On the number of solutions to systems of Pell equations

Journal of Number Theory 125 (2007) 356–392 www.elsevier.com/locate/jnt On the number of solutions to systems of Pell equations Mihai Cipu a,1 , Maurice Mignotte b,∗ a Institute of Mathematics of the Romanian Academy, PO Box 1-764, RO-014700, Bucharest, Romania b Université Louis Pasteur, UFR de Mathématiques, 7, rue René Descartes, 67084 Strasbourg Cedex, France Received 31 March 2006; revised 29 August 2006 Available online 23 January 2007 Communicated by Michael A. Bennett Abstract We prove that, for positive integers a, b, c and d with c = d, a > 1, b > 1, the number of simultaneous solutions in positive integers to ax 2 − cz2 = 1, by 2 − dz2 = 1 is at most two. This result is the best possible one. We prove a similar result for the system of equations x 2 − ay 2 = 1, z2 − bx 2 = 1.  2006 Elsevier Inc. All rights reserved. Keywords: Simultaneous Pell equations; Linear forms in logarithms 1. Introduction The number of solutions of the simultaneous Diophantine equations ax 2 − cz2 = δ1 , by 2 − dz2 = δ2 (1) was a question of constant interest in the last century. It is known already since A. Thue [26] and C.L. Siegel [25] that Eqs. (1) have finitely many solutions when cδ2 = dδ1 . Their works do not provide estimations for the number of solutions. Using the methods developed by W.M. Schmidt [23,24], H.P. Schlickewei [22] proved that the number of integer solutions to x 2 − cz2 = 1, y 2 − dz2 = 1 (2) * Corresponding author. E-mail addresses: [email protected] (M. Cipu), [email protected] (M. Mignotte). 1 Partially supported by the CEEX Program of the Romanian Ministry of Education and Research, contract CEx05- D11-11/2005. 0022-314X/$ – see front matter  2006 Elsevier Inc. All rights reserved. doi:10.1016/j.jnt.2006.09.016 M. Cipu, M. Mignotte / Journal of Number Theory 125 (2007) 356–392 357 78 is at most 4 × 82 . D.W. Masser and J.H. Rickert [15] improved considerably the bound, proving that these equations have at most sixteen solutions (x, y, z) in positive integers. The proof uses the hyper-geometric method. The same approach was followed by M.A. Bennett [6], who lowered the bound to three. Since there is no known pair of Pell equations with three solutions, he conjectured that Eqs. (2) have at most two solutions for any c = d. A recent result of P. Yuan [29] shows that there are at most finitely many exceptions to this conjecture. To be precise, if max{ c, d}  1.4 × 1057 , the system of Eqs. (2) has at most two integer solutions (x, y, z) with x, y, z > 0. M.A. Bennett, R. Okazaki and the present authors have given an unconditional proof of the conjecture in [7]. (The reader willing to see a thorough description of the approach, with more details than in the printed version of the paper, may consult [10].) The result is best possible because there are families of (c, d) for which the system (2) has two positive solutions. For l and √ m integers greater than 1, set α = m + m2 − 1 and α 2l − α −2l . n(l, m) = √ 4 m2 − 1 (3) Then and   (x1 , y1 , z1 ) = m, n(l, m), 1 (x2 , y2 , z2 ) =   α 2l + α −2l 2 , 2n(l, m) − 1, 2 n(l, m) 2 (4) (5) are two positive integral solutions to (2) with c = m2 − 1 and d = n(l, m)2 − 1. The proofs in [29,7,10] are based on A. Baker’s theory [3] on bounds for linear forms in logarithms of algebraic numbers. This method combined with techniques from computational Diophantine approximation is instrumental in obtaining all solutions of instances of (1). Since the ground-breaking paper of A. Baker and H. Davenport [4], such a combination has been repeatedly successfully employed (see, for instance, [11,12]). Using this approach, W.S. Anglin showed [2] (see, also, [1]) that the system of Eqs. (2) has at most one positive solution for 2  c < d  200. In [10] it is reported that for coefficients in the range 2  c < d  2000, the system has two solutions if and only if these solutions have the form described in (4) and (5). This is the most extensive numerical confirmation of a stronger conjecture of Yuan [30, Conjecture 1.1] recalled below. A fruitful way to study the solutions of (1) is by connecting this system to the elliptic equation ab(xy)2 = (cz2 + δ1 )(dz2 + δ2 ). One can show that each non-trivial solution to (1) gives rise to a point of infinite order on the elliptic curve Y 2 = X 3 + (c + d)X 2 + cdX. K. Ono [20] dealt with several infinite families of such systems and deduced the lack of non-trivial solutions by simply computing the number of representations of certain integers by pairs of suitable ternary quadratic forms. N. Tzanakis, in a very well written exposition [27], advocates the use of linear forms in elliptic logarithms. The same paper contains an ample bibliography, with pointers to other works based on this idea. Several elementary methods for solving specific pairs of generalized Pell equations have been devised (see, for instance, [8,13,19]). There are also conditional results, assuming the ABC conjecture [28]. 358 M. Cipu, M. Mignotte / Journal of Number Theory 125 (2007) 356–392 It is very hard to have an idea on the precise number of solutions of a system of simultaneous Pell equations of the most general type (1) just by looking at the coefficients. Indeed, Masser and Rickert [15] have devised a method to produce simultaneous Pell equations with any prescribed number of solutions. However, Yuan [30] put forward a conjecture aiming to describe when there are at least two solutions in positive integers to ax 2 − cz2 = 1, by 2 − dz2 = 1. (6) Yuan claims that the coefficients of systems with two solutions are obtained as follows. For integers l > 1, m > 1, and a > 1, put √ √ (m + m2 − 1 )2l − (m − m2 − 1 )2l n(l, m) = √ 4 m2 − 1 and 4b(l, a) − 1 = √ √ √ √ ( a + a − 1 )l − ( a − a − 1 )l , √ 2 a−1 l≡3 (mod 4). It is easily seen that n(l, m) and b(l, a) are positive integers and each of the systems of Pell equations     x 2 − m2 − 1 z2 = y 2 − n(l, m)2 − 1 z2 = 1 and   ax 2 − (a − 1)z2 = b(l, a)y 2 − b(l, a) − 1 z2 = 1 has two solutions in positive integers, given by   (x, y, z) = m, n(l, m), 1 , √ √   (m + m2 − 1 )2l + (m − m2 − 1 )2l 2 , 2n(l, m) − 1, 2n(l, m) (x, y, z) = 2 and (x, y, z) = (1, 1, 1), √ √ √  √  ( a + a − 1 )l + ( a − a − 1 )l (x, y, z) = , 4b(l, a) − 3, 4b(l, a) − 1 , √ 2 a respectively. Call (a1 , b1 , c1 , d1 ) an equivalent form of (a, b, c, d) if there are positive integers a0 , b0 , c0 , d0 such that (a1 , b1 , c1 , d1 ) = (a/a02 , b/b02 , c/c02 , d/d02 ). Conjecture 1. [30, Conjecture 1.1] If the system of simultaneous Pell equations (6) has at least two solutions in positive integers, then their coefficients are given by     (a, b, c, d) = 1, 1, m2 − 1, n(l, m)2 − 1 , a, b(l, a), a − 1, b(l, a) − 1 or equivalent forms thereof. M. Cipu, M. Mignotte / Journal of Number Theory 125 (2007) 356–392 359 The first aim of this paper is to study the number of solutions for the system of Pell equations (6) when a > 1, b > 1. Yuan [30] proved that if c = d and max{a, b, c, d} > 1.16 × 1059 then the system has at most two solutions in positive integers. Here we prove that the same result holds regardless of the size of the coefficients. Theorem 2. If a > 1 and b > 1 are distinct positive integers, then the system of Pell equations (6) has at most two integer solutions with x, y, z > 0. In order to establish this result, we adapt the approach developed in [10]. Roughly speaking, the proof has three parts, devoted respectively to a theoretical study of the properties of the solutions, to applications of general results on lower bounds for linear forms in logarithms, and to computer-aided search for a third solution. Our study of systems of the type (6) involves proving certain gap principles. Such results show that consecutive solutions must be at a significant distance from one another. In fact, the gap between consecutive solutions is sufficiently large that one can then apply lower bounds for linear forms in the logarithms of three algebraic numbers in a classical manner. Specialization of a general theorem of E.M. Matveev [16], valid for any number of logarithms, yields the fact that three solutions for (6) can exist only if max{a, b} < 3.1 × 1051 . In order to significantly reduce this bound, we apply a suitable instance of a theorem of M. Laurent, M. Mignotte and Yu. Nesterenko [14]. The main idea of the second part of the proof is to use lower bounds for linear forms in two logarithms. As a result of these considerations, we obtain in Section 4 that max{a, b} < 4 × 1038 . To complete the proof, we use computers to perform two types of computations. On the one hand, standard techniques from computational Diophantine approximation yield the conclusion that small values of max{a, b}, say, smaller than 2000, are not compatible with the existence of three positive solutions for Eqs. (6). On the other hand, various verifications eliminate the possibility that the third solution exists when max{a, b} is in the domain not excluded for other reasons. The proof we give to Theorem 2 is essentially self-contained. In its structure, computations are interspersed with inequalities relating various numerical characteristics associated to putative solutions. A useful feature of this interplay is that by repeating a reasoning with good bounds results in even better bounds. Starting the game simultaneously for small and big values of max{a, b}, we gradually shrink the search domain, eventually arriving to exhaust the range not excluded in the previous steps of the proof. Another feature of the approach developed to study the number of solutions for simultaneous Pell equations is its flexibility. The second aim of the present paper is to deal with the system of Diophantine equations x 2 − ay 2 = 1, z2 − bx 2 = 1. (7) In the second part of the paper we establish a bound for the number of positive solutions for these equations, which is almost sharp. Theorem 3. Let a and b be positive integers. Then the system of simultaneous Pell equations (7) has at most two common solutions with x, y, z > 0. Unlike Theorem 2, this result is not assuredly best possible. We know of no system (7) having more than one solution in positive integers, and presumably none exists. 360 M. Cipu, M. Mignotte / Journal of Number Theory 125 (2007) 356–392 Conjecture 4. [31, Conjecture 1.1] For any positive integers a, b, the system of Diophantine equations (7) has at most one solution in positive integers. Conjecture 4 has been confirmed for several classes of coefficients, see [31] and [9]. The methods of this paper are not only capable of proving sharp results on the types of systems of simultaneous Pell equations discussed so far, but can be used to deal with other families of equations, as already seen in [10]. This will be the subject of future work. 2. Properties of solutions The results obtained here will eventually lead to strong gap principles (see Section 3). The properties we shall prove in this section hold for all systems (6), regardless of the number of solutions, as long as there exists at least one solution in positive integers. First we note that, when dealing with simultaneous Diophantine equations of the type (6), it is sufficient to consider only coefficients c and d which are one less than a and b, respectively. The explanation is given by Lemma 5 below. In its proof (and also later on), divisibility properties of Lehmer numbers are invoked. Lemma 5. Assume that the system (6) has at least one solution in positive integers and a > 1, b > 1. Let (x0 , y0 , z0 ) be a solution with z0 the smallest positive value taken by the third component of a solution (x, y, z) of (6). (a) For any solution (x, y, z) with x, y, z > 1, one has x0 divides x, y0 divides y, and z0 divides z. (b) The system (6) has the same number of solutions as a system of the form AX 2 − (A − 1)Z 2 = BY 2 − (B − 1)Z 2 = 1. √ √ √ √ Proof. Let α = u a + v c, respectively β = u′ b + v ′ d, be the fundamental solution of the Pell equation ax 2 − cz2 = 1, respectively by 2 − dz2 = 1. For j and k odd, put Uj := α j − α −j , α − α −1 Uk′ := β k − β −k , β − β −1 Vj := α j + α −j , α + α −1 Vk′ := β k + β −k . β + β −1 Since a  2 and b  2, there are odd integers j0  1, k0  1, such that x0 = uVj0 , y0 = u′ Vk′0 , z0 = vUj0 = v ′ Uk′ 0 . For any solution (x, y, z) of (6) one has x = uVj , y = u′ Vk′ , z = vUj = v ′ Uk′ for certain odd integers j  j0 , k  k0 . Suppose, by way of contradiction, that z0 does not divide z. Then Uj0 > 1 and, having in view parts (a) and (b) of Lemma 6, by Euclidean division one gets j = 2qj0 ± r, k = 2tk0 ± s, 0 < r < j0 , 0 < s < k0 . From the identity stated in part (c) of the next lemma one obtains Uj ∓ Ur = 2Vqj0 ±r Uqj0 , ′ Uk′ ∓ Us′ = 2Vtk′ 0 ±s Utk . 0 Having in view Lemma 6(a), one gets z = vUj ≡ ±vUr (mod z0 ) and z = v ′ Uk′ ≡ ±v ′ Us′ (mod z0 ). M. Cipu, M. Mignotte / Journal of Number Theory 125 (2007) 356–392 361 ′ Since 2vUr < vUr+1  vUj0 = z0 and 2v ′ Us′ < v ′ Us+1  v ′ Uk′ 0 = z0 , one concludes that vUr = v ′ Us′ . Hence, (uVr , u′ Vs′ , vUr ) is a positive solution of (6) whose third component is smaller than z0 . This contradicts the choice of (x0 , y0 , z0 ). One concludes that z0 divides z. Therefore, there is a positive integer g for which z = gz0 , and furthermore, both j0 divides j and k0 divides k. Lemma 6(a) implies x = ex0 , y = fy0 for suitable positive integers e, f . Hence,   1 = ax 2 − cz2 = ae2 x02 − cg 2 z02 = ax02 e2 − g 2 ax02 − 1 . One concludes that (e, f, g) is a positive solution for the simultaneous Pell equations ax02 X 2 − (ax02 − 1)Z 2 = by02 Y 2 − (by02 − 1)Z 2 = 1. ✷ Lemma 6. (a) If Uk = 1, then Uk | Un if and only if k | n. (b) If k  1, then Vk | Vn if and only if n/k is an odd integer. (c) If n = 2tm ± r, with 0  r  m and t  0, then Un ∓ Ur = 2Vtm±r Utm . Proof. The properties are well known. They are proved in various places, for instance, in [21] and [17]. ✷ By Lemma 5, we may restrict our attention to the case that c = a − 1 and d = b − 1 for some integers b > a  2, and thereby study the system of Diophantine equations ax 2 − (a − 1)z2 = 1, by 2 − (b − 1)z2 = 1. (8) Put α= √ √ a + a − 1, β= √ √ b+ b−1 (9) and consider (x, y, z) a positive integer solution to (8). Then z appears in two linear recurrent sequences z = Uj = Uk′ , (10) for some positive odd integers j and k, where α j − α −j Uj = √ 2 a−1 and Uk′ = β k − β −k , √ 2 b−1 j, k ∈ N. (11) Note that j k α <β <α Indeed, from α, α −1 , β, β −1 > 0 it follows that j  b−1 . a−1 (12) 362 M. Cipu, M. Mignotte / Journal of Number Theory 125 (2007) 356–392 √ √ β k − α j = 2z b − 1 + β −k − 2z a − 1 − α −j √ √ 2z > 2z( b − 1 − a − 1 ) − α −j  − α −j > 0. α This in turn implies 1 − α −2j βk = αj 1 − β −2k  b−1 < a−1  b−1 . a−1 From (12) one infers that the linear form in logarithms   b−1 + j log α 2 − k log β 2 Λ = log a−1 (13) is positive and bounded from above by   2α 2−2j −2 log 1 − α −2j < 2 . α −1 Hence, log Λ < −2j log α + log 2α 2 < −2j log α + 0.882. α2 − 1 (14) (15) The observation stated in the next lemma allows one to obtain a companion inequality to (12). Lemma 7. In the above notation and hypotheses we have  1+  2  2  β β b−1 1 3 1 1 < . − − < 1+ 2 2 2(a − 1) 2(b − 1) 16(a − 1) α a−1 2(a − 1) α 2 √ Proof. We shall use the elementary inequality 2 t 2 + t < 2t + 1 valid for positive t. The left inequality in our lemma is implied by     (b − 1) 16a 2 − 24a + 5 − 8a 2 + 16a − 8 (2b − 1)    < 8(b − 1)2 (a − 1) 2a − 1 + 2 a 2 − a . Expanding and squaring, one finds the equivalent form     0 < 128a 2 − 16a − 4 b4 + −256a 3 + 224a 2 + 80a + 60 b3   + 256a 4 − 128a 3 − 448a 2 + 448a − 245 b2   + −512a 4 + 1152a 3 − 720a 2 − 16a + 150 b + 192a 4 − 512a 3 + 432a 2 − 96a − 25 =: f (b). Routine computations yield f ′′ (b) > 0, f ′ (b) > f ′ (a) > 0, f (b) > f (a) > 0 for a  2. M. Cipu, M. Mignotte / Journal of Number Theory 125 (2007) 356–392 363 The right inequality is equivalent to       2(b − 1) 2a − 1 + 2 a 2 − a < (2a − 1) 2b − 1 + 2 b2 − b , √ which follows from 4(b − 1) < 2b − 1 + 2 b2 − b. ✷ From (12) and this lemma, the following inequality holds, and plays an important role in the rest of the paper: β 2k−2  1 α 2j −2 . < 1+ 2(a − 1)  (16) A consequence of Eq. (16) is that j  k if both j and k are greater than 1. Since the application t → (t j − t −j )/(t − t −1 ) (j > 1) is increasing for t > 1, one concludes from Eq. (10) that, if k > 1 and j > 1, then j > k. The explicit computations described in Section 5 will benefit from the results below. Lemma 8. Let (x, y, z) be a solution of the simultaneous Diophantine equations (8). If z = Uj = Uk′ , with j > k, then j and k are congruent modulo 4. If j ≡ ±3 (mod 8), then 8 divides a(j − k). If j ≡ ±1 (mod 8), then 8 divides b(j − k). Proof. The values Uj , Uk′ are generated by recurrence sequences of the type wt+2 = (4s − 2)wt+1 − wt , w0 = 1, w1 = 4s − 1. Looking at this recurrence modulo 4 suffices to conclude j ≡ k (mod 4). The second part of the conclusion is derived by reducing modulo 8. ✷ Lemma 9. For any t  2, (4a − 3)t < U2t+1 < (4a − 1)t . Consequently,  1/t U2t+1 = 4a − 2 or 4a − 3. Proof. The first inequality results from U2t+1 = α 2t + α 2t−1 + · · · + α −2t , because both α and α −1 are positive, and α 2 > 4a − 3. The upper bound for Ut+1 is obtained by induction, since U3 = 4a − 1 and U2t+3 = (4a − 2)U2t+1 − U2t−1 < (4a − 2)U2t+1 for any positive index t. ✷ 3. Gap principles The aim of this section is to prove that the second and the third solution of a system of simultaneous Diophantine equations of the type (6) are distant of each other. We shall follow Yuan’s strategy from [30], paying considerable attention to the details of the reasoning. Consequently, for k2  7, our gap principle is much stronger than that used by Yuan. In [30] it is proved that the quotient j3 /j2 is bounded from below by an expression which is linear in β, whereas our bound is of the type β (k2 −3)/2 . This new gap principle allows for certain computations to be completed which were not previously achievable. This connection will be pursued in a subsequent section. 364 M. Cipu, M. Mignotte / Journal of Number Theory 125 (2007) 356–392 Suppose that the system (8) has three positive integer solutions (xi , yi , zi ) (i = 1, 2, 3). Then zi = β ki − β −ki α ji − α −ji = √ √ 2 a−1 2 b−1 for odd integers ji and ki (i = 1, 2, 3) with 1 = k1 < k2 < k3 and 1 = j1 < j2 < j3 . We first note that either j2 divides j3 and k2 divides k3 , or else j3 = 2qj2 ± 1 and k3 = 2q1 k2 ± 1 for some positive integers q and q1 . The reasoning is similar to that used in the last paragraph of the proof of part (a) of Lemma 5. After one concludes that (Vr , Vs′ , Ur ) is a positive solution of (8) whose third component is smaller than z2 , it follows that Ur = z1 = 1, whence r = 1, and therefore Vs′ = 1 = V1′ , so that s = 1. As seen above, there exist integers q, q1  2 and σ , σ1 ∈ {−1, 0, 1} such that j3 = qj2 + σ , k3 = q1 k2 + σ1 and both qσ and q1 σ1 are even. As a preparation for the proof of our gap principle we establish the following result. Lemma 10. σ = σ1 . Proof. If σ = 0, then j2 divides j3 , so that z3 = Uj3 = Uk′ 3 is multiple of z2 = Uj2 = Uk′ 2 . Thus, Uk′ 2 divides Uk′ 3 , whence k2 divides k3 . This means σ1 = 0, as asserted. By symmetry, one obtains that σ = 0 whenever σ1 = 0. Suppose now that both σ and σ1 are non-zero. From Uj3 − σ = 2Vqj2 +σ Uqj2 (cf. Lemma 6(c)) and Uj2 | Uqj2 it follows that z3 = Uj3 ≡ σ (mod z2 ). Similarly, Uk′ 3 ≡ σ1 (mod z2 ). The claim follows from |σ − σ1 |  2 < 4a − 3 < z2 (cf. Lemma 9). ✷ Another observation needed later on is that the relationship between q and q1 is the same as between j2 and k2 . Lemma 11. q > q1 . Proof. If j3 is multiple of j2 , then k3 is divisible by k2 and we have √ √ √ √ (x2 a + z2 a − 1 )q − (x2 a − z2 a − 1 )q z3 = Uqj2 = √ 2z2 a − 1   √ q (x2 a )q−2t−1 z22t (a − 1)t = 2t + 1 t0 and z3 = Uq1 k2 = t0   √ q1 (y2 b )q1 −2t−1 z22t (b − 1)t . 2t + 1 √ √ The desired conclusion follows because y2 b > x2 a and b > a. It remains to examine the case j3 = qj2 + σ and k3 = q1 k2 + σ with σ = 0. We shall implicitly use the well-known inequalities t− valid for positive t. t2 < log(1 + t) < t 2 M. Cipu, M. Mignotte / Journal of Number Theory 125 (2007) 356–392 365 Let us consider the linear forms in logarithms   b−1 Λ2 = log + j2 log α 2 − k2 log β 2 , a−1   b−1 + j3 log α 2 − k3 log β 2 . Λ3 = log a−1 By relations (12) and (16), we have     b−1 b−1 − Λ2 = 2(k2 log β − j2 log α) < log , 0 < log a−1 a−1   b−1 β 1 − Λ2 < 2 log + log . a−1 α 2(a − 1) Replacing in Λ3 the coefficients j3 and k3 in terms of j2 and k2 , one gets  β b−1 + 2σ log − q1 Λ2 a−1 α       β b−1 b−1 − 2 log + q1 log − Λ2 .  Λ3 − log a−1 α a−1 2(q − q1 )j2 log α = Λ3 + (q1 − 1) log  (17) From Lemma 7 one obtains   β 1 b−1 log < 2 log + , a−1 α 2a − 2 so that the rightmost term in the chain of inequalities (17) is greater than     b−1 1 β Λ3 − 4 log − − Λ2 . + q1 log α 2a − 2 a−1 From the left inequality in Lemma 7 one gets for all (a, b) = (2, 3)   b−1 1 1 β β log > 2 log + > 2 log + . 2 a−1 α 9(a − 1) α 9(a − 1)2 (18) (19) In the remaining case, a numerical verification shows that the leftmost term in Eq. (19) is bigger than the rightmost one. On the other hand, by (14) we have Λ2 < 1 2α −12 . < 2 α − 1 (4a − 4)6 Therefore, the expression in (18) is bounded from below by   1 1 β 1 − . 2(q1 − 2) log + q1 − 2 6 α 2a − 2 9(a − 1) (4a − 4) (20) 366 M. Cipu, M. Mignotte / Journal of Number Theory 125 (2007) 356–392 Since 2 log 3 β 1 1 > − 2 , α a 2a 4a we have 2(q − q1 )j2 log α >   1 1 3(q1 − 2) 1 + q1 . − − 2 6 4a 2a − 2 9(a − 1) (4a − 4) As the expression in the right side of the previous inequality is certainly positive when q1  4, we have obtained the conclusion of the lemma, unless q = q1 = 2. In this situation, it follows from Lemma 7, (17) and (14) that   2 β 1 b−1 − 2 log > > 2Λ − Λ  log , 2 3 a−1 α 9(a − 1)2 (4a − 4)6 which is false for a > 1. The proof is complete. ✷ The following result will be used in the proof of a gap principle in a manner which is similar to Lemma 11. Lemma 12. qj2 log α < q1 k2 log β. Proof. By Lemma 7 and relations (13)–(14) we have   b−1 β2 (k3 − 1) log β 2 − (j3 − 1) log α 2 = log − Λ3 − log 2 a−1 α > 1 1 1 − Λ3 > − > 0, 9(a − 1)2 9(a − 1)2 (4a − 4)6 so that (j3 − 1) log α < (k3 − 1) log β. Hence, (qj2 + σ − 1) log α < (q1 k2 + σ − 1) log β, or qj2 log α < q1 k2 log β + (σ − 1) log(β/α). Since σ is at most 1 and β > α, the result follows. ✷ Corollary 13. aq 2 < bq12 . Proof. We argue by contradiction. Supposing that aq 2  bq12 , then  q log β b k2 log β  < . < a q1 j2 log α log α M. Cipu, M. Mignotte / Journal of Number Theory 125 (2007) 356–392 367 √ √ √ It√is straightforward to establish that the mapping t → log( t + t − 1 )/ t, defined for t  2, is increasing for t  t0 ≃ 3.276 and decreasing for t  t0 . Thus, for a > 2 or a = 2 and b > 6, from the assumption aq 2  bq12 one gets the contradiction a > b. For a = 2 < b  6 one sees that the inequalities (12) are not satisfied. ✷ We can now prove the two main results of this section. Proposition 14. Assume k2 = 3. Then for all β (respectively for β  1000) one has j3 > 1.7j2 β 2/3 j3 > 1.7j2 β 4/5 j3 > 1.7j2 β 6/7   respectively j3 > 1.98j2 β 2/3   respectively j3 > 1.90j2 β 4/5   respectively j3 > 1.83j2 β 6/7 for j2 = 7, for j2 = 11, for j2  15. Proof. We first examine the situation when k3 = 3q1 . Then j3 = qj2 , for some odd integers q and q1 . From ax22 ≡ by22 ≡ 1 (mod z22 ) and  (q−1)/2  (q −1)/2 z3 ≡ q ax22 ≡ q1 by22 1   mod z22 , one gets q ≡ q1 (mod z22 ). Since z2 β −2 > β 2 and q > q1 (see Lemma 11), one obtains q > z22 > β 4 , which yields j3 > j2 β 4 , a much stronger result than the desired one. Suppose now that k3 = 3q1 + σ , j3 = qj2 + σ , with σ ∈ {−1, 1} and even q, q1 . This time one has z3 ≡ aqx2 z2 + σ ≡ bq1 y2 z2 + σ whence bq1 y2 ≡ aqx2   mod z22 , (mod z2 ). From z2 = 4b − 1, j2  7 and ax22 ≡ by22 ≡ 1 (mod z22 ) one infers 4aq 2 ≡ 4bq12 ≡ q12 (mod z2 ), so that q2  1 + Uj 2 . 4a We shall give all the details for the rest of the proof in the case j2 = 7. The same idea, with obvious modifications, works for the other cases. Since U7 = 64a 3 − 80a 2 + 24a − 1, one has q 2  16a 2 − 20a + 6. We shall determine a positive K such that 16a 2 − 20a + 6  K 2 β 4/3 . It is sufficient to have K6  2(8a 2 − 10a + 3)3 (16a 2 − 20a + 6)3 = . 4(2b − 1)2 (32a 3 − 40a 2 + 12a − 1)2 368 M. Cipu, M. Mignotte / Journal of Number Theory 125 (2007) 356–392 Denote by f (a) the rational function in the right side of the last relation, and note that β  1000 implies a  26. As f is increasing for a  2, it is sufficient to have K 6  f (2) (respectively K 6  f (26)). We verify that K = 0.883 (respectively K = 0.991) works. Indeed, for this value of K we have q > 0.883β 2/3 and j3  2qj2 − 1 > 1.766j2 β 2/3 − 1 > 1.7j2 β 2/3 . ✷ The gap principle for higher values of k2 is much stronger due to the presence of a greater exponent of β. Proposition 15. If k2  5, then j3 > 3.96j2 β (k2 −3)/2 . Proof. Recall that j3 = qj2 + σ and k3 = q1 k2 + σ for certain integers q > q1  2 and σ ∈ {−1, 0, 1} such that qσ and q1 σ are even. The details of the proof are slightly different in the three cases corresponding to the value of σ , so we will discuss them separately. To begin with, let us consider σ = 0. Then z3 = Uqj2 = t0   √ √ q (x2 a )q−2t−1 z22t (a − 1)t ≡ q(x2 a )q−1 2t + 1   mod z22 . In a similar manner, one obtains √ z3 ≡ q1 (y2 b )q1 −1 so that   mod z22 , √ √ q(x2 a )q−1 ≡ q1 (y2 b )q1 −1   mod z22 . Since ax22 ≡ by22 ≡ 1 (mod z22 ), and q and q1 are odd integers, this yields q ≡ q1 (mod z22 ). In view of Lemma 11, one concludes that q > z22 > β 2(k2 −1) and therefore j3 = qj2 > j2 β 2(k2 −1) , which is stronger than the desired conclusion. If σ = 0, then both q and q1 are even. From it follows that  α 2j2 = 2z22 (a − 1) + 1 + 2x2 z2 a 2 − a z3 = Uj3 ≡ aqx2 z2 + σ With the same reasoning one obtains z3 = Uk′ 3 ≡ bq1 y2 z2 + σ Therefore bq1 y2 ≡ aqx2   mod 2z22 .   mod 2z22 . (mod 2z2 ) (21) M. Cipu, M. Mignotte / Journal of Number Theory 125 (2007) 356–392 369 and  q1 b 2 2  2 q ≡a 2 (mod z2 ). As aq 2 < bq12 and q > q1 (see Corollary 13 and Lemma 11), one obtains whence   bq 2  4z2 + 16a + 12b > 4 β k2 −1 + β k2 −3 + β k2 −5 + 12b, 2 q > 4β k2 −3   1 4− + 12. b Since b  60 for each system (8) with at least three positive solutions, one has j3 > 3.96j2 β (k2 −3)/2 . ✷ 4. Linear forms in logarithms The linear form in three logarithms (13) naturally attached to a term common to two secondorder linear recurrent sequences made its appearance in Section 2. In the previous sections we made use of such linear forms in the proofs of several theoretical results. From now on, these forms are employed especially with a view towards explicit computations. It was easy to bound from above the form in three logarithms Λ (cf. (15)). It is much harder to obtain good lower bounds. First we use the special case of three logarithms of a theorem of E.M. Matveev [16]; thus we quote his result. This theorem enables us to get a rough bound on the coefficients a and b for which the system of Eqs. (8) has three solutions. This bound will be subject to further improvements by using suitable results on linear forms in two logarithms. As will be shown, dealing with a linear form in two more complicated logarithms will prove to be a superior approach rather than using a linear form in three logarithms. Below h(γ ) denotes the absolute logarithmic height of the algebraic number γ . Recall that if the minimal polynomial of γ has degree d, leading coefficient a and complex roots γ1 , γ2 , . . . , γd , then h(γ ) = 1 log |a| + d d j =1   max 0, log |γj | . Theorem A. [16] Let λ1 , λ2 , λ3 be Q-linearly independent logarithms of non-zero algebraic numbers and let b1 , b2 , b3 be rational integers with b1 = 0. Define αj = exp(λj ) for j = 1, 2, 3 and Λ = b1 λ1 + b2 λ2 + b3 λ3 . Let D be the degree of the number field Q(α1 , α2 , α3 ) over Q. Put  χ = R(α1 , α2 , α3 ) : R . 370 M. Cipu, M. Mignotte / Journal of Number Theory 125 (2007) 356–392 Let A1 , A2 , A3 be positive real numbers, which satisfy Assume that Define also   Aj  max Dh(αj ), |λj |, 0.16 (1  j  3).   B  max |bj |Aj /A1 : 1  j  3 .  χ   3e  5 × 165 3 e (7 + 2χ) 20.2 + log 35.5 D 2 log(eD) . C1 = 6χ 2 Then   log |Λ| > −C1 D 2 A1 A2 A3 log 1.5eDB log(eD) . We first verify that the hypothesis of linear independence is fulfilled in the case of multiple solutions, the only case of interest for us. Lemma 16. If α, β and (b − 1)/(a − 1) are multiplicatively dependent, then the system of Eqs. (8) has at most one solution in positive integers. Proof. Suppose, by way of contradiction, that the system (8) has at least one solution given by relation (11) with j > 1, k > 1. Since α and β are algebraic units, while (b − 1)/(a − 1) is not, it follows that α and β are multiplicatively dependent. Then there exists a quadratic unit γ > 1 such that α 2 and β 2 are powers of γ . Therefore,     b−1 b−1 2 2 + j log α − k log β = log − t log γ Λ = log a−1 a−1 for some positive integer t. As b − 1 − (a − 1)γ t is a non-zero algebraic integer, (b − 1 − (a − 1)γ t )(b − 1 − (a − 1)γ −t ) is a positive rational integer. Hence, 3 , 4(b − 1) b−1 3 , −1 (a − 1)γ t 4(b − 1)2 b − 1 − (a − 1)γ t  and Λ> 2 . 3(b − 1)2 Comparison with the upper bound for Λ given in (14) results in   β 4 > 16(b − 1)2 > 16 1 − α −2 α 2j . 371 M. Cipu, M. Mignotte / Journal of Number Theory 125 (2007) 356–392 Recall that, according to (16), one has (k − 1) log β 2 < (j − 1) log α 2 + 0.5. Hence,    (j − 1) log α 2 + 0.5 > j log α 2 + log 4 1 − α −2 , wherefrom we obtain the contradiction log α 2 < 0.5 − log(4(1 − α −2 )) < 0. ✷ Let (x, y, z) be a solution in positive integers for the simultaneous Pell equations (8). As explained above, if z > 1, then z is given by Eq. (10) for suitable odd integers j > k > 1. We apply Matveev’s result for the linear form   b−1 + 2j3 log α − 2k3 log β, Λ = log a−1 which we rewrite as    2 b−1 Λ = log + r log α 2 − k3 log β/α h , a−1 where j3 = hk3 + r, We have D = 4 and   b−1 h = log(b − 1), a−1 0  r < k3 .   h α 2 = log α,  2  h β/α h = log β. A2 = D log α, A3 = D log β, Therefore we can take A1 = 2D log(β/2), and  B = max 1,  k3 log β k3 log β r log α = , . 2 log(β/2) 2 log(β/2) 2 log(β/2) With these choices it follows C1 < 1.925 × 1010 , so that Matveev’s theorem yields the inequality   k3 log β . log Λ > −3942.4 × 10 log α log β log(β/2) log 19.46 log(β/2) 10 372 M. Cipu, M. Mignotte / Journal of Number Theory 125 (2007) 356–392 For reasons that will be explained in Section 5 (see Lemma 19), we may suppose that b  2000. Under this assumption we get 19.46 k3 log β < 24j3 . log(β/2) Therefore, by (15) we have j3 < 1.9712 × 1013 log β log(β/2) log(24j3 ). (22) This relation, together with Propositions 14–15, and the inequalities √ √ 2 b − 1 < β < 2 b, (23) yield upper bound for integers b for which the system of simultaneous Pell equations (6) can have (at least) three positive solutions: k2 = 3: k2 = 5: b < 3.1 × 1051 b < 3.1 × 1032 , for j2 = 7, k2  7: b < 1.1 × 1042 for j2  11, b < 1.6 × 1015 . The admissible range for b shrinks very fast as k2 increases. However, for k2 at most 5, the domain search is far too big to be exhausted by a brute-force computer search. In order to place ourselves in a reasonably sized search domain, we employ the idea of using a linear form in two logarithms, for which we use the main result of [14]. This approach works quite well in the present circumstances. For later reference, we state the result in the form that it will be used. Theorem B. [14] Consider the linear form Λ = b2 log α2 − b1 log α1 , with b1 and b2 positive integers. Let K  3, L  2, R1 , R2 , S1 , S2 be positive integers and ρ a real number greater than 1. Put R = R1 + R2 − 1, S = S1 + S2 − 1, N = KL, 1 N , g= − 4 12RS ((R − 1)b2 + (S − 1)b1 ) c= 2 K−1  −2/(K 2 −K) k! . k=1 Let a1 , a2 be positive real numbers such that ai  ρ|log αi | − log |αi | + 2Dh(αi ), for i = 1, 2. Suppose that:   Card α1r α2s ; 0  r < R1 , 0  s < S1  L, Card{rb2 + sb1 ; 0  r < R2 , 0  s < S2 } > (K − 1)L, (I) (II) M. Cipu, M. Mignotte / Journal of Number Theory 125 (2007) 356–392 373 K(L − 1) log ρ − (D + 1) log N − D(K − 1) log c − gL(Ra1 + Sa2 ) > 0. (III) and Then,   LSeLS|Λ|/(2b2 ) LReLR|Λ|/(2b1 ) . where Λ′ = Λ · max , 2b2 2b1 |Λ′ |  ρ −KL+(1/2) , In the case when the number α1 is not a root of unity we shall deduce the following result from Theorem B, which is a variant of Théorème 2 of [14], similar to Theorem 1.5 of [18]. Proposition 17. Consider the linear form Λ = b2 log α2 − b1 log α1 , where b1 and b2 are positive integers. Suppose that α1 is not a root of unity. Put   D = Q(α1 , α2 ) : Q R(α1 , α2 ) : R . Let a1 , a2 , h, t be real positive numbers, and ρ be a real number greater than 1. Put λ = log ρ and suppose that       b1 b2 + log λ + f (K) + ε , h  max 1, 1.5λ, D log + a2 a1   ai  max 4, 2.7λ, ρ|log αi | − log |αi | + 2Dh(αi ) (i = 1, 2), ε = 0.0262, a1 a2  20λ2 , where f (x) = log (1 + √ √ x log x 3 3 log x−1 x − 1) x + + + log + x −1 6x(x − 1) 2 4 x −1 and L = 2 + ⌊2h/λ⌋  5, K = 1 + ⌊tLa1 a2 ⌋. Then we have the lower bound √     log |Λ|  −λtL2 a1 a2 − max λ(L − 0.5) + log L2 (1 + t )a2 , D log 2 , √ provided that t satisfies t  2.2λ−2 and tΩ − L t − W  0, with Ω = 3(L − 1)λ − 3h, W=   1 3 L . + 4 a2 a1 374 M. Cipu, M. Mignotte / Journal of Number Theory 125 (2007) 356–392 We postpone the proof of the above result to the last section of the paper. Now we apply Proposition 17 to the system ax 2 − (a − 1)z2 = by 2 − (b − 1)z2 = 1 with three positive solutions, the second of which is obtained for k2 = 3 and j2 = 7 or 11. In the present study we have two linear forms   b−1 Λi = log + ji log α 2 − ki log β 2 , a−1 i = 2, 3. As a consequence of Lemma 7 we have  log 1 + 1 8(a − 1)2      β b−1 β 1 + 2 log < log + 2 log , < log 1 + α a−1 2a − 2 α so that (cf. the proof of Lemma 11) 1 1 < Λ2 − (j2 − 1) log α 2 + (k2 − 1) log β 2 < . a 9a 2 Define the integer l by k2 − 1 j2 − 1 j2 − k2 j3 = k3 + l − . 2 2 2 Having in view that the term in the left side and the first term in the right side are odd, while the last term is even by Lemma 8, it follows that l is even. We apply the estimate for the linear form in two logarithms Λ := b2 log α2 − b1 log α1 , with α1 = β k2 −1 , α j2 −1 α2 =  b−1 a−1 (k2 −1)/2 α 2l−j2 +k2 . Here D = 4 and b1 = k3 , b2 = 1. Using inequality (16), one obtains h(α1 ) = j2 − 1 1 j2 − 1 k2 − 1 log β  log α + < log(4a). 2 2 4a 4 We claim that 0 < log α1  Indeed, by Lemma 7 and Eqs. (13)–(14) we have 1 . 4(a − 1) M. Cipu, M. Mignotte / Journal of Number Theory 125 (2007) 356–392 2 log α1 = (k2 − 1) log β 2 − (j2 − 1) log α 2 = log >  375  β2 b−1 − Λ2 − log 2 a−1 α 1 1 1 − Λ2 > − > 0, 2 2 9(a − 1) 9(a − 1) (4a − 4)6 and   β2 1 b−1 2 log α1 < log − log 2 < , a−1 2(a − 1) α as claimed. The next immediate concern is to show that l is non-negative. The equality (k2 − 1)Λ3 = log β k2 −1 (b − 1)(k2 −1)/2 2 + l log α − k log 3 α j2 −1 (a − 1)(k2 −1)/2 α j2 −k2 is equivalent to log α2 = (k2 − 1)Λ3 + k3 log α1 . Both terms in the right side are positive. Therefore, α2 > 1. Further, Lemma 7 and inequality (16) imply   (k2 − 1) b−1 − log α (j2 −k2 ) log 2 a−1 < (k2 − 1) log k2 − 1 β + − log α (j2 −k2 ) α 4(a − 1) = (k2 − 1) log β − (j2 − 1) log α + k2 k2 − 1 < , 4(a − 1) 4(a − 1) from which it follows that 0 < log α2 < l log α 2 + k2 . 4(a − 1) Since k2 = 3 and a  2, one concludes that l is non-negative. In fact, it is positive. We prove this claim par reduction to absurd. When l = 0, we have (k2 − 1)(j3 − 1) = (j2 − 1)(k3 − 1), and therefore Λ2 − Λ3 = (j2 − j3 ) log α 2 − (k2 − k3 ) log β 2 = (k3 − k2 ) log By relations (19) and (20), one obtains 2 log   2  β b−1 β k2 −1 − Λ − log = log 2 a−1 α j2 −1 α2 > 1 1 7 − > , 2 6 9(a − 1) (4a − 4) 64(a − 1)2 β k2 −1 . α j2 −1 376 M. Cipu, M. Mignotte / Journal of Number Theory 125 (2007) 356–392 whence the contradiction 1 7(k3 − 3) > Λ2 > Λ2 − Λ3 > . 6 (4a − 4) 128(a − 1)2 Thus l  2. In fact, for j2 = 11 one has l  4. Indeed, for this value of j2 , l is given by j3 = 5k3 + l − 4. From Lemma 8 we know that j3 ≡ k3 (mod 4), so that l is multiple of 4 and our claim follows. Note that the inequalities l  2 for j2 = 7 and l  4 for j2 = 11 are equivalent to 2l − j2 + k2  0. Now we proceed to bound from above the logarithm and the height of α2 for a  1000. Lemma 7 yields   b−1 log α2 = log + (2l − j2 + k2 ) log α a−1 < 2 log β 1 + + (2l − j2 + k2 ) log α. α 2(a − 1) We also obtain h(α2 ) < 2l − j2 + k2 k2 − 1 log(b − 1) + log(4a)  2 4   l + 3 log(4a) 2 and 0 < log α2  (l + 1) log(4a). The hypothesis of Proposition 17 requires a1  ρ −1 + 2(j2 − 1) log(4a), 4(a − 1)   a2  (ρ − 1)(l + 1) + 4l + 24 log(4a), and we will choose a1 := 2j2 log(4a),   a2 := (ρ − 1)(l + 1) + 4l + 24 log(4a), which is admissible if ρ < 1 + 8(a − 1) log(4a). The specialization ρ = 27 yields 2 1 1 a2 = (30l + 50) log(4a), t= + + 0.003 , λ h   1 k3 + 4 log λ + 4f (K) + 0.0262. + h = 4 log (30l + 50) log(4a) 2j2 log(4a)  The estimate for linear forms in two logarithms from Proposition 17 implies M. Cipu, M. Mignotte / Journal of Number Theory 125 (2007) 356–392 √   log |Λ3 |  −λtL2 a1 a2 − λ(L − 0.5) − log L2 (1 + t )a2   h 2 a1 a2 , > −0.7 1 + λ 377 (24) whereas from (15) we know that log |Λ3 | < −j3 log α 2 + 0.9. In order to exploit these bounds for log Λ3 , we need to relate k3 to a and l. We show that for 2l = j2 − k2 and a > 1000 one has k3  1.998al log α 2 . From (12) and (16) it follows that α (j2 −1)k3 +2l−j2 +k2 < β 2k3  and β 4 < 1 +  1 α 2(j2 −1) , 2(a − 1) whence α 2(2l−j2 +k2 )/k3 < 1 + 1 . 2(a − 1) Therefore, (2l − j2 + k2 ) log α 2 < k3 . 2(a − 1) Since 2(a − 1)  1.998a and l  j2 − k2 , it follows that k3 > 2(a − 1)(2l − j2 + k2 ) log α 2  1.998al log α 2 . Therefore k3  1.997al log(4a) for a > 1000. We conclude that either a is at most 1000 or   h 2 j3 log α < 0.7 1 + a1 a2 + 0.9 λ   h 2 < 1.41 1 + j2 (30l + 50) log2 (4a), λ 2 (25) whence   4 0.998(j2 − 1)a log(4a) + 1 − log α 2 l     50 h 2 log2 (4a). < 1.41 1 + j2 30 + λ l (26) 378 M. Cipu, M. Mignotte / Journal of Number Theory 125 (2007) 356–392 Similar inequalities are valid in the case 2l = j2 − k2 not covered by the assumptions under which relation (26) has been derived. Using the two methods just described, we conclude our study of linear forms in logarithms with comparatively small upper bounds for the values of a and b, under the assumption that the system of Eqs. (8) has at least three solutions in positive integers. The bounds for a are derived by using Proposition 9. Lemma 18. Suppose that Eqs. (8) have three solutions given by (10) for j1 = 1 < j2 < j3 and k1 = 1 < k2 < k3 . Then the following statements hold: k2 = 3: a < 22 000 for j2 = 7, k2 = 3: a < 105 000 for j2 = 15, k2 = 5: k2 = 5: a < 8.9 × 1015 a < 3.5 × 107 k2 = 7: a < 40 000 for j2 = 19, a < 1.7 × 106 k2 = 11: a < 57 000, k2 = 15: a < 1800, a < 3 × 1010 a < 7.5 × 105 forj2 = 17, a < 7.4 × 108 k2 = 9: a < 23 000 for j2  19, for j2 = 9, k2 = 7: for j2 = 11, a < 1.3 × 106 b < 7.7 × 106 , k2 = 17: b < 3.4 × 1032 , b < 6.8 × 1031 , for j2  21, for j2 = 15, a < 7000, a < 600, b < 1.6 × 1015 , b < 8.6 × 1014 , a < 30 000 for j2  17, k2 = 13: b < 4 × 1038 , for j2 = 13, a < 4300 for j2  23, for j2 = 13, b < 20 000, b < 3.9 × 1023 , a < 15 000 for j2 = 11, b < 4.1 × 109 , b < 2 × 105 , b < 3000. 5. In the search of a third solution Since the ground-breaking paper of Baker and Davenport [4], techniques of computational Diophantine approximation are systematically used in solving Diophantine equations. Usually, variants of Davenport’s lemma allows one to eliminate putative large solutions. In the case at hand, such a result serves to conclude that the second solution of a system with three solutions appears soon (k2  17), but not immediately after the first solution (k2  5). For the last part of our proof of Theorem 2 we need large, computer-aided computations. Throughout this section we suppose that our system has three positive solutions. The idea underlying our approach is as follows. Matveev’s theorem for three logarithms gives an upper bound on j3 . Using the suitable gap principle (either Proposition 14 or Proposition 15), one obtains an upper bound for b. From α j3 < β k3 it follows that a is below a certain small value. Then one performs a search for a third solution. However, when k2 = 3 and j2  11, the domain where the third solution may exist is well beyond the reach of a direct search, so we are forced to modify the approach. We rewrite Λ3 as a linear form in two logarithms. Proposition 17 yields a comparatively small upper bound on a, which forces b to remain confined to a domain defined by the inequality (16). At this point, the entire search space can be fully checked. In the previous section we established that no system (6) has three positive solutions if b = max{a, b} is very large (see Lemma 18). Here we first show that the same statement is true if b is very small. Lemma 19. There is no pair of integers (a, b) with 1 < a < b  2000 for which the system of equations ax 2 − (a − 1)z2 = by 2 − (b − 1)z2 = 1 has three solutions in positive integers. M. Cipu, M. Mignotte / Journal of Number Theory 125 (2007) 356–392 379 Proof. We shall argue by contradiction. Suppose that there exists a pair (a, b) of integers such that 1 < a < b  2000, and that the system of Diophantine equations ax 2 − (a − 1)z2 = by 2 − (b − 1)z2 = 1 has three positive integer solutions. Keeping the notation used up to now, we have  b−1 + j3 log α 2 − k3 log β 2 < −2j3 log α + 0.882. Λ3 = log a−1  (27) A variant of Davenport’s lemma, due to Pethő [12], is useful for the present context. We record it for reference. Lemma C. [12] Let A, B, θ , µ be positive real numbers and M be a positive integer. Suppose that P /Q is a convergent of the continued fraction expansion of θ such that Q > 6M. Put ε = µQ − Mθ Q, where  ·  denotes the distance from the nearest integer. If ε > 0, then there exist no solutions to the inequalities 0 < j θ − k + µ < A · B −j in integers j and k with log(AQ/ε)  j  M. log B We have applied this result for Λ3 / log α 2 , choosing M = 1020 as an upper bound for j3 . The output of a code written in PARI/GP [5] is that the inequality (27) does not hold for j3 > 167. Our gap principles and the obvious inequality 4b − 3 < β 2 imply that   (a, b) ∈ (2, 60), (3, 270), (4, 728), (5, 1530) . However, none of the systems corresponding to these values of the coefficients a, b has three solutions in positive integers. ✷ Now it is easy to prove that the existence of three solutions forces the second one to appear very soon in the recurrent sequence (Uk′ )k , precisely, k2  17. Namely, from Matveev’s theorem, Proposition 15 and inequalities (23) it readily follows that b < 1000 if k2  19, so we may conclude that k2  17 by Lemma 19. We employ Lemma C to show that k2 = 3 may occur only for systems with two positive solutions. Since z2 = 4b − 1 = Uj2 with a in a range that shrinks very rapidly with increasing j2 , we can compute the continued fraction expansion of θ = log β/ log α for the relevant values of a. Reasoning as in the proof of Lemma 19, we reach the conclusion that if Eqs. (8) have three solutions in positive integers, given by relation (10) for j1 = 1 < j2 < j3 and k1 = 1 < k2 < k3 , then k2 = 3. To complete the proof of Theorem 2, we have to show that k2 cannot have an odd value between 4 and 18. The volume of explicit computations needed to confirm that k2 = 5 is not possible is largely reduced by the theoretical result below. Lemma 20. Suppose that the system of Diophantine equations (8) has three positive solutions given by (10) for j1 = 1 < j2 < j3 and k1 = 1 < k2 < k3 . If k2 = 5, then j2  21. 380 M. Cipu, M. Mignotte / Journal of Number Theory 125 (2007) 356–392 Proof. The present hypotheses yield z2 = 16b2 − 12b + 1. We verify that for j2 = 9, 13 and 17, 4z2 + 5 = (8b − 3)2 is between two consecutive perfect squares. Namely, we show that for these values of j2 there exist equalities (8b − 3)2 = A2j2 + Pj2 = (Aj2 + 1)2 − Qj2 with Aj2 , Pj2 and Qj2 certain polynomials in a with positive leading coefficient, which take integral values for integral a, and all real roots of Pj2 and Qj2 are smaller than 2. Here are the relevant polynomials: A9 = 32a 2 − 28a + 2, P9 = 48a 2 − 48a + 5, A13 = 128a 3 − 176a 2 + 59a − 3, Q9 = 16a 2 − 8a, P13 = 32a 3 − 57a 2 + 18a, Q13 = 224a 3 − 295a 2 + 100a − 5,  for a even, 512a 4 − 960a 3 + 556a 2 − 101.5a + 2 A17 = 4 3 2 512a − 960a + 556a − 101.5a + 2.5 for a odd,  for a even, 816a 4 − 1564a 3 + 913.75a 2 − 170a + 5 P17 = 304a 4 − 604a 3 + 357.75a 2 − 68.5a + 2.75 for a odd,  for a even, 208a 4 − 356a 3 + 198.25a 2 − 33a Q17 = 4 3 2 720a − 1316a + 754.25a − 134.5a + 3.25 for a odd. ✷ The rest of the proof consists of computer calculations of an other kind. Essentially, we look for solutions of the Diophantine equation Uj2 (a) = Uk2 (b), where Un (x) are Chebyshev polynomials (cf. relation (10)). For each of the seven possible values of k2 we are examining, the corresponding j2 is subject to restrictions given by Lemma 8 (and Lemma 20 if k2 = 5), while a and b are confined to domains described by Lemma 18. Moreover, an upper bound for j2 is obtained from Lemma 19 in conjunction with Lemma 9. The procedure we used is as follows. For each admissible pair of indices (k2 , j2 ) the search of solutions starts by putting a = 2 and ua := Uj2 (a). If there exists a positive integer b > a such that ua = Uk2 (b), from Lemma 9 it follows that b is either (b∗ + 2)/4 or (b∗ + 3)/4, with b∗ := ⌊ua 2/(k2 −1) ⌋. First check whether b∗ is congruent to 1 or 2 modulo 4. If this is the case, check if Uk2 (b) coincides with ua. Then increase b and compute ub := Uk2 (b). Do the same if b∗ ≡ 1, 2 (mod 4). By Lemma 9, the only possible values of a such that ub = Uj2 (a) are either (a ∗ + 2)/4 or (a ∗ + 3)/4, with a ∗ := ⌊ub2/(j2 −1) ⌋. If one of these values is integral, check whether ub and Uj2 (a) are equal. Then increase a and resume the loop until the value of a or b becomes bigger than the bounds obtained previously. The program executing the search described above found no integers a, b, k2 , j2 such that Uj2 (a) = Uk2 (b) and 5  k2  17. The computations allow us to conclude the proof of Theorem 2. 6. Proof of Theorem 3 As indicated in the Introduction, we shall conveniently modify the strategy developed in the proof of Theorem 2. The theoretical study of properties of solutions yields conclusions quite similar to those established in Sections 2 and 3. Since here the reasonings are essentially the same as (but easier than) in the case of system (6), we shall avoid repeating many arguments. However, we shall point out the specific details of the proofs for the results needed to establish Theorem 3. The explicit computations which permit to conclude the proof will be described at length because they involve new ideas. M. Cipu, M. Mignotte / Journal of Number Theory 125 (2007) 356–392 381 Let us consider the system of Diophantine equations x 2 − ay 2 = 1, z2 − bx 2 = 1, a, b  2, having a solution in positive integers. By a reasoning similar to the proof of Lemma 5 one sees that the smallest solution has the property that each component divides the corresponding component of any other solution of (7). Hence, this system has the same number of positive solutions as the system   x 2 − m2 − 1 y 2 = 1, z2 − bx 2 = 1, m, b  2. (28) Let (m, 1, n) be the smallest solution in positive integers for this system, and denote α := m +  m2 − 1, √ β := n + m b. (29) Then any positive solution (x, y, z) of Eqs. (28) satisfies x= α j + α −j β k − β −k = √ 2 2 b (30) for some positive odd integers j and k. Since α, α −1 , β and β −1 are positive, it follows that √ β k > α j b. (31) Hence, if j > 1 and k > 1, then j > k. From inequality (31), it also follows that the linear form in logarithms √ Λ := k log β − j log α − log b (32) is positive, while Eq. (30) implies     β 2−2k < 1.01α −2j . Λ < log 1 + α −2j − log 1 − β −2j < α −2j + 2 β −1 Hence, log Λ < −2j log α + 0.01. (33) Despite its innocuous appearance, the forthcoming result allows for a significant reduction on the amount of computation needed to complete the proof of Theorem 3. Lemma 21. Assume (x, y, z) is a positive solution of the Diophantine equations (28) for which relation (30) holds with j , k > 1. Then k ≡ 1 (mod 4) and j + k  4m2 . Moreover, if j ≡ 1 (mod 4), then j  4m2 + k  4m2 + 5. 382 M. Cipu, M. Mignotte / Journal of Number Theory 125 (2007) 356–392 Proof. It is easy to see that β k − β −k ≡k β − β −1   mod 4m2 , α j + α −j ≡ (−1)(j −1)/2 j α + α −1   mod 4m2 , and therefore k ≡ (−1)(j −1)/2 j (mod 4m2 ). For j ≡ 1 (mod 4) one gets (−1)(j −1)/2 j = j and j ≡ k (mod 4m2 ), so that k ≡ 1 (mod 4) and j  4m2 + k  4m2 + 5. For j ≡ −1 (mod 4) one has (−1)(j −1)/2 j ≡ 1 (mod 4) and k ≡ −j (mod 4m2 ), whence 2j > j + k  4m2 . ✷ Suppose that the system (28) has three positive solutions (xi , yi , zi ) (i = 1, 2, 3). Then xi = β ki − β −ki α ji + α −ji = √ 2 2 b (34) for some positive odd integers ji and ki (i = 1, 2, 3) with 1 = j1 < j2 < j3 and 1 = k1 < k2 < k3 . By a reasoning similar to that employed in Section 3 we get that either j2 divides j3 and k2 divides k3 , or else j3 = 2qj2 + σ and k3 = 2q1 k2 + σ for some positive integers q > q1 , and σ either 1 or −1. Here is the gap principle for the system of Eqs. (28). Lemma 22. Suppose that the system of simultaneous Pell equations (28) has three solutions (xi , yi , zi ) (i = 1, 2, 3) in positive integers, given by relation (34) for some odd integers ji and ki (i = 1, 2, 3) with 1 = j1 < j2 < j3 and 1 = k1 < k2 < k3 . Then j3 > τj2 β (k2 −3)/2 , with τ = 3.93 for m = 2, τ = 4.87 for m = 3, and τ = 5.63 for m  4. Proof. Let us first examine the case j2 divides j3 . Then x3 is multiple of x2 and there exist positive odd integers q  3 and q1  3 such that j3 = qj2 and k3 = q1 k2 . As in the proof of Lemma 21 one finds x3 ≡ q1 x2   mod 4x22 , x3 ≡ (−1)(q−1)/2 q x2 so that q1 ≡ (−1)(q−1)/2 q (mod 4x22 ) and q > 2x22 . Since x2 =   mod 4x22 ,  (β k2 − β −k2 )m  4 > β + β 2 + 1 β k2 −5 m > 2β k2 −1 , −1 β −β one gets j3 > 16j2 β 2k2 −2 , a much stronger inequality that the desired one. In the case j3 = 2qj2 + σ , k3 = 2q1 k2 + σ , with σ = ±1, one finds on the one hand       mod 2x22 , x3 ≡ (−1)q m − 2σ m2 − 1 qx2 y2 (35) M. Cipu, M. Mignotte / Journal of Number Theory 125 (2007) 356–392 383 while on the other hand,   mod 2x22 . x3 ≡ σ m + 2nq1 x2 z2 As 2m < x2 , it follows that σ = (−1)q , and   nq1 z2 ≡ −q m2 − 1 y2 (mod x2 ). This, along with the congruences z22 ≡ 1 (mod x22 ), (m2 − 1)y22 ≡ −1 (mod x22 ), imply that   n2 q12 ≡ − m2 − 1 q 2 (mod x2 ). Thus, by inequality (35), one has     2 max n2 q12 , m2 q 2 > x2 > β 4 + β 2 + 1 β k2 −5 m. For mq  nq1 one gets √ (β 4 + β 2 + 1)β k2 −5 (β 4 + β 2 + 1)β k2 −5 2  q > . 2m β2 − 1 2 Since the rightmost term in this chain of inequalities is greater than 8β k2 −3 , one deduces that √ j3  2qj2 − 1 > 4 2j2 β (k2 −3)/2 > 5.63j2 β (k2 −3)/2 . It remains to examine the possibility that mq < nq1 . By direct computation one finds q12 > m(β 4 + β 2 + 1)β k2 −5 2m(β 4 + β 2 + 1)β k2 −3 = > γβ k2 −3 , 2n2 (β 2 + 1)2 with γ = 3.88 for m = 2, γ = 5.94 for m = 3, and γ = 7.94 for m  4. This implies that j3  2qj2 − 1  2(q1 + 2)j2 − 1 > τj2 β (k2 −3)/2 , with τ as specified in the statement of the lemma. ✷ The second phase of the proof of Theorem 3 consists of applying Matveev’s theorem for the linear form in logarithms √ Λ := k3 log β − j3 log α − log b, √ with the choice α1 := β, α2 := α, α3 := b, b1 := k3 , b2 := −j3 , b3 := −1. If these algebraic numbers are multiplicatively dependent, then the system of Eqs. (28) has at most one common solution in positive integers (cf. Lemma 16). In the notation introduced in the statement of Theorem A we have D = 4, χ = 1, h(β) = 1 log β, 2 h(α) = 1 log α, 2 √ h( b ) = log b, 384 M. Cipu, M. Mignotte / Journal of Number Theory 125 (2007) 356–392 Table 1 Necessary condition in order that Eqs. (28) have three positive solutions m= log β < log j3 < j2  j2  n< 5 2 3 4 5 36.40 35.08 34.22 33.72 40.25 40.09 40.02 40.00 11 31 59 95 125 97 81 73 3.3 × 1015 8.6 × 1014 3.7 × 1014 2.2 × 1014 9 2 3 4 11.31 10.91 10.58 37.67 37.57 37.45 11 27 55 71 55 45 41 000 28 000 20 000 13 2 3 4 6.48 6.33 6.10 36.42 36.29 36.14 15 23 51 59 45 37 350 300 230 k2 = and therefore one may choose √ A1 := 2 log β, A2 := 2 log α, A3 := 4 log b. Since one has k3 log β > max{j3 log α, log b } (by inequality (31)) and k3  11 (by Lemma 21), the choice B := k3 is admissible. As C1 < 1.925 × 1010 , the conclusion of Matveev’s theorem is log Λ > −4.928 × 1012 log α log β log b log(38.92k3 ). Comparison of this inequality with relation (33) written for the current Λ results in j3 < 2.464 × 1012 log β log b log(38.92k3 ). √ √ Recall that j3 > k3 and β > 2m b  4 b, so that   β log(38.92j3 ). j3 < 4.928 × 10 log β log 2m 12 (36) Table 1 gives information on numerical characteristics associated to putative solutions of a Diophantine system of the type (28) with three positive solutions. Data referring to log β and log j3 are derived by combining inequality (36) with Lemma 22. The lower bounds on j2 and the upper bounds on m result from the information contained in the previous columns of Table 1 and Lemma 21. Once β is bounded from above, one may use relation (31) to obtain upper bound on j2 . The estimation for n is obtained with the help of the inequality 2n < β + 1. From β < 2n < β + 1 and log β < 6.5, it follows that n < 350. Thus, for k2  17 one has n < 350. With the help of Lemma C, we find that for n  1000 one has j3 < 46 for any system (28) with √ three positive solutions. Then Lemma 22 implies that β < 1.1, in contradiction with β > 4 2. At this point of the proof we know that, if the system (28) has at least three solutions in positive integers, then either k2 = 5 and m = 2, 3, 4, or k2 = 9 and m = 2, 3. If k2 = 9, then we run again our program implementing Davenport’s lemma, this time for m = 2, n odd and n < 41 000, m = 3, n ≡ ±1 (mod 9) and n < 28 000. M. Cipu, M. Mignotte / Journal of Number Theory 125 (2007) 356–392 385 The outcome is that the index j3 of the third solution is at most 49. This in conjunction with our gap principle rejects all candidates. To conclude the proof of Theorem 3, we still have to examine the possibility that k2 = 5. Let us denote Tt (m) := 4(α t + α −t ) +5 α + α −1 for positive odd t. Since   β 5 − β −5 = m 16n4 − 12n2 + 1 , √ 2 b Eq. (30) written for j2 and k2 is equivalent to  2 Tj2 (m) = 8n2 − 3 . (37) Thus, the proof of Theorem 3 is complete as soon as we show that none of the numbers Tt (m) (t ≡ 1 (mod 4), m = 2, 3, 4) is a perfect square. The values Tt (m) (t odd) are given by the recurrence relation    Tt+2 (m) = 4m2 − 2 Tt (m) − 5 − Tt−2 (m) + 10 with the initial terms T1 (m) = 1, T3 (m) = 2m2 − 3. For j2 an odd integer in the range specified in Table 1 we examine for each m = 2, 3, 4 the values Tj2 (m) modulo several primes p. We remove from the list of admissible j2 those values for which the Legendre symbol of Tj2 (m) with respect to p is equal to −1. After a few changes of the prime considered, we are left with the empty list, which means none of the numbers Tt (m) (t ≡ 1 (mod 4), m = 2, 3, 4) is a perfect square. 7. Proof of Proposition 17 In order to prove Proposition 17, we will compute lower and upper bounds for the parameter t. Put ∆ = L2 + 4ΩW , the condition on t implies t  t0 , where  √ 2 2 L L2 L W 4W W L2 L + ∆ + 2L ∆ +  2+ , = + + t0 = 2 2 2 Ω 2Ω Ω Ω Ω 4Ω 2Ω Ω with λ−1 (2h + λ) L 1 L 1 8   = 9λ 3 (2h + λ) − (h + λ) Ω 3 λL − (h + λ)  λ−1 2(h + λ) 2 1 = , 3 2(h + λ) − (h + λ) 3λ since ∂(L/Ω)/∂L < 0 and 1 + 2h/λ  L  2(1 + h/λ) when h  1.5λ. Moreover, W satisfies   W 1 L 1 1 1 1 =  + , + Ω 4 a2 a1 λL − λ − h 4a1 (λL − λ − h) 2a2 λ 386 M. Cipu, M. Mignotte / Journal of Number Theory 125 (2007) 356–392 and also     1 1 1 1 W 1 1 + 2h/λ 1 1 + +  = + Ω 4 a2 a1 h 2a2 λ a1 a2 4h  1 + 1 , if λ  1,  2λ2 12λ 2 + 16.2λ2 , if λ  1, 2.7λ2 because of our hypotheses on a1 , a2 , and h. Thus we always have 7 W  . Ω 8.1λ2 It is easy to check that the previous inequalities imply √ t0  1.48 . λ Hence, t0 < 2.2λ−2 and we can always choose t satisfying 4 2.2 t  2 . 2 9λ λ Then we have tLa1 a2    1 L 4 a2 + a1 La1 a2 , + 9λ2 4(λL − λ − h) so that tLa1 a2  4a1 a2 L a1 L a2 + = ψ(L), + 2λ 2λ 9λ2 say. Clearly, ψ increases with L and, using the fact that a1 a2  20λ2 , it is easy to check that ψ(5) > 54. We suppose that α1 is not a root of unity, and we apply Theorem B with a suitable choice of the parameters. The proof follows the proof of Théorème 2 of [14]. For the convenience of the reader we keep the numbering of the formulas as in [14], except that when there is some change in the formula labelled (i) the new formula is denoted by (7.i)′ . Put L = 2 + ⌊2h/λ⌋, K = 1 + ⌊tLa1 a2 ⌋, then L  5 and K  55, and R1 = L, S1 = 1, By Liouville inequality, √ R2 = 1 + ⌊ tLa2 ⌋, √ S2 = 1 + ⌊ tLa1 ⌋. (7.1)′ M. Cipu, M. Mignotte / Journal of Number Theory 125 (2007) 356–392 387 1 log |Λ|  −D log 2 − Db1 h(α1 ) − Db2 h(α2 )  −D log 2 − (b1 a1 + b2 a2 ) 2 1 ′ = −D log 2 − b a1 a2 , 2 where b′ = b1 b2 + . a2 a1 We consider separately the two possible cases: b′  2λtL2 , or b′ > 2λtL2 . In the first case, Liouville inequality implies log |Λ|  −D log 2 − λtL2 a1 a2 and Proposition 17 holds. Remark 23. The condition b′ > 2tλL2 implies         3 3 λL 2h > 8.626,   2 log 2tλ2 L2 + f (K)  2 log 8L2 /9 + + log D D 2 4 using L  5 and the above estimates on t and ψ. Suppose for the moment that condition (III) from Theorem B is satisfied. We show that under this assumption Proposition 17 holds. Theorem B implies log |Λ′ |  −KLλ + λ/2, where   LSeLS|Λ|/(2b2 ) LReLR|Λ|/(2b1 ) . Λ′ = Λ · max , 2b2 2b1 Notice that R = R1 + R2 − 1  L + √ √ tLa2 and S = S1 + S2 − 1  1 + tLa1 . This shows that √   max{LR, LS}  L2 (1 + a2 t ) < L2 1 + 1.5λ−1 a2   a 1 a 2 L2 1.5 2 1 a2 < . =L + a2 λ 2λ As we may, suppose that log |Λ|  −λtL2 a1 a2 . From L  5, a1  4 and a2  2.7λ we get L2 a1 a2 /λ  270, so that √   (1 + a2 t )L2 |Λ| L2 a1 a2 −4L2 a1 a2 /(9λ) LR|Λ| LS|Λ|   e , , max 2b2 2b1 2 4λ 388 M. Cipu, M. Mignotte / Journal of Number Theory 125 (2007) 356–392 because (4/9)λ−2  t  2.2λ−2 . Therefore we have   LR|Λ| LS|Λ| < 10−10 . , max 2b2 2b1 √ Thus, |Λ′ |  |Λ| × L2 (1 + a2 t ), which implies √   log |Λ|  −λtL2 a1 a2 − λ(L − 0.5) − log L2 (1 + a2 t ) and Proposition 17 follows. Now we verify that condition (III) is indeed satisfied. We have to prove that Φ0 = K(L − 1) log ρ − (D + 1) log N − D(K − 1) log b − gL(Ra1 + Sa2 ) is positive when b′ > 2λtL2 . Notice that the condition b′ > 2λtL2 implies    2     3 3 8L + + log > 4.313D. h  D log 2λ2 tL2 + f (K)  D log 9 2 4 We replace this condition by the two conditions Φ > 0, Θ > 0, where Φ0  Φ + Θ. The term Φ is the main one, Θ is a sum of residual terms. As indicated in [14], the condition Φ > 0 leads to the choice of the parameters in (7.1)′ , whereas Θ > 0 is a secondary condition, which leads to assume some technical hypotheses on h, a1 , and a2 . As in [14, Lemme 8] we get √   log(2πK/ e ) b1 b2 + log λ − + f (K) + log b  log a2 a1 K −1 √ ε log(2πK/ e ) h ,  − − D D K −1 (7.17)′ which follows from the condition h  D(log b′ + log λ + f (K)) + ε. Here we have   KL 1 L(Ra1 + Sa2 ) − gL(Ra1 + Sa2 )  4 12RS   L(Ra1 + Sa2 ) KL2 a1 a2 , − + = 4 12 S R which implies √ L KL (a1 L + a2 + 2a1 a2 L t ) − √ 4 6 t √ 2 tL a1 a2 L .  (a1 L + a2 ) + 4 3 gL(Ra1 + Sa2 )  (7.18)′ Put Φ = K(L − 1)λ − Kh − √ 2 tL a1 a2 L(a1 L + a2 ) − 3 4 (7.21)′ 389 M. Cipu, M. Mignotte / Journal of Number Theory 125 (2007) 356–392 and  √  Θ = ε(K − 1) + h − D log L e/(2π) − log(KL). (7.22)′ By (7.17)′ and (7.18)′ we see that Φ0  Φ + Θ, where tLa1 a2 < K  1 + tLa1 a2 , hence √ Φ > tΩ − L t − W, La1 a2 where Ω = 3(L − 1)λ − 3h, W=   1 3 L . + 4 a2 a1 √ This proves that Φ > 0 provided that tΩ − L t − W  0. We have    √  Θ  h − log tL2 a1 a2 − D log L e/(2π) + ε(tLa1 a2 − 1). To prove that Θ  0, rewrite (7.22)′ as Θ = Θ0 (D − 1) + Θ1 , where   2π ′ Θ0 = log(λb ) + f (K) − log L + log √ , e and  2π Θ1 = εK − log K − 2 log L + log √ + log(λb′ ) + f (K). e  We conclude by proving that Θ0 and Θ1 are both positive. Since b′ > 2tλL2 and t  4/(9λ2 ), we have   log(λb′ ) > log 2tλ2 L2 > log(8/9) + 2 log L, and this implies that √ 3 2π 8L 3 + + log + log √ > 0. Θ0 > log(8L/9) + f (K) + log(2π/ e ) > log 9 2 4 e This also implies that Θ1  εK − log K + log 2π 8 + log √ + f (K). 9 e Thus,   16π Θ1  0.0262K − log K + log √ + f (K) 9 e and an elementary numerical verification shows that Θ1 is positive for K  55, which holds as we previously saw. 390 M. Cipu, M. Mignotte / Journal of Number Theory 125 (2007) 356–392 Remark 24. We have proved that, under the hypotheses of our result, we can choose ε = 0.0262. More generally the condition on ε is  16π εK − log K + log √ + f (K)  0 9 e  for all K  K0 , where K0 = ⌈t0 La1 a2 ⌉. We apply the Corollary of Theorem 2 of [18]: Proposition 25. Consider the linear form in two logarithms Λ = b2 log α2 − b1 log α1 , where b1 and b2 are positive integers. Suppose that α1 and α2 are multiplicatively independent. Put   D = Q(α1 , α2 ) : Q R(α1 , α2 ) : R . Let a1 , a2 , h, t be real positive numbers, and ρ be a real number, e3/2  ρ  e3 . Put λ = log ρ, χ = h/λ, v = 4χ + 4 + 1/χ , A = max{a1 , a2 } and suppose that χ  χ0 for some number χ0  0 and that       b1 b 2 h  max 7.5, 3λ, D log + log λ + 1.285 + 0.023 , + a2 a1   ai  max 4, λ, ρ|log αi | − log |αi | + 2Dh(αi ) (i = 1, 2), a1 a2  100. Then we have the lower bound log |Λ|  −(C0 + c1 + c2 )(λ + h)2 a1 a2 , where 1 C0 = 3 λ  2+ 1 2χ(χ + 1)  1 + 3  √   2 32 2(1 + χ)3/2 1 4λ 1 1 + + + √ 9 3v a1 a2 3v 2 a1 a2 and c1 = λ(1.5λ + 2h) , (λ + h)2 a1 a2 c2 = 1.11λ log(A(2λ + 2h)2 ) . (λ + h)2 a1 a2 Proof. The only difference with Theorem 2 of [18] is the definition of the term h. Put 1 K0 := λ  √ √    4λ 2 + χ0 2 2 + 2χ0 2(1 + χ0 ) 2λ 1 1 + + + a1 a2 + √ 3 9 3 a1 a2 3 a1 a2 M. Cipu, M. Mignotte / Journal of Number Theory 125 (2007) 356–392 391 and f (x) = log (1 + √ √ x log x 3 3 log x−1 x − 1) x + + + log + . x −1 6x(x − 1) 2 4 x −1 Then the condition on h in Theorem 2 of [18] is       b1 b 2 + log λ + f ⌈K0 ⌉ + 0.023. + h  D log a2 a1 Here we can take χ0 = 3 and it is easy to check that our present hypotheses imply K0 > 195. Since f (x) < 1.285 for x  195, we get the result. ✷ We notice √ that c1 in the statement of Proposition 25 is a decreasing function of χ , so for χ  1 + 3 we have c1  1 . 2a1 a2 We also have c2  log(a1 a2 (λ + h)2 ) 1.11λ 1.11λ log(a1 a2 (λ + h)2 ) = , √ √ (λ + h)2 a1 a2 ((λ + h) a1 a2 )9/5 ((λ + h) a1 a2 )1/5 whence c2  11.1λ 11.1 = < 0.177 · (a1 a2 )−9/10 √ 9/5 e((λ + h) a1 a2 ) e(1 + χ)(λ + h)4/5 (a1 a2 )9/10 (notice that the hypotheses of Proposition 25 imply χ  3 and λ + h  9). Having in view these remarks and simplifying the expression of C0 using v  16 we get a simpler estimate. Corollary 26. With the notation and hypotheses of the above proposition, we have the lower bound   log |Λ|  − C0′ + c1′ + c2′ (λ + h)2 a1 a2 , where 1 C0′ = 3 λ  1 2+ 2χ(χ + 1)  1 + 3  √   2 1 2 1 λ 1 + √ + + 9 12 a1 a2 3 a1 a2 and c1′ = 1 , 2a1 a2 c2′ = 0.177 · (a1 a2 )−9/10 . 392 M. Cipu, M. Mignotte / Journal of Number Theory 125 (2007) 356–392 Acknowledgment The authors thank the referee for many detailed comments and suggestions. References [1] [2] [3] [4] [5] [6] [7] [8] [9] [10] [11] [12] [13] [14] [15] [16] [17] [18] [19] [20] [21] [22] [23] [24] [25] [26] [27] [28] [29] [30] [31] W.S. Anglin, The Queen of Mathematics: An Introduction to Number Theory, Kluwer, Dordrecht, 1995. W.S. Anglin, Simultaneous Pell equations, Math. Comp. 65 (1996) 355–359. A. Baker, Linear forms in the logarithms of algebraic numbers, IV, Mathematika 15 (1968) 204–216. A. Baker, H. Davenport, The equations 3x 2 − 2 = y 2 and 8x 2 − 7 = z2 , Quart. J. Math. Oxford Ser. (2) 20 (1969) 129–137. C. Batut, K. Belabas, D. Bernardi, H. Cohen, M. Olivier, User’s guide to PARI/GP, version 2.1.6, http://www.parigphome.de/. M.A. Bennett, On the number of solutions of simultaneous Pell equations, J. Reine Angew. Math. 498 (1998) 173– 199. M.A. Bennett, M. Cipu, M. Mignotte, R. Okazaki, On the number of solutions of simultaneous Pell equations, II, Acta Arith. 122 (2006) 407–417. E. Brown, Sets in which xy + k is always a square, Math. Comp. 45 (172) (1985) 613–620. M. Cipu, Pairs of Pell equations having at most one common solution in positive integers, submitted for publication. M. Cipu, M. Mignotte, On the number of solutions of simultaneous Pell equations, http://www-irma.u-strasbg. fr/irma/publications/2006/06004.pdf.gz. A. Dujella, Complete solution of a family of simultaneous Pellian equations, Acta Math. Inform. Univ. Ostraviensis 6 (1998) 59–67. A. Dujella, A. Pethő, A generalization of a theorem of Baker and Davenport, Quart. J. Math. Oxford Ser. (2) 49 (1998) 291–306. C.M. Grinstead, On a method of solving a class of Diophantine equations, Math. Comp. 32 (143) (1978) 936–940. M. Laurent, M. Mignotte, Yu. Nesterenko, Formes linéaires en deux logarithmes et déterminants d’interpolation, J. Number Theory 55 (1995) 285–321. D.W. Masser, J.H. Rickert, Simultaneous Pell equations, J. Number Theory 61 (1996) 52–66. E.M. Matveev, An explicit lower bound for a homogeneous rational linear form in logarithms of algebraic numbers, II, Izv. Ross. Akad. Nauk Ser. Mat. 64 (2000) 125–180 (in Russian). English translation in: Izv. Math. 64 (2000) 1217–1269. W.L. McDaniel, The g.c.d. in Lucas sequences and Lehmer number sequences, Fibonacci Quart. 29 (1991) 24–29. M. Mignotte, A corollary to a theorem of Laurent–Mignotte–Nesterenko, Acta Arith. 86 (1998) 101–111. S.P. Mohanty, M.S. Ramasamy, The simultaneous Diophantine equations 5y 2 −20 = X 2 and 2y 2 +1 = Z 2 , J. Number Theory 18 (3) (1984) 356–359. K. Ono, Euler’s concordant forms, Acta Arith. 78 (1996) 101–123. P. Ribenboim, The Book of Prime Number Records, Springer, New York, 1989. H.P. Schlickewei, The number of subspaces occurring in the p-adic subspace theorem in Diophantine approximation, J. Reine Angew. Math. 406 (1990) 44–108. W.M. Schmidt, Norm form equations, Ann. of Math. 96 (1972) 526–551. W.M. Schmidt, Diophantine Approximation, Lecture Notes in Math., vol. 785, Springer, 1980. C.L. Siegel, Über einige Anwendungen diophantischer Approximationen, Abh. Preuss. Akad. Wiss. 1 (1929) 1–70. A. Thue, Über Annäherungswerte algebraischer Zahlen, J. Reine Angew. Math. 135 (1909) 284–305. N. Tzanakis, Effective solution of two simultaneous Pell equations by the elliptic logarithm method, Acta Arith. 103 (2002) 119–135. P.G. Walsh, On two classes of simultaneous Pell equations with no solutions, Math. Comp. 68 (225) (1999) 385– 388. P. Yuan, On the number of solutions of simultaneous Pell equations, Acta Arith. 101 (2002) 215–221. P. Yuan, Simultaneous Pell equations, Acta Arith. 115 (2004) 119–131. P. Yuan, On the number of solutions of x 2 − 4m(m + 1)y 2 = y 2 − bz2 = 1, Proc. Amer. Math. Soc. 132 (2004) 1561–1566.