , for i = 2, 3. Since λ2 and λ3 have been chosen so that ~λ2 and ~λ3 form a reduced basis for the lattice Λ, we get m| ~λ2|1 , n| ~λ3|1 ≤∣∣∣∣(log |(ρ− ρ′′)(x− ρ′y)||(ρ′ − ρ)(x− ρ′′y)| , log |(ρ′ − ρ)(x− ρ′′y)||(ρ′ − ρ′′)(x− ρy)| , log |(ρ′ − ρ′′)(x− ρy)||(ρ− ρ′′)(x− ρy)| ) ∣∣∣∣ 1 + ∣∣∣∣(log ∣∣∣∣u1u2 ∣∣∣∣ , log ∣∣∣∣u2u3 ∣∣∣∣ , log ∣∣∣∣u3u1 ∣∣∣∣)∣∣∣∣ 1 . Therefore, by (7.15) and (7.17) m| ~λ2|1, n| ~λ3|1 ≤ 3(|~u|2 + |~u(x0, y0)|2). (7.19) Theorem 7.5.2. Let F be a cubic binary equation with positive discriminant. For all pairs of solution (a, b) to the equation (7.1), except possibly one of them, we have D ≤ 64e2 √ 6t, where D is the discriminant of F (x, y) and t = t(a, b), for the function t defined in Section 7.3. Moreover, when t ≥ 5 D ≤ 1 2 e2 √ 6t. Proof. By (7.3) |H| = |ξη| = |p+ iq√ 2 . p− iq√ 2 | = p 2 + q2 2 . By (7.5), q = √ 6 2 D1/6u3 and |p| = | √ 2 2 (u1 − u2)|D1/6 ≤ √ 2 2 D1/6(|u1|+ |u2|). Therefore, by (7.14) |H| ≤ 1 2 D1/3(e2t/ √ 6(e2s/ √ 2 + e−2s/ √ 2 + 2)/2 + 3 2 e−4t/6). (7.20) Therefore, by Lemma 7.2.1, for all solutions (a, b) to (7.1), except at most one of them 1 2 D1/2 √ 3 ≤ 1 2 D1/3(e2t/ √ 6(e2s/ √ 2 + e−2s/ √ 2 + 2)/2 + 3 2 e−4t/6). (7.21) 161 Chapter 7. Geometry of Binary Cubic Forms Since t > 0, part (ii) of Lemma 7.1 says that |s| < e− √ 6t/2/ √ 2. Hence, D1/6 ≤ 2e2t/ √ 6, which proves the theorem for general t. When t ≥ 5, we note that 3 2e −4t/6 < 0.054 and |s| < 0.0016 by Theorem 7.4.3 . Since ~u = (t, s) and ‖~u‖ = √t2 + s2, from Theorem 7.4.3, we deduce that |~u|2 is an increasing function of t. So we can assume that Theorem 7.5.2 is satisfied for all solutions, except possibly (x0, y0), where ‖~u(x0, y0)‖ = min (x,y)∈S ‖~u(x, y)‖, S is the set of all solutions to (7.1) and ~u = (log |u1|, log |u2|, log |u3|). Suppose that three distinct solutions (x, y), (x′, y′) and (x′′, y′′) of (7.1) are related to (ξ, η) and t′′ = t(x′′, y′′) > t′ = t(x′, y′) > t = t(x, y). First, we recall that ~u = (t, s) and |~u|2 = √ t2 + s2. By Theorem 7.4.3, if we take t ≥ 5 (Theorem 7.5.2 enables us to assume that t is large), we get |~u|2 = √ t2 + e− √ 6t/2 ≤ 1.0016t. Therefore, by (7.18) and theorem 7.5.2, we can take A1 = 1.0016( 3 2 t+ √ 6t). Inequality (7.19) suggests the value 3(1.0016)(t′′ + t) for B1. But by theorem 7.4.5, for large discriminant D, t′′ > 4t . So we take B1 = 3(1.0016)(t′′ + t′′/4). (7.22) So Matveev ’s lower bound gives us: log |L| > −1.5036× 1011A1A2A3 log(25.6708(3.0048)(1.25)t′′/A3), where L = log |u1| − log |u2|. On the other hand, by Theorem 7.4.3, we have log |L| = log √ 2|s′′| ≤ − √ 6t′′/2. We conclude that t′′ ≤ 1.2276× 1011A1A2A3 log(96.2751t′′/A3), 162 Chapter 7. Geometry of Binary Cubic Forms or 96.2751t′′/A3 log(96.2751t′′/A3) ≤ 1.1892× 1013A1A2. Therefore, log(96.2751t′′/A3) ≤ e e− 1 log(1.1892× 10 13A1A2). Recalling that A2A3 ≤ 2/ √ 3Vol(Λ), we obtain the following upper bound for t′′: t′′ ≤ e e− 11.2276× 10 11( 2√ 3 )Vol(Λ)A1 log(1.1892× 1013A1A2). (7.23) 7.6 Proof Of The Main Results Let ‖~u(x0, y0)‖ = min(x,y)∈S‖~u(x, y)‖. Suppose that (x, y), (x′, y′) and (x′′, y′′), with none of them equal to (x0, y0), are three distinct solutions to (7.1), and related to a fixed choice of resolvent form. Let t = t(x, y) < t′ = t(x′, y′) < t′′ = t(x′′, y′′). By (7.23) and Theorem 7.4.7, we get e e− 11.2276× 10 11A1( 2√ 3 )Vol(Λ) log(1.1892× 1013A1A2) ≥ √ 2Vol(Λ)exp( √ 6t/2) 1 + exp(−√6(t′ − t)/√5) , (7.24) where A1 = (3/2 + √ 6)(1.006)t and A2 = |τ(λ2)|2. Without loss of generality, we can assume that |τ(λ2)|2 ≤ |τ(λ3)|2. Therefore, √ 3 2 |τ(λ2)|22 ≤ √ 3 2 |τ(λ2)|2|τ(λ3)|2 ≤ Vol(Λ). We have log |u1(x, y)| − log |u2(x, y)| = log λ1 +m′ log λ2 + n′ log λ3, where m′ and n′ are integers. Since (x, y) 6= (x0, y0), at least one of m′ or n′ is a nonzero integer. So by (7.19) we have |τ(λ2)|2 ≤ 6.01t . Using Theorem 7.4.5, we get 163 Chapter 7. Geometry of Binary Cubic Forms e 1− e4.8484× 10 11( 2√ 3 )Vol(Λ)t log(2.8184× 1014t2) ≥ √ 2Vol(Λ)exp( √ 6t/2) 1 + exp(−√6t/√5) . Therefore, t < 27.91 and by equation (7.21), D < 5.31× 1059; i.e. we have proven that there are at most 2 pairs of solutions (x, y) 6= (1, 0) and (x′, y′) 6= (1, 0) related to a resolvent form (ξ, η) , when D ≥ 5.31× 1059. If we suppose that D < 5.31× 1059, by (7.18), we can take A1 = 3 2 t+ 1 2 log(5.31× 1059). By substituting this new value of A1 in (7.23), we get t ≤ 27.5321, and therefore, D < 1.4× 1057. Since we have three pairs of resolvent forms, Theorem (7.1.1) is proven. As we mentioned in the remark after the proof of Lemma 7.2.1, the solution (1, 0) needs to be treated separately, only if F is equivalent to a monic reduced form. Otherwise, (x0, y0) 6= (1, 0) and Lemma 7.2.1 and therefore Lemma 7.5.2 will hold for all solutions without any exception. By the analytic class number formula and Louboutin’s upper bound (which can be found in [3]) : Vol(Λ) ≤ √ 3 8 √ D log2D, we have that A2 ≤ 14D 1/4 logD. By Theorem 7.5.2, A2 ≤ 14e √ 6t/2(log 1 2 + 2 √ 6t). Now, having appropriate values of A1 and A2 in hand, we solve inequality (7.24) to get t ≤ 28.38 and consequently by (7.21), D ≤ 9× 1058; i.e. we have proven that there are at most 2 pairs of solutions (x, y) and (x′, y′) related to a resolvent form (ξ, η) , when D > 9× 1058. Therefore, we get Theorem 7.1.2. 164 Chapter 7. Geometry of Binary Cubic Forms In [2], it is proven that if D ≥ 2400, related to a fixed pair of resolvent form, there are at most 3 different pairs of solutions (x, y) to (7.1) with H(x, y) ≥ 12 √ 3D, where H is the Hessian of F . This together with lemma 7.2.1 leads to the main theorem of [2], that is, the equation F (x, y) = 1 has at most 10 solutions in integer x and y . For 0 < D < 2400, equation F (x, y) = 1 is completely solved for representatives of every equivalent class of binary cubic forms. This computations show that the equation (7.1) with discriminant 0 < D < 106 has at most 9 solutions in integers x and y. The complete result of these computations are tabulated in section 9 of [2]. 165 Bibliography [1] B.N. Delone and D.K. Fadeev. The Theory of Irrationalities of the Third Degree. Translation of math. Monographs, AMS 10 (1964). [2] M.A. Bennett. On the representation of unity By binary Cubic Forms. Trans. Amer. Math. Soc. 353 (2001), 1507-1534. [3] J.H.E. Cohn. The Diophantine equation x2 + C = yn, II, Acta Arith. 109.2 (2003), 205-206. [4] V.I. Baulin. On an intermediate equation of the third degree with least positive discriminant (Russian). Tul’sk Gos.Ped.Inst. Ucen. Zap. Fiz. Math. Nauk. Vip. 7 (1960), 138-170. [5] J.H. Evertse. On the representation of integers by binary cubic forms of positive discriminant. Invent. Math.73(1983), 117-138. [6] I. Gall and N. Schulte. Computing all power integral of cubic fields. Math. Comp. 53 (1989), 689-696. [7] E. Lee. Studies on Diophantine Equations. PhD thesis, Cambridge University, 1992. [8] F. Lippok. On the representation of 1 by binary cubic forms of positive discriminant. J. Symbolic Computation 15 (1993), 297-313. [9] E.M. Matveev, An explicit lower bound for a homogeneous rational linear form in logarithms of algebraic numbers, Izv. Ross. Akad. Nauk Ser. Mat. 62 (1998), 81-136, translation in Izv. Math. 62 (1998), 723-772. [10] E.M. Matveev, An explicit lower bound for a homogeneous rational linear form in logarithms of algebraic numbers, Izv. Ross. Akad. Nauk Ser. Mat. 64 (2000), 125-180, translation in Izv. Math. 64 (2000), 1217-1269. 166 Bibliography [11] M. Mignotte. Verification of a conjecture of E. Thomas . J. Number Theory 44 (1993), 172-177. [12] M. Mignotte. Petho’s cubics. Dedicated to Professor Kalman Győry on the occasion of his 60th birthday. Publ. Math. Debrecen 56 (2000), 481-505. [13] M. Mignotte and N. Tzanakis. On a family of cubics. J. Number Theory 39(1991), 41-49. [14] W. Ljunggren. Einige Bemerkungen uber die Darstellung ganzer zahlen durch bibare kubische formen mit positiver Diskriminante. Acta Math. 75(1942), 1-21. [15] R. Okazaki. Geometry of a cubic Thue equation, Publ. Math. Debrecen. 61 (2002),267-314. [16] C.L. Siegel. Uber einige Anwendungen diophanticher Approximationen. Abh.Preuss.Akad.Wiss. (1929), Nr.1. [17] E. Thomas, Complete solutions to a family of cubic Diophantine equations, J. Number Theory 34 (1990), 235-250. 167 Chapter 8 Conclusion In Chapter 2, we develop general machinery based on the Thue-Siegel Method to find an upper bound upon the number of integral solutions to a family of quartic Thue inequalities. This specific family turned out to be a very important one. In Chapters 3, 4 and 5, upper bounds are given upon the number of positive integral solutions to Diophantine equations aX4 − bY 2 = 1 and aX4 − bY 2 = 2 by reducing them to quartic Thue inequalities belonging to the family of inequalities studied in Chapter 2. We believe that there are other interesting questions we could answer using the method of Chapter 2. Particularly, we are hoping to use our method to obtain a good upper bound for the number of integer points on the curve x2 − dy4 = k. Let a and b be positive integers. In chapter 3, we show that the equation aX4 − bY 2 = 1 has at most two solutions in positive integers (X,Y ). This result is sharp, for let m be a positive integer, then the positive integral solutions to the equation (m2 +m+ 1)X4 − (m2 +m)Y 2 = 1 are given by (X,Y ) = (1, 1) and (X,Y ) = (2m+ 1, 4m2 + 4m+ 3). In fact, these are the only examples known to have as many as two positive solutions. This suggest a stronger version of Theorem 3.1.1. In Chapters 6 and 7, we study two different methods, one from Diophantine analysis and the other one from the geometry of numbers. 168 Chapter 8. Conclusion There is another method invented by Silverman for computing the number of solutions of Thue equations. He relates the number of these solutions to the rank of Mordell-Weil group on the Jacobian of a related genus 1 curve. Because of its different nature, Silverman’s method might be applied to strengthen our results when combined with our approximations. 169 Bibliography [1] S. Akhtari, The method of Thue-Siegel for binary quartic forms. submitted (2007). [2] M.A. Bennett. On the representation of unity by binary cubic forms. Trans. Amer. Math. Soc. 353 (2001), 1507-1534. [3] E. Bombieri and W.M. Schmidt. On Thue’s equation. Invent. Math. 88 (1987), 69-81. [4] G.V. Chudnovsky. On the method of Thue-Siegel. Ann. of Math. II Ser. 117 (1983), 325-382. [5] W. Ljunggren. On the representation of integers by certain binary cubic and biquadratic forms. Acta. Arith. XVII (1971). [6] L.J. Mordell. On the rational solutions of indeterminate equations of the 3rd and 4th degrees. Proc. Camb. Phil. Soc.21 (1922), 179-192. [7] W.M. Schmidt, Diophantine Approximations and Diophantine Equations, Lecture Notes in Mathematics, Springer Verlag (2000). [8] A. Thue. Über Annäherungenswerte algebraischen Zahlen. J. reine angew. Math. 135 (1909), 284-305. 170

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