Mihai Cipu
email: mihai.cipu@imar.ro

Maurice Mignotte
email: mignotte@math.u-strasbg.fr

ON THE NUMBER OF SOLUTIONS OF SIMULTANEOUS HYPERBOLIC DIOPHANTINE EQUATIONS

Keywords: Mathematics Subject Classification 2000 Primary: 11D09, Secondary: 11D45, 11J20, 11J86 simultaneous Pell equations, linear forms in logarithms

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 2cz 2 = 1, by 2dz 2 = 1 is at most two. This result is the best possible one. We prove a similar result for the equations x 2 -ay 2 = 1, z 2bx 2 = 1.

Introduction

The number of solutions of the simultaneous Diophantine equations [START_REF] Anglin | Simultaneous Pell equations[END_REF] ax 2 -cz 2 = δ 1 , by 2 -dz 2 = δ 2 was a question of constant interest in the last century. It is known already since A. Thue [START_REF] Thue | Über Annäherungswerte algebraischer Zahlen[END_REF] and C.L. Siegel [START_REF] Siegel | Über einige Anwendungen diophantischer Approximationen[END_REF] that the equations (1) have finitely many solutions when cδ 2 = dδ 1 . Their works do not provide estimations for the number of solutions. Using the methods developed by W.M. Schmidt [START_REF] Schmidt | Norm form equations[END_REF][START_REF] Schmidt | Diophantine Approximation[END_REF], H.P. Schlickewei [START_REF] Schlickewei | The number of subspaces occurring in the p-adic subspace theorem in Diophantine approximation[END_REF] proved that the number of integer solutions to

(2)

x 2 -cz 2 = 1, y 2 -dz 2 = 1
is at most 4 × 8 2 78 . Masser and Rickert [START_REF] Masser | Simultaneous Pell equations[END_REF] improved considerably the bound, proving that these equations have at most sixteen solutions (x, y, z) in positive integers. The proof uses the hypergeometric method. The same approach was followed by M. Bennett [START_REF] Bennett | On the number of solutions of simultaneous Pell equations[END_REF], who lowered the bound to three. Since there is no known pair of Pell equations with three solutions, he conjectured that equations (2) have at most two solutions for any c = d. A recent result of Yuan [START_REF] Yuan | On the number of solutions of simultaneous Pell equations[END_REF] shows that there are at most finitely many exceptions to this conjecture. To be precise, if max{ c, d} ≥ 1.4 × 10 57 , the equations (2) have at most two integer solutions (x, y, z) with x, y, z > 0. M. Bennett, R. Okazaki and the present authors have given an unconditional proof of the conjecture in [START_REF] Bennett | On the number of solutions of simultaneous Pell equations[END_REF]. (The reader willing to see a thorough description of the approach, with more details than in the printed version of the paper, may consult [START_REF] Cipu | On the number of solutions of simultaneous Pell equations[END_REF].) 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 + √ m 2 -1 and

(3) The proofs in [START_REF] Yuan | On the number of solutions of simultaneous Pell equations[END_REF][START_REF] Bennett | On the number of solutions of simultaneous Pell equations[END_REF][START_REF] Cipu | On the number of solutions of simultaneous Pell equations[END_REF] are based on A. Baker's theory [START_REF] Baker | Linear forms in the logarithms of algebraic numbers[END_REF] on bounds for linear forms in logarithms of algebraic numbers. This method combined with techniques from computational Diophantine approximation is very instrumental in obtaining all solutions of instances of [START_REF] Anglin | Simultaneous Pell equations[END_REF]. Since the ground-braking paper of A. Baker and H. Davenport [START_REF] Baker | The equations 3x 2 -2 = y 2 and 8x 2 -7 = z[END_REF], such a combination has been repeatedly successfully employed (see, for instance, [START_REF] Dujella | Complete solution of a family of simultaneous Pellian equations[END_REF][START_REF] Dujella | A generalization of a theorem of Baker and Davenport[END_REF]). Using this approach, W.S. Anglin showed [START_REF] Anglin | Simultaneous Pell equations[END_REF] (see, also, [2]) that ( 2) have at most one positive solution for 2 ≤ c < d ≤ 200. In [START_REF] Cipu | On the number of solutions of simultaneous Pell equations[END_REF] 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 [START_REF] Baker | Logarithmic forms and group varieties[END_REF]. This is the most extensive numerical confirmation of a stronger conjecture of Yuan [START_REF] Yuan | Simultaneous Pell equations[END_REF]Conjecture 1.1] recalled below.

n(l, m) = α 2l -α -2l 4 √ m 2 -1 .
A fruitful way to study the solutions of ( 1) is by connecting this system to the elliptic equation ab(xy) 2 = (cz 2 + δ 1 )(dz 2 + δ 2 ). One can show that each non-trivial solution to [START_REF] Anglin | Simultaneous Pell equations[END_REF] gives rise to a point of infinite order on the elliptic curve Y 2 = X 3 + (c + d)X 2 + cdX. K. Ono [START_REF] Ono | Euler's concordant forms[END_REF] 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 [START_REF] Tzanakis | Effective solution of two simultaneous Pell equations by the Elliptic Logarithm Method[END_REF], advocates the use of linear forms in elliptic logarithms. The same paper contains an ample bibliography, with pointers to works based on this idea. Several elementary methods for solving specific pairs of generalized Pell equations have been devised (see, for instance, [START_REF] Brown | Sets in which xy + k is always a square[END_REF][START_REF] Grinstead | On a method of solving a class of Diophantine equations[END_REF][START_REF] Mohanty | The simultaneous Diophantine equations 5y 2 -20 = X 2 and 2y 2 + 1 = Z 2[END_REF]). Other papers, like [START_REF] Pinch | Simultaneous Pellian equations[END_REF][START_REF] Walsh | On two classes of simultaneous Pell equations with no solutions[END_REF], to name but a few, deal in an ad hoc manner with systems [START_REF] Anglin | Simultaneous Pell equations[END_REF] or [START_REF] Batut | User's guide to PARI/GP, version 2.1[END_REF] whose coefficients are subject to quite special restrictions. There are also conditional results, assuming the ABC conjecture [START_REF] Walsh | On two classes of simultaneous Pell equations with no solutions[END_REF].

It is very hard to have an idea on the precise number of solutions of a pair of hyperbolic Diophantine equations of the most general type [START_REF] Anglin | Simultaneous Pell equations[END_REF] just by looking at the coefficients. Indeed, Masser and Rickert [START_REF] Masser | Simultaneous Pell equations[END_REF] have devised a method to produce simultaneous Pell equations with any prescribed number of solutions. However, Yuan [START_REF] Yuan | Simultaneous Pell equations[END_REF] put forward a conjecture aiming to describe when there are at least two solutions in positive integers to [START_REF] Batut | User's guide to PARI/GP, version 2.1[END_REF] ax 2 -cz 2 = 1, by 2 -dz 2 = 1.

Yuan claims that the coefficients of systems with two solutions are obtained as follows.

For integers l > 1, m > 1, and a > 1, put

n(l, m) = (m + √ m 2 -1 ) 2l -(m - √ m 2 -1 ) 2l 4 √ m 2 -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 the simultaneous Pell equations

(a, b, c, d) = 1, 1, m 2 -1, n(l, m) 2 -1 , a, b(l, a), a -1, b(l, a) -1
or equivalent forms thereof.

The first aim of this paper is to study the number of solutions for the pair of hyperbolic Diophantine equations ( 6) when a > 1, b > 1. Yuan [START_REF] Yuan | Simultaneous Pell equations[END_REF] proved that if c = d and max{ a, b, c, d } > 1.16 × 10 59 then the system has at most two positive solutions. Here we prove that the same result holds regardless of the size of the coefficients. In order to establish this result, we adapt the approach developed in [START_REF] Cipu | On the number of solutions of simultaneous Pell equations[END_REF]. 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 having three putative solutions provides strong gap principles, according to which consecutive solutions are to be found by looking at rather distant terms of second-order recurrent sequences. One can pass from recurrent sequences to linear forms in the logarithms of three algebraic numbers in a classical manner. Specialization of a general theorem of E.V. Matveev [START_REF] Matveev | An explicit lower bound for a homogeneous rational linear form in logarithms of algebraic numbers, II (Russian)[END_REF], valid for any number of logarithms, yields the fact that three solutions for (6) can exist only if max{ a, b } < 3.1 × 10 51 . In order to significantly reduce this bound, we apply a suitable instance of a theorem of M. Laurent, M. Mignotte and Yu. Nesterenko [START_REF] Laurent | Formes linéaires en deux logarithmes et déterminants d'interpolation[END_REF]. 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 our deliberations, in Section 4 one obtains that max{ a, b } < 4 × 10 38 . 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 equations [START_REF] Batut | User's guide to PARI/GP, version 2.1[END_REF]. On the other hand, various verifications eliminate the possibility that a third solution exist when max{ a, b } is in the domain not excluded for other reasons.

The proof we give to Theorem 1.1 is essentially self-contained. It has a non-linear 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 Diophantine equations [START_REF] Bennett | On the number of solutions of simultaneous Pell equations[END_REF] x 2 -

ay 2 = 1, z 2 -bx 2 = 1.
In the second part of the paper we establish a tight bound for the number of positive solutions for these equations.

Theorem 1.2. Let a and b be positive integers. Then the simultaneous Pell equations (7) have at most two common solutions with x, y, z > 0.

Unlike Theorem 1.1, this result is not assuredly best possible. We know no system [START_REF] Bennett | On the number of solutions of simultaneous Pell equations[END_REF] having more than one solution in positive integers and we think that none exists. Conjecture 2. For any positive integers a, b the system of Diophantine equations [START_REF] Bennett | On the number of solutions of simultaneous Pell equations[END_REF] has at most one solution in positive integers.

Our results illustrate the power of the method we developed for the study of numbers of solutions to simultaneous hyperbolic Diophantine equations. The approach is quite adequate for the investigation of other families of equations (as already seen in [START_REF] Cipu | On the number of solutions of simultaneous Pell equations[END_REF]). We leave it on another occasion to present the details for the systems (1) not covered yet.

Properties of solutions

In this section we start to study the positive solutions of the equations [START_REF] Batut | User's guide to PARI/GP, version 2.1[END_REF]. 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 compatible systems [START_REF] Batut | User's guide to PARI/GP, version 2.1[END_REF], regardless of the number of solutions.

First we note that, when dealing with simultaneous Diophantine equations of the type [START_REF] Batut | User's guide to PARI/GP, version 2.1[END_REF], it is sufficient to consider only coefficients c and d one less than a and b, respectively. The explanation is given by Lemma 2.1 below. In its proof (and also later on), divisibility properties of Lehmer numbers are invoked.

Lemma 2.1. Assume the system (6) has at least one solution in positive integers and a > 1, b > 1. Let (x 0 , y 0 , z 0 ) be a solution with z 0 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 x multiple of x 0 , y multiple of y 0 , and z multiple of z 0 .

b) The equations (6) have as many positive 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, resp. β = u √ b + v √ d,
be the fundamental solution of the Pell equation ax 2 -cz 2 = 1, resp. by 2 -dz 2 = 1. Put for odd j and k

U j := α j -α -j α -α -1 , U k := β k -β -k β -β -1 , V j := α j + α -j α + α -1 , V k := β k + β -k β + β -1 .
Since a ≥ 2 and b ≥ 2, there are odd integers

j 0 ≥ 1, k 0 ≥ 1, such that x 0 = u V j0 , y 0 = u V k0 , z 0 = v U j0 = v U k0 .
For any solution (x, y, z) of ( 6) one has

x = u V j , y = u V k , z = v U j = v U k for certain odd integers j ≥ j 0 , k ≥ k 0 .
Suppose, by way of contradiction, that z 0 does not divide z. Then U j0 > 1 and parts a) and b) of Lemma 2.2 yield j 0 j, k 0 k. By Euclidean division one gets j = 2qj 0 ± r, k = 2tk 0 ± s, 0 < r < j 0 , 0 < s < k 0 .

From the identity stated in part c) of the next lemma one obtains

U j ∓ U r = 2 V qj0±r U qj0 , U k ∓ U s = 2 V tk0±s U tk0 .
Having in view Lemma 2.2 a), one gets

z = v U j ≡ ±v U r (mod z 0 ) and z = v U k ≡ ±v U s (mod z 0 ). Since 2 v U r < v U r+1 ≤ v U j0 = z 0 and 2 v U s < v U s+1 ≤ v U k0 = z 0 , one concludes that v U r = v U s . Hence, (u V r , u V s , v U r )
is a positive solution of (6) whose third component is smaller than z 0 . This contradicts the choice of (x 0 , y 0 , z 0 ).

So it is true that z 0 divides z. Then z = gz 0 for a certain positive integer g and j 0 | j, k 0 | k. Lemma 2.2 a) implies x = ex 0 , y = f y 0 for suitable positive integers e, f . Hence,

1 = a x 2 -c z 2 = a e 2 x 2 0 -c g 2 z 2 0 = a x 2 0 e 2 -g 2 (a x 2 0 -1). One concludes that (e, f, g) is a positive solution for the simultaneous Pell equations a x 2 0 X 2 - (a x 2 0 -1)Z 2 = b y 2 0 Y 2 -(b y 2 0 -1)Z 2 = 1. Lemma 2.2. a) If U k = 1, then U k | U n if and only if k | n. b) If k ≥ 1, then V k | V n if and only if n/k is an odd integer. c) If n = 2 t m ± r, with 0 ≤ r ≤ m and t ≥ 0, then U n ∓ U r = 2 V tm±r U tm .
Proof. The properties are well-known. They are proved in various places, for instance, in [START_REF] Ribenboim | The book of prime number records[END_REF] and [START_REF] Mcdaniel | The g.c.d. in Lucas sequences and Lehmer number sequences[END_REF].

So we will consider from now on c = a -1 and d = b -1 for some integers b > a ≥ 2 and study the system of Diophantine equations

(8) ax 2 -(a -1)z 2 = 1, by 2 -(b -1)z 2 = 1. Put (9) α = √ a + √ a -1 , β = √ b + √ b -1
and consider (x, y, z) a positive integer solution to [START_REF] Bennett | On the number of solutions of simultaneous Pell equations[END_REF]. Then z appears in two linear recurrent sequences

(10) z = U j = U k ,
for some positive odd integers j and k, where

U j = α j -α -j 2 √ a -1 and U k = β k -β -k 2 √ b -1 , j, k ∈ N. Note that (12) α j < β k < α j b -1 a -1 . (11)
Indeed, from α, α -1 , β, β -1 > 0 it results

β k -α j = 2 z √ b -1 + β -k -2 z √ a -1 -α -j > 2 z √ b -1 - √ a -1 -α -j ≥ 2 z α -α -j > 0.
This in turn implies

β k α j = 1 -α -2j 1 -β -2k b -1 a -1 < b -1 a -1 .
From [START_REF] Cipu | On the number of solutions of simultaneous Pell equations[END_REF] one infers that the linear form in logarithms

(13) Λ = log b -1 a -1 + j log α 2 -k log β 2
is positive and bounded from above by ( 14)

-2 log (1 -α -2j ) < 2 α 2-2j α 2 -1 . Hence, ( 15
) log Λ < -2j log α + log 2 α 2 α 2 -1 < -2j log α + 0.882.
The observation stated in the next lemma allows one to obtain a companion inequality to [START_REF] Dujella | A generalization of a theorem of Baker and Davenport[END_REF].

Lemma 2.3. 1 + 1 2(a -1) - 1 2(b -1) - 3 16(a -1) 2 β 2 α 2 < b -1 a -1 < 1 + 1 2(a -1) β 2 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)b 4 + (-256a 3 + 224a 2 + 80a + 60)b 3 + (256a 4 -128a 3 -448a 2 + 448a -245)b 2 + (-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.
The right inequality is equivalent to

2(b -1) 2a -1 + 2 a 2 -a < (2a -1) 2b -1 + 2 b 2 -b , which follows from 4(b -1) < 2b -1 + 2 √ b 2 -b.
From [START_REF] Dujella | A generalization of a theorem of Baker and Davenport[END_REF] and this lemma it follows an inequality which has an important role in the rest of the paper:

(16) β 2k-2 < 1 + 1 2(a -1) α 2j-2 .
A direct consequence of it 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 equation (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 2.4. Let (x, y, z) be a solution of the simultaneous Diophantine equations [START_REF] Bennett | On the number of solutions of simultaneous Pell equations[END_REF]. If z = U j = U k , 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 U j , U k are generated by recurrence sequences of the type

w t+2 = (4s -2) w t+1 -w t , w 0 = 1, w 1 = 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 2.5. For any t ≥ 2 one has (4a -3) t < U 2t+1 < (4a -1) t . Consequently,

U 1/t 2t+1 = 4a -2 or 4a -3. Proof.
The first inequality results from U 2t+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 U t+1 is obtained by induction, since

U 3 = 4a-1 and U 2t+3 = (4a -2) U 2t+1 -U 2t-1 < (4a -2) U 2t+1 for any positive index t.

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 rather distant. We shall follow Yuan's strategy from [START_REF] Yuan | Simultaneous Pell equations[END_REF], paying considerable attention to the details of the reasoning. Consequently, for k 2 ≥ 7, our gap principle is much stronger than that used by Yuan. In [START_REF] Yuan | Simultaneous Pell equations[END_REF] it is proved that the quotient j 3 /j 2 is bounded from below by an expression linear in β, whereas our bound is of the type β (k2-3)/2 . Our result made possible explicit computations to be described in subsequent sections.

Suppose that the system (8) has three positive solutions (x i , y i , z i ) (i = 1, 2, 3). Then

z i = α ji -α -ji 2 √ a = β ki -β -ki 2 √ b for odd integers j i and k i (i = 1, 2, 3) with 1 = k 1 < k 2 < k 3 and 1 = j 1 < j 2 < j 3 .
We first note that either j 2 divides j 3 and k 2 divides k 3 or j 3 = 2qj 2 ± 1 and k 3 = 2q 1 k 2 ± 1 for some positive integers q and q 1 . The reasoning is similar to that used in the last paragraph of the proof of part a) of Lemma 2.1. After one concludes that (V r , V s , U r ) is a positive solution of (8) whose third component is smaller than z 2 , it follows that U r = z 1 = 1, whence r = 1, and therefore

V s = 1 = V 1 , so that s = 1.
As seen above, there exist integers q, q 1 ≥ 2 and σ, σ 1 ∈ { -1, 0, 1 } such that j 3 = qj 2 + σ, k 3 = q 1 k 2 + σ 1 and both qσ and q 1 σ 1 are even. As a preparation for the proof of our gap principle we establish first the next result.

Lemma 3.1. σ = σ 1 . Proof. If σ = 0, then j 2 divides j 3 , so that z 3 = U j3 = U k3 is multiple of z 2 = U j2 = U k2 . Thus, U k2 divides U k3 , whence k 2 divides k 3 .
This means σ 1 = 0, as asserted. By symmetry, one obtains that σ = 0 whenever σ 1 = 0.

Suppose now that both σ and σ

1 are non-zero. From U j3 -σ = 2V qj2+σ U qj2 (cf. Lemma 2.2 c)) and U j2 | U qj2 it results z 3 = U j3 ≡ σ (mod z 2 ). Similarly, U k3 ≡ σ 1 (mod z 2 ). The claim follows from |σ -σ 1 | ≤ 2 < 4 a -3 < z 2 (cf. Lemma 2.5).
Another observation needed later on is that the relationship between q and q 1 is the same as between j 2 and k 2 . Lemma 3.2. q > q 1 . Proof. If j 3 is multiple of j 2 , then k 3 is divisible by k 2 and we have

z 3 = U qj2 = (x 2 √ a + z 2 √ a -1 ) q -(x 2 √ a -z 2 √ a -1 ) q 2z 2 √ a -1 = t≥0 q 2t + 1 x 2 √ a q-2t-1 z 2t 2 (a -1) t and z 3 = U q1k2 = t≥0 q 1 2t + 1 y 2 √ b q1-2t-1 z 2t 2 (b -1) t .
The desired conclusion follows because

y 2 √ b > x 2 √
a and b > a. It remains to examine the case j 3 = qj 2 + σ and k 3 = q 1 k 2 + σ with σ = 0. We shall implicitly use the well-known inequalities

t - t 2 2 < log(1 + t) < t valid for positive t.
Let us consider the linear forms in logarithms

Λ 2 = log b -1 a -1 + j 2 log α 2 -k 2 log β 2 , Λ 3 = log b -1 a -1 + j 3 log α 2 -k 3 log β 2 .
By relations ( 12) and ( 16), we have

0 < log b -1 a -1 -Λ 2 = 2(k 2 log β -j 2 log α) < log b -1 a -1 , log b -1 a -1 -Λ 2 < 2 log β α + 1 2(a -1)
.

Replacing in Λ 3 the coefficients j 3 and k 3 in terms of j 2 and k 2 , one gets

2(q -q 1 )j 2 log α = Λ 3 + (q 1 -1) log b -1 a -1 + 2σ log β α -q 1 Λ 2 (17) ≥ Λ 3 -log b -1 a -1 -2 log β α + q 1 log b -1 a -1 -Λ 2 . From Lemma 2.3 one obtains log b -1 a -1 < 2 log β α + 1 2a -2 ,
so that the rightmost term in the chain of inequalities ( 17) is greater than

(18) Λ 3 -4 log β α - 1 2a -2 + q 1 log b -1 a -1 -Λ 2 .
From the left inequality in Lemma 2.3 one gets for all (a, b) = (2, 3)

log b -1 a -1 > 2 log β α + log 1 + 1 8(a -1) 2 > 2 log β α + 1 9(a -1) 2 .
Numerical verification shows that always one has

(19) log b -1 a -1 > 2 log β α + 1 9(a -1) 2 .
On the other hand, by [START_REF] Laurent | Formes linéaires en deux logarithmes et déterminants d'interpolation[END_REF] we have

(20) Λ 2 < 2 α -12 α 2 -1 < 1 (4a -4) 6 .
Therefore, the expression in ( 18) is bounded from below by

2(q 1 -2) log β α + q 1 1 9(a -1) 2 - 1 (4a -4) 6 - 1 2a -2 . Since β α 2 > b a ≥ 1 + 1 a ,
we have

2 log β α > 1 a - 1 2a 2 ≥ 3 4a . Thus 2(q -q 1 )j 2 log α > 3(q 1 -2) 4a + q 1 1 9(a -1) 2 - 1 (4a -4) 6 - 1 2a -2 .
As the expression in the right side of the previous inequality is certainly positive when q 1 ≥ 4, we have obtained the conclusion of the Lemma, unless q = q 1 = 2. In this special situation, from equation ( 17), Lemma 2.3 and relation (14

) it results 2 (4a -4) 6 > 2Λ 2 -Λ 3 ≥ log b -1 a -1 -2 log β α > 1 9(a -1) 2 ,
which is false for a > 1.

The proof is complete.

Here is a statement which will play in the proof of our gap principle a similar role as the lemma above.

Lemma 3.3. qj 2 log α < q 1 k 2 log β.
Proof. By Lemma 2.3 and relations ( 13) -( 14) we have

(k 3 -1) log β 2 -(j 3 -1) log α 2 = log b -1 a -1 -Λ 3 -log β 2 α 2 > 1 9(a -1) 2 -Λ 3 > 1 9(a -1) 2 - 1 (4a -4) 6 > 0, so that (j 3 -1) log α < (k 3 -1) log β. Hence, (qj 2 + σ -1) log α < (q 1 k 2 + σ -1) log β, or qj 2 log α < q 1 k 2 log β + (σ -1) log(β/α).
Since σ is at most 1 and β > α, the result follows. Proof. We argue by contradiction. Supposing that

aq 2 ≥ bq 2 1 , it results b a ≤ q q 1 < k 2 log β j 2 log α < log β log α .
It is straightforward to establish that the mapping t → log( √ t+ √ t -1 )/ √ t, defined for t ≥ √ 2, is increasing for t ≤ t 0 3.276 and decreasing for t ≥ t 0 . Thus, for a > 2 or a = 2 and b > 6, from the assumption aq 2 ≥ bq 2 1 one gets the contradiction a > b. For a = 2 < b ≤ 6 one sees that the inequalities [START_REF] Dujella | A generalization of a theorem of Baker and Davenport[END_REF] are not satisfied.

We can now prove the main results of this section. Proposition 3.5. Assume k 2 = 3. Then for all β (resp. for β ≥ 1000) one has

j 3 > 1.7 j 2 β 2/3 (resp. j 3 > 1.98 j 2 β 2/3 ) for j 2 = 7, j 3 > 1.7 j 2 β 4/5 (resp. j 3 > 1.90 j 2 β 4/5 ) for j 2 = 11, j 3 > 1.7 j 2 β 6/7 (resp. j 3 > 1.83 j 2 β 6/7 ) for j 2 ≥ 15.
Proof. We first examine the situation when k 3 = 3q 1 . Then j 3 = qj 2 , for some odd integers q and q 1 . From ax 2 2 ≡ by 2 2 ≡ 1 (mod z 2 2 ) and z 3 ≡ q(ax 2

2 ) (q-1)/2 ≡ q 1 (by 2 2 ) (q1-1)/2 (mod z 2 2 ) one gets q ≡ q 1 (mod z 2 2 ). Since z 2 = β 2 + 1 + β -2 > β 2 and q > q 1 (see Lemma 3.2), one obtains q > z 2 2 > β 4 , which yields j 3 > j 2 β 4 , a much stronger result than the desired one. Suppose now that k 3 = 3q 1 + σ, j 3 = qj 2 + σ, with σ ∈ { -1, 1 } and even q, q 1 . This time one has

z 3 ≡ aqx 2 z 2 + σ ≡ bq 1 y 2 z 2 + σ (mod z 2 2 ), whence bq 1 y 2 ≡ aqx 2 (mod z 2 ).
From z 2 = 4b -1, j 2 ≥ 7 and ax 2 2 ≡ by 2 2 ≡ 1 (mod z 2 2 ) one infers 4aq 2 ≡ 4bq 2 1 ≡ q 2 1 (mod z 2 ).

Note that q > q 1 implies 4aq 2 > q 2 1 , so that

q 2 ≥ 1 + U j2 4a .
We shall give all the details for the rest of the proof in the case j 2 = 7. The same idea, with obvious modifications, works for the other cases.

Since U 7 = 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 K 6 ≤ (16a 2 -20a + 6) 3 4(2b -1) 2 = 2(8a 2 -10a + 3) 3 (32a 3 -40a 2 + 12a -1) 2 .
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) (resp. K 6 ≤ f (26)). We verify that K = 0.883 (resp. K = 0.991) works. Indeed, then we have q > 0.883 β 2/3 and

j 3 ≥ 2 q j 2 -1 > 1.766 j 2 β 2/3 -1 > 1.7 j 2 β 2/3 .
The gap principle for higher values of k 2 is much stronger due to the presence of a greater exponent of β.

Proposition 3.6. If k 2 ≥ 5, then j 3 > 3.96 j 2 β (k2-3)/2 .
Proof. Recall that j 3 = qj 2 +σ and k 3 = q 1 k 2 +σ for certain integers q > q 1 ≥ 2 and σ ∈ { -1, 0, 1 } such that qσ and q 1 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

z 3 = U qj2 = t≥0 q 2t + 1 x 2 √ a q-2t-1 z 2t 2 (a -1) t ≡ q x 2 √ a q-1 (mod z 2 2 ).
One gets similarly

z 3 ≡ q 1 y 2 √ b q1-1 (mod z 2 2 ), so that q x 2 √ a q-1 ≡ q 1 y 2 √ b q1-1 (mod z 2 2 ). Since ax 2 2 ≡ by 2 2 ≡ 1 (mod z 2
2 ), and q and q 1 are odd integers, this yields q ≡ q 1 (mod z 2 2 ). Having in view Lemma 3.2, one concludes that q > z 2 2 > β 2(k2-1) and therefore j 3 = q j 2 > j 2 β 2(k2-1) , which is stronger than the desired conclusion.

If σ = 0, then both q and q 1 are even. From

α 2j2 = 2z 2 2 (a -1) + 1 + 2x 2 z 2 a 2 -a it results z 3 = U j3 ≡ aqx 2 z 2 + σ (mod 2z 2 2
). With the same reasoning one obtains

z 3 = U k3 ≡ bq 1 y 2 z 2 + σ (mod 2z 2
2 ). Therefore [START_REF] Ono | Euler's concordant forms[END_REF] bq

1 y 2 ≡ aqx 2 (mod 2z 2 ) and b q 1 2 2 ≡ a q 2 2 (mod z 2 ).
As aq 2 < bq 2 1 and q > q 1 (see Corollary 3.4 and Lemma 3.2), one obtains bq 2 ≥ b(q 1 + 2) 2 ≥ aq 2 + 4z 2 + 4bq 1 + 4b

≥ 4z 2 + 16a + 12b > 4 β k2-1 + β k2-3 + β k2-5 + 12b, whence q 2 > 4β k2-3 4 - 1 b + 12.
Since b ≥ 60 for each system (8) with at least three positive solutions, one has

j 3 > j 2 4β k2-3 4 - 1 b + 12 -1 > 3.96 j 2 β (k2-3)/2 .

Linear forms in logarithms

The linear form in three logarithms naturally attached to a term common to two second-order linear recurrent sequences made its appearance in Section 2. In the previous sections we made good 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 [START_REF] Matveev | An explicit lower bound for a homogeneous rational linear form in logarithms of algebraic numbers, II (Russian)[END_REF]; thus we quote his result. This theorem enables us to get a first bound on coefficients a and b for which the equations ( 8) have three solutions. This bound will be subject to further improvements by using suitable results on linear forms in two logarithms. As we shall see in due time, the idea of trading three self-imposing logarithms by two complicated logarithms pays well the extra work needed. Below h(γ) denotes the absolute logarithmic height of the algebraic number γ.

Theorem A ( [START_REF] Matveev | An explicit lower bound for a homogeneous rational linear form in logarithms of algebraic numbers, II (Russian)[END_REF]). Let λ 1 , λ 2 , λ 3 be Q-linearly independent logarithms of non-zero algebraic numbers and let b 1 , b 2 , b 3 be rational integers with b 1 = 0. Define α j = exp(λ j ) for j = 1, 2, 3 and

Λ = b 1 λ 1 + b 2 λ 2 + b 3 λ 3 .
Let D be the degree of the number field

Q(α 1 , α 2 , α 3 ) over Q. Put χ = [R(α 1 , α 2 , α 3 ) : R].
Let A 1 , A 2 , A 3 be positive real numbers, which satisfy

A j ≥ max D h(α j ), |λ j |, 0.16 (1 ≤ j ≤ 3). Assume that B ≥ max |b j |A j /A 1 : 1 ≤ j ≤ 3 .
Define also

C 1 = 5 × 16 5 6χ e 3 (7 + 2χ) 3e 2 χ 20.2 + log 3 5.5 D 2 log(eD) . Then log |Λ| > -C 1 D 2 A 1 A 2 A 3 log 1.5 eDB 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. Proof. Suppose, by way of contradiction, that the system (8) has at least one solution given by relation [START_REF] Dujella | Complete solution of a family of simultaneous Pellian equations[END_REF]

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,

Λ = log b -1 a -1 + j log α 2 -k log β 2 = log b -1 a -1 -t log γ for some positive integer t. As b -1 -(a -1)γ t is a nonzero algebraic integer, b -1 -(a -1)γ t b -1 -(a -1)γ -t is a positive rational integer. Hence, b -1 -(a -1)γ t ≥ 3 4(b -1) , b -1 (a -1)γ t -1 ≥ 3 4(b -1) 2 , and Λ > 2 3(b -1) 2 .
Comparison with the upper bound for Λ given in [START_REF] Laurent | Formes linéaires en deux logarithmes et déterminants d'interpolation[END_REF] results in

(b -1) 2 > (1 -α -2 )α 2j .
Using the obvious inequality β 2 > 4(b -1), one gets β 4 > 16(1 -α -2 )α 2j . But we have seen (cf. ( 16))

(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 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 [START_REF] Bennett | On the number of solutions of simultaneous Pell equations[END_REF]. As explained above, if z > 1, then z is given by relation [START_REF] Cipu | On the number of solutions of simultaneous Pell equations[END_REF] for suitable odd integers j > k > 1. We apply Matveev's result for the linear form

Λ = log b -1 a -1 + 2 j 3 log α -2 k 3 log β,
which we rewrite as

Λ = log b -1 a -1 + r log α 2 -k 3 log(β/α h ) 2
, where j 3 = hk 3 + r, 0 ≤ r < k 3 .

We have D = 4 and

h b -1 a -1 = log(b -1), h(α 2 ) = log α, h (β/α h ) 2 = log β.
Therefore we can take

A 1 = 2 D log(β/2), A 2 = D log α, A 3 = D log β, and
B = max 1, r log α 2 log(β/2) , k 3 log β 2 log(β/2) = k 3 log β 2 log(β/2)
.

With these choices it follows Matveev's theorem yields in our specific situation the inequality This relation together with Propositions 3.5-3.6 and the obvious inequalities

C 1 = 5 × 16 5 6 e 3 • 9 • 3e 2
log Λ > -C 1 • 16A 1 A 2 A 3 log 6eB log(4e) = -C 1 • 16 • 128 log α log β log(β/2)
(23) 2 √ b -1 < β < 2 √ b
yield upper bound for integers b for which the simultaneous Pell equations [START_REF] Batut | User's guide to PARI/GP, version 2.1[END_REF] can have (at least) three positive solutions:

k 2 = 3 : b < 3.1 × 10 51 for j 2 = 7, b < 1.1 × 10 42 for j 2 ≥ 11, k 2 = 5 : b < 3.1 × 10 32 , k 2 ≥ 7 : b < 1.6 × 10 15 .
The admissible range for b shrinks very fast with growing k 2 . However, for k 2 at most 5, it is far too big to be exhausted by a 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 [START_REF] Laurent | Formes linéaires en deux logarithmes et déterminants d'interpolation[END_REF] (which works quite well in the present circumstances). For later reference, we state the result in the form we actually used.

Theorem B ([14]

). Consider the linear form

Λ = b 2 log α 2 -b 1 log α 1 , with b 1 and b 2 positive integers. Let K ≥ 3, L ≥ 2, R 1 , R 2 , S 1 , S 2 be positive integers and ρ a real number greater than 1. Put R = R 1 + R 2 -1, S = S 1 + S 2 -1, N = KL, g = 1 4 - N 12RS , c = (R -1)b 2 + (S -1)b 1 2 K-1 k=1 k! -2/(K 2 -K)
.

Let a 1 , a 2 be positive real numbers such that

a i ≥ ρ | log α i | -log |α i | + 2D h(α i ),
for i = 1, 2. Suppose that:

(I) Card α r 1 α s 2 ; 0 ≤ r < R 1 , 0 ≤ s < S 1 ≥ L, (II) Card rb 2 + sb 1 ; 0 ≤ r < R 2 , 0 ≤ s < S 2 > (K -1)L, and
(III) K(L -1) log ρ -(D + 1) log N -D(K -1) log c -gL (Ra 1 + Sa 2 ) > 0.
Then,

|Λ | ≥ ρ -KL+(1/2) , where Λ = Λ • max LSe LS|Λ|/(2b2) 2b 2 , LRe LR|Λ|/(2b1) 2b 1 .
In the case when the number α 1 is not a root of unity we shall deduce the following result from Theorem B, which is a variant of Théorème 2 of [START_REF] Laurent | Formes linéaires en deux logarithmes et déterminants d'interpolation[END_REF], close to Theorem 1.5 of [START_REF] Mignotte | A corollary to a theorem of Laurent-Mignotte-Nesterenko[END_REF]. Proposition 4.2. Consider the linear form

Λ = b 2 log α 2 -b 1 log α 1 ,
where b 1 and b 2 are positive integers. Suppose that α 1 is not a root of unity. Put

D = [Q(α 1 , α 2 ) : Q] / [R(α 1 , α 2 ) : R].
Let a 1 , a 2 , h, t be real positive numbers, and ρ a real number greater than 1. Put λ = log ρ and suppose that

h ≥ max 1, 1.5 λ, D log b 1 a 2 + b 2 a 1 + log λ + f (K) + ε , ε = 0.0262, a i ≥ max 4, 2.7 λ, ρ | log α i | -log |α i | + 2D h(α i ) , (i = 1, 2), a 1 a 2 ≥ 20λ 2 ,
where

f (x) = log 1 + √ x -1 √ x x -1 + log x 6x(x -1) + 3 2 + log 3 4 + log x
x-1

x -1 and

L = 2 + 2h/λ ≥ 5, K = 1 + tLa 1 a 2 .
Then we have the lower bound

log |Λ| ≥ -λtL 2 a 1 a 2 -max λ(L -0.5) + log L 2 (1 + √ t )a 2 , 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 = 3 4 L a 2 + 1 a 1 .
We postpone the proof of the above result to the last section of the paper. Now we apply Proposition 4.2 to the system ax 2 -(a -1)z 2 = by 2 -(b -1)z 2 = 1 with three solutions, the second of which is obtained for k 2 = 3 and j 2 = 7 or 11.

In the present study we have two linear forms

Λ i = log b -1 a -1 + j i log α 2 -k i log β 2 , i = 2, 3.
As a consequence of Lemma 2.3 we have

log 1 + 1 8(a -1) 2 + 2 log β α < log b -1 a -1 < log 1 + 1 2a -2 + 2 log β α ,
so that (cf. the proof of Lemma 3.2)

1 9a 2 + (j 2 -1) log α 2 -(k 2 -1) log β 2 < Λ 2 < (j 2 -1) log α 2 -(k 2 -1) log β 2 + 1 a .
Define the integer l by

k 2 -1 2 j 3 = j 2 -1 2 k 3 + l - j 2 -k 2 2 .
It is easily seen that l is even.

We apply the estimate for the linear form in two logarithms

Λ := b 2 log α 2 -b 1 log α 1 , with α 1 = β k2-1 α j2-1 , α 2 = b -1 a -1 (k2-1)/2 α 2l-j2 +k2 .
Here D = 4 and b

1 = k 3 , b 2 = 1.
Using inequality [START_REF] Matveev | An explicit lower bound for a homogeneous rational linear form in logarithms of algebraic numbers, II (Russian)[END_REF], one obtains

h(α 1 ) = k 2 -1 2 log β ≤ j 2 -1 2 log α + 1 4a < j 2 -1 4 log(4a).
We claim that 0 < log α 1 ≤ 1 4(a -1) .

Indeed, by Lemma 2.3 and relations ( 13) -( 14) we have

2 log α 1 = (k 2 -1) log β 2 -(j 2 -1) log α 2 = log b -1 a -1 -Λ 2 -log β 2 α 2 > 1 9(a -1) 2 -Λ 2 > 1 9(a -1) 2 - 1 (4a -4) 6 > 0, and 2 log α 1 < log b -1 a -1 -log β 2 α 2 < log 1 + 1 2(a -1) < 1 2(a -1) , as claimed.
The next immediate concern is to show that l is non-negative. The equality

(k 2 -1)Λ 3 = log (b -1) (k2-1)/2 (a -1) (k2-1)/2 α j2-k2 + l log α 2 -k 3 log β k2-1 α j2-1 is equivalent to log α 2 = (k 2 -1)Λ 3 + k 3 log α 1 .
Both terms in the right side are positive. Therefore, α 2 > 1. Further, Lemma 2.3 and inequality ( 16) imply

(k 2 -1) 2 log b -1 a -1 -log α (j2-k2) < (k 2 -1) log β α + k 2 -1 4(a -1) -log α (j2 -k2) = (k 2 -1) log β -(j 2 -1) log α + k 2 -1 4(a -1) < k 2 4(a -1) , whence it follows 0 < log α 2 < l log α 2 + k 2 4(a -1)
.

Since k 2 = 3 and a ≥ 2, one concludes that l is non-negative. In fact, it is positive. We prove this claim par reductio ad absurdum.

When l = 0, we have (k 2 -1)(j 3 -1) = (j 2 -1)(k 3 -1), and therefore

Λ 2 -Λ 3 = (j 2 -j 3 ) log α 2 -(k 2 -k 3 ) log β 2 = (k 3 -k 2 ) log β k2-1 α j2-1 .
By relations ( 19) and ( 20), one obtains

2 log β k2 -1 α j2 -1 = log b -1 a -1 -Λ 2 -log β 2 α 2 > 1 9(a -1) 2 - 1 (4a -4) 6 > 7 64(a -1) 2 , whence the contradiction 1 (4a -4) 6 > Λ 2 > Λ 2 -Λ 3 > 7(k 3 -3) 128(a -1) 2 .
Thus l ≥ 2. In fact, for j 2 = 11 one has l ≥ 4. Indeed, for this value of j 2 , l is given by j 3 = 5k 3 + l -4. From Lemma 2.4 we know that j 3 ≡ k 3 (mod 4), so that l is multiple of 4 and our claim follows. Note that the inequalities l ≥ 2 for j 2 = 7 and l ≥ 4 for j 2 = 11 are equivalent to 2l -j 2 + k 2 ≥ 0. Now we proceed to bound from above the logarithm and the height of α 2 for a bigger than 1000. Lemma 2.3 yields

log α 2 = log b -1 a -1 + (2l -j 2 + k 2 ) log α < 2 log β α + 1 2(a -1) + (2l -j 2 + k 2 ) log α.
We also obtain

h(α 2 ) < k 2 -1 2 log(b -1) + 2l -j 2 + k 2 4 log(4a) ≤ l 2 + 3 log(4a)
and 0 < log α 2 ≤ (l + 1) log(4a).

The hypothesis of Proposition 4.2 requires

a 1 ≥ ρ -1 4(a -1)
+ 2(j 2 -1) log(4a), a 2 ≥ (ρ -1)(l + 1) + 4l + 24 log(4a), and we will choose

a 1 := 2 j 2 log(
| ≥ -λ t L 2 a 1 a 2 -λ (L -0.5) -log L 2 (1 + √ t )a 2 (24) > -0.21016λ 1 + h λ 2 a 1 a 2 -λ (L -0.5) -log L 2 (1 + √ t )a 2 > -0.7 1 + h λ 2 a 1 a 2 ,
whereas from [START_REF] Masser | Simultaneous Pell equations[END_REF] we know that log |Λ 3 | < -j 3 log α 2 + 0.9.

In order to exploit these bounds for log Λ 3 , we need to relate k 3 to a and l. We show that for 2l = j 2 -k 2 and a > 1000 one has k 3 ≥ 1.998a l log α 2 .

From ( 12) and ( 16) it follows

α (j2 -1)k3+2l-j2+k2 < β 2k3 , β 4 < 1 + 1 2(a -1) α 2(j2-1) , whence α 2(2l-j2 +k2)/k3 < 1 + 1 2(a -1)
.

Therefore,

(2l -j 2 + k 2 ) log α 2 < k 3 log 1 + 1 2(a -1) < k 3 2(a -1)
.

Since 2(a -1) ≥ 1.998a and l ≥ j 2 -k 2 (recall that 2l > j 2 -k 2 and l is multiple of 4 if j 2 = 11), it results

k 3 > 2(a -1)(2l -j 2 + k 2 ) log α 2 ≥ 1.998a l log α 2 .
Therefore k 3 ≥ 1.997 al log(4a) for a > 1000. We conclude that either a ≤ 1000 or (25) j 3 log α 2 < 0. Similar inequalities are valid in the case 2l = j 2 -k 2 not covered by the assumptions under which relation [START_REF] Schmidt | Diophantine Approximation[END_REF] has been derived.

Using the two methods just described, we conclude our study of linear forms in logarithms with tight upper bounds for the values of a and b for which the equations ( 8) have at least three solutions in positive integers. The bounds for a are derived by using Proposition 2.5. Lemma 4.3. Suppose that equations (8) have three solutions given by (10) for j 1 = 1 < j 2 < j 3 and k 1 = 1 < k 2 < k 3 . Then the following statements hold.

In the search of a third solution

Since the epoch-making paper of Baker and Davenport [START_REF] Baker | The equations 3x 2 -2 = y 2 and 8x 2 -7 = z[END_REF], 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 (k 2 ≤ 17), but not immediately after the first solution (k 2 ≥ 5).

For the last part of our proof of Theorem 1.1 we need large, explicit 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 j 3 . Using the suitable gap principle (either Proposition 3.5 or Proposition 3.6), one bounds from above 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 k 2 = 3 and j 2 ≤ 11, the domain where a third solution may exist is well beyond the reach of a direct search, so we modify the approach. We rewrite Λ 3 as a linear form in two logarithms. Proposition 4.2 yields a tight upper bound on a, which forces b to remain confined to a domain defined by the inequality [START_REF] Matveev | An explicit lower bound for a homogeneous rational linear form in logarithms of algebraic numbers, II (Russian)[END_REF]. It turns out that now we may successfully proceed to explore the entire search domain.

In the previous section we have established that no system (6) has three positive solutions if b = max{ a, b } is very large (see Lemma 4.3). Here we first show that the same statement is true if b is very small.

Λ 3 = log b -1 a -1 + j 3 log α 2 -k 3 log β 2 < -2j 3 log α + 0.882.
A variant of Davenport's lemma due to Pethő [START_REF] Dujella | A generalization of a theorem of Baker and Davenport[END_REF] proved itself most suitable in our context. We record it for handy reference.

Lemma C ([12]

). Let A, B, θ, µ be positive real numbers and M a positive integer. Suppose that P/Q is a convergent of the continued fraction expansion of θ such that Q > 6M . Put ε = ||µQ|| -M ||θQ||, where || • || denotes the distance from the nearest integer. If ε > 0, then there exists no solutions to the inequalities

0 < jθ -k + µ < A • B -j in integers j and k with log(AQ/ε) log B ≤ j ≤ M.
We have applied this result for Λ 3 / log α 2 , choosing M = 10 20 as an upper bound for j 3 . The output of a code written in Pari/GP [START_REF] Batut | User's guide to PARI/GP, version 2.1[END_REF] is that the inequality [START_REF] Siegel | Über einige Anwendungen diophantischer Approximationen[END_REF] does not hold for j 3 > 167. Our gap principles and the obvious inequality 4b -3 < β 2 imply that (a, b) ∈ { (2, 60), [START_REF] Baker | Linear forms in the logarithms of algebraic numbers[END_REF]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 (U k ) k , precisely, k 2 ≤ 17. Namely, from Matveev's theorem, Proposition 3.6 and inequalities [START_REF] Ribenboim | The book of prime number records[END_REF] it readily follows that b < 1000 if k 2 ≥ 19, so we may conclude by Lemma 5.1.

We employ Lemma C to show that k 2 = 3 may occur only for systems with two positive solutions. Since z 2 = 4b -1 = U j2 with a in a range that shrinks very rapidly with increasing j 2 , 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 5.1, we reach the conclusion that if the equations ( 8) have three solutions in positive integers, given by relation [START_REF] Cipu | On the number of solutions of simultaneous Pell equations[END_REF] for

j 1 = 1 < j 2 < j 3 and k 1 = 1 < k 2 < k 3 , then k 2 = 3.
To complete the proof of Theorem 1.1, we have to show that k 2 cannot have an odd value between 4 and 18. The volume of explicit computations needed to confirm that k 2 = 5 is not possible is largely reduced by the theoretical result below. Lemma 5.2. Suppose that equations (8) have three solutions given by [START_REF] Cipu | On the number of solutions of simultaneous Pell equations[END_REF]

for j 1 = 1 < j 2 < j 3 and k 1 = 1 < k 2 < k 3 . If k 2 = 5, then j 2 ≥ 21.
Proof. The present hypotheses yield z 2 = 16b 2 -12b + 1. We verify that for j 2 = 9, 13 and 17, 4z 2 + 5 = (8b -3) 2 is between two consecutive perfect squares. Namely, we show that for these values of j 2 there exist equalities (8b -3) 2 = A 2 j2 + P j2 = (A j2 + 1) 2 -Q j2 with A j2 , P j2 and Q j2 certain polynomials in a with positive leading coefficient, which take integral values for integral a, and all real roots of P j2 and Q j2 are smaller than 2.

Here are the relevant polynomials:

A 9 = 32a 2 -28a + 2, P 9 = 48a 2 -48a + 5, Q 9 = 16a 2 -8a, A 13 = 128a 3 -176a 2 + 59a -3, P 13 = 32a 3 -57a 2 + 18a, Q 13 = 224a 3 -295a 2 + 100a -5,
A 17 = 512a 4 -960a 3 + 556a 2 -101.5a + 2 for a even, 512a 4 -960a 3 + 556a 2 -101.5a + 2.5 for a odd,

P 17 = 816a 4 -1564a 3 + 913.75a 2 -170a + 5
for a even, 304a 4 -604a 3 + 357.75a 2 -68.5a + 2.75 for a odd,

Q 17 = 208a 4 -356a 3 + 198.25a 2 -33a
for a even, 720a 4 -1316a 3 + 754.25a 2 -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 U j2 (a) = U k2 (b), where U n (x) are Chebyshev polynomials (cf. relation [START_REF] Cipu | On the number of solutions of simultaneous Pell equations[END_REF]). For each of the seven possible values of k 2 we are examining, the corresponding j 2 is subject to restrictions given by Lemma 2.4 (and Lemma 5.2 if k 2 = 5), while a and b are confined to domains described by Lemma 4.3. Moreover, an upper bound for j 2 is obtained from Lemma 5.1 in conjunction with Lemma 2.5. The procedure we used is as follows. For each admissible pair of indices (k 2 , j 2 ) the search of solutions starts by putting a = 2 and ua := U j2 (a). If there exists a positive integer b > a such that ua = U k2 (b), from Lemma 2.5 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 mod 4. If this is the case, check if U k2 (b) coincides with ua. Then increase b and compute ub := U k2 (b). Do the same if b * ≡ 1, 2 (mod 4). By Lemma 2.5, the only possible value of a such that ub = U j2 (a) is either (a * + 2)/4 or (a * + 3)/4, with a * := ub 2/(j2-1) . If one of these values is integral, check whether ub and U j2 (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, k 2 , j 2 such that U j2 (a) = U k2 (b) and 5 ≤ k 2 ≤ 17. The computations allow us to conclude the proof of Theorem 1.1.

Proof of Theorem 1.2

As indicated in the Introduction, we shall conveniently modify the strategy developed in the proof of Theorem 1.1.

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 equations ( 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 1.2. The explicit computations which permit to conclude the proof will be described at length because they involve new ideas.

Let us consider the pair of Diophantine equations

x 2 -ay 2 = 1, z 2 -bx 2 = 1, a, b ≥ 2,
having a solution in positive integers. By a reasoning similar to the proof of Lemma 2.1 one sees that the smallest solution has the property that each component divides the corresponding component of any other solution of [START_REF] Bennett | On the number of solutions of simultaneous Pell equations[END_REF]. Hence, these equations have as many positive solutions as the system

(28) x 2 -(m 2 -1)y 2 = 1, z 2 -bx 2 = 1, m, b ≥ 2.
Let (m, 1, n) be the smallest solution in positive integers for these equations and denote (29)

α := m + m 2 -1, β := n + m √ b.
Then any positive solution (x, y, z) of the equations ( 28) satisfies

(30) x = α j + α -j 2 = β k -β -k 2 √
b for some positive odd integers j and k.

Since α, α -1 , β and β -1 are positive, it readily follows (31)

β k > α j √ b.
Hence, if j > 1 and k > 1, then j > k.

From inequality [START_REF] Walsh | On two classes of simultaneous Pell equations with no solutions[END_REF] it also results that the linear form in logarithms

(32) Λ := k log β -j log α -log √ b
is positive, while equation [START_REF] Walsh | On integer solutions to x 2dy 2 = 1, z 2 -2dy 2 = 1[END_REF] implies

Λ < log(1 + α -2j ) -log(1 -β -2j ) < α -2j + β 2-2k β 2 -1 < 1.01α -2j .
Hence, [START_REF] Yuan | Simultaneous Pell equations[END_REF] log Λ < -2j log α + 0.01.

Despite its innocuous appearance, the forthcoming result has spectacular consequences on the volume of explicit computations needed to complete the proof of Theorem 1.2. Lemma 6.1. Assume (x, y, z) is a positive solution of the Diophantine equations [START_REF] Thue | Über Annäherungswerte algebraischer Zahlen[END_REF] for which relation [START_REF] Walsh | On integer solutions to x 2dy 2 = 1, z 2 -2dy 2 = 1[END_REF] holds with j, k > 1. Then k ≡ 1 (mod 4) and j + k ≥ 4m 2 . Moreover, if j ≡ 1 (mod 4), then j ≥ 4m 2 + k ≥ 4m 2 + 5.

Proof. It is easy to see that

β k -β -k β -β -1 ≡ k (mod 4m 2 ),
α j + α -j α + α -1 ≡ (-1) (j-1)/2 j (mod 4m 2 ), and therefore k ≡ (-1) (j-1)/2 j (mod 4m 2 ). For j ≡ 1 (mod 4) one gets (-1) (j-1)/2 j = j and j ≡ k (mod 4m 2 ), so that k ≡ 1 (mod 4) and j ≥ 4m 2 + k ≥ 4m 2 + 5. For j ≡ -1 (mod 4) one has (-1) (j-1)/2 j ≡ 1 (mod 4) and k ≡ -j (mod 4m 2 ), whence 2j > j + k ≥ 4m 2 . Suppose that the system (28) has three positive solutions (x i , y i , z i ) (i = 1, 2, 3). Then (34)

x i = α ji + α -ji 2 = β ki -β -ki 2 √
b for some positive odd integers j i and k i (i = 1, 2, 3) with 1 = j 1 < j 2 < j 3 and 1 = k 1 < k 2 < k 3 . By a reasoning similar to that employed in Section 3 we get that either j 2 | j 3 and k 2 | k 3 or j 3 = 2qj 2 + σ and k 3 = 2q 1 k 2 + σ for some positive integers q > q 1 , and σ either 1 or -1.

Here is the gap principle for the equations ( 28).

Lemma 6.2. Suppose that the simultaneous Pell equations (28) have three solutions (x i , y i , z i ) (i = 1, 2, 3) in positive integers, given by relation (34) for some odd integers j i and k i (i = 1, 2, 3) with 1 = j 1 < j 2 < j 3 and 1 = k 1 < k 2 < k 3 . Then Proof. Let us first examine the case j 2 divides j 3 . Then x 3 is multiple of x 2 and there exist positive odd integers q ≥ 3 and q 1 ≥ 3 such that j 3 = qj 2 and k 3 = q 1 k 2 . As in the proof of Lemma 6.1 one finds x 3 x 2 ≡ q 1 (mod 4x 2 2 ),

x 3

x 2 ≡ (-1) (q-1)/2 q (mod 4x 2 2 ), so that q 1 ≡ (-1) (q-1)/2 q (mod 4x 2 2 ) and q > 2x 2 2 . Since (35)

x 2 = (β k2 -β -k2 )m β -β -1 > (β 4 + β 2 + 1)β k2-5 m > 2β k2-1 ,
one gets j 3 > 16 j 2 β 2k2-2 , a much stronger inequality that the desired one.

In the case j 3 = 2qj 2 + σ, k 3 = 2q 1 k 2 + σ, with σ = ±1, one finds on the one hand

x 3 ≡ (-1) q m -2σ(m 2 -1)qx 2 y 2 (mod 2x 2
2 ), and on the other hand

x 3 ≡ σm + 2nq 1 x 2 z 2 (mod 2x 2
2 ). As 2m < x 2 , it results σ = (-1) q and nq 1 z 2 ≡ -q(m 2 -1)y 2 (mod x 2 ). This and z 2 2 ≡ 1 (mod x 2 2 ), (m 2 -1)y 2 2 ≡ -1 (mod x 2 2 ) imply n 2 q 2 1 ≡ -(m 2 -1)q 2 (mod x 2 ). Thus, by inequality (35), one has 2 max{ n 2 q 2 1 , m 2 q 2 } > x 2 > (β 4 + β 2 + 1)β k2-5 m. For mq ≥ nq 1 one gets

q 2 > (β 4 + β 2 + 1)β k2-5 2m = (β 4 + β 2 + 1)β k2-5 √ b β 2 -1 ≥ (β 4 + β 2 + 1)β k2-5 √ 2 β 2 -1 .
Since the rightmost term in this chain of inequalities is greater than 8β k2-3 (because the largest real root of the polynomial X 4 -4 √ 2X 3 +X 2 +4 √ 2X +1 is less than 5.5, while β ≥ 3+2 √ 2 > 5.8), one deduces j 3 ≥ 2 q j 2 -1 > 4 √ 2j 2 β (k2-3)/2 > 5.63j 2 β (k2-3)/2 . It remains to examine the possibility that mq < nq 1 . By direct computation one finds

q 2 1 > m(β 4 + β 2 + 1)β k2-5 2n 2 = 2m(β 4 + β 2 + 1)β k2-3 (β 2 + 1) 2 > γβ k2-3 ,
with γ = 3.88 for m = 2, γ = 5.94 for m = 3, and γ = 7.94 for m ≥ 4. This implies at once

j 3 ≥ 2 q j 2 -1 ≥ 2(q 1 + 2)j 2 -1 > τ j 2 β (k2-3)/2 ,
with τ as specified in the statement of our gap principle.

The second phase of the proof of Theorem 1. Table 1 gives information on numerical characteristics associated to putative solutions of a Diophantine system of the type [START_REF] Thue | Über Annäherungswerte algebraischer Zahlen[END_REF] with three positive solutions. Data referring to log β and log j 3 are derived by combining inequality (36) with Lemma 6.2. The lower bounds on j 2 and the upper bounds on m result from the information contained in the previous columns of Table 1 and the powerful Lemma 6.1. Once β is bounded from above, one may use relation [START_REF] Walsh | On two classes of simultaneous Pell equations with no solutions[END_REF] to obtain upper bound on j 2 . The estimation for n is gotten with the help of the inequality 2n < β + 1. 1. Necessary condition in order that equations [START_REF] Thue | Über Annäherungswerte algebraischer Zahlen[END_REF] have three positive solutions From β < 2n < β + 1 and log β < 6.5 it follows n < 350. Thus, for k 2 ≥ 17 one has n < 350. With the help of Lemma C we find that for n at most 1000 one has j 3 < 46 for any system [START_REF] Thue | Über Annäherungswerte algebraischer Zahlen[END_REF] with three positive solutions. Then Lemma 6.2 implies β < 1.1, in contradiction with β > 4 √ 2. At this point of the proof we know that, if a system (28) has at least three positive solutions, then either k 2 = 5 and m = 2, 3, 4, or k 2 = 9 and m = 2, 3.

k 2 = m = log β < log j 3 < j 2 ≥ j 2 ≤ n < 5
If k 2 = 9, then we run again our program implementing Davenport's lemma, this time for m = 2, n odd and n < 41000, m = 3, n ≡ ±1 (mod 9) and n < 28000.

The outcome is that the index j 3 of the third solution is at most 49. This in conjunction with our gap principle rejects all candidates.

To conclude the proof of Theorem 1.2, we still have to examine the possibility that k 2 = 5.

Let us denote

T t (m) := 4(α t + α -t ) α + α -1 + 5 for positive odd t. Since β 5 -β -5 2 √ b = m(16n 4 -12n 2 + 1),
the equation ( 30) written for j 2 and k 2 is equivalent to

(37) T k2 (m) = (8n 2 -3) 2 .
Thus, the proof of Theorem 1.2 is complete as soon as we show that none of the numbers T t (m) (t ≡ 1 (mod 4), m = 2, 3, 4) is a perfect square. The values T t (m) (t odd) are given by the recurrence relation

T t+2 (m) = (4m 2 -2) T t (m) -5 -T t-2 (m) + 10, T 1 (m) = 1, T 3 (m) = 2m 2 -3.
For k 2 an odd integer in the range specified in Table 1 we examine for each m = 2, 3, 4 the values T k2 (m) modulo several primes p. We remove from the list of admissible k 2 those values for which the Lagrange symbol of T k2 (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 T t (m), (t ≡ 1 (mod 4), m = 2, 3, 4) is a perfect square.

Proof of Proposition 4.2

Now we are ready to prove Proposition 4.2. Before proceeding to the proof, we need to compute upper and lower bounds for the parameter t.

Put ∆ = L 2 + 4ΩW , the condition on t implies t ≥ t 0 , where

√ t 0 = L + √ ∆ 2Ω , t 0 = L 2 + ∆ + 2L √ ∆ 4Ω 2 = L 2 2Ω 2 + W Ω + L 2Ω L 2 Ω 2 + 4W Ω ≥ L 2 Ω 2 + W Ω , with 8 9λ ≥ 1 3 λ -1 (2h + λ) (2h + λ) -(h + λ) ≥ L Ω = 1 3 L λL -(h + λ) ≥ 1 3 λ -1 2(h + λ) 2(h + λ) -(h + λ) = 2 3λ , since ∂(L/Ω)/∂L < 0 and 1 + 2h/λ ≤ L ≤ 2(1 + h/λ) when h ≥ 1.5λ. Moreover, W satisfies W Ω = 1 4 L a 2 + 1 a 1 1 λL -λ -h ≥ 1 4a 1 (λL -λ -h) + 1 2a 2 λ ,
and also

W Ω ≤ 1 4 1 + 2h/λ a 2 + 1 a 1 1 h = 1 2a 2 λ + 1 a 1 + 1 a 2 1 4h ≤ 1 2λ + 1 12λ , if λ ≤ 1, 2
+ L a2 + 1 a1 4(λL -λ -h) La 1 a 2 , so that tLa 1 a 2 ≥ 4a 1 a 2 L 9λ 2 + a 1 L 2λ + a 2 2λ = ψ(L),
say.

Clearly, ψ increases with L and, using the fact that a 1 a 2 ≥ 20λ 2 , it is easy to check that ψ(5) > 54.

We suppose that α 1 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 [START_REF] Laurent | Formes linéaires en deux logarithmes et déterminants d'interpolation[END_REF]. For the convenience of the reader we keep the numbering of the formulas as in [START_REF] Laurent | Formes linéaires en deux logarithmes et déterminants d'interpolation[END_REF], except that when there is some change the new formula is denoted by (7.i) .

Put

L = 2 + 2h/λ , K = 1 + tLa 1 a 2 , thus L ≥ 5 and K ≥ 55, (7.1) R 1 = L, S 1 = 1, R 2 = 1 + √ tLa 2 , S 2 = 1 + √ tLa 1 . By Liouville inequality, log |Λ| ≥ -D log 2 -D b 1 h(α 1 ) -D b 2 h(α 2 ) ≥ -D log 2 -1 2 (b 1 a 1 + b 2 a 2 ) = -D log 2 -1 2 b a 1 a 2 , where b = b 1 a 2 + b 2 a 1 .
We consider separately the two possible cases:

b ≤ 2λtL 2 , or b > 2λtL 2 .
In the first case, Liouville inequality implies Notice that

R = R 1 + R 2 -1 ≤ L + √ tLa 2 and S = S 1 + S 2 -1 ≤ 1 + √ tLa 1 . This shows that max{LR, LS} ≤ L 2 (1 + a 2 √ t ) < L 2 (1 + 1.5λ -1 a 2 ) = L 2 1 a 2 + 1.5 λ a 2 < a 1 a 2 L 2 2λ .
As we may, suppose that log |Λ| ≤ -λtL 2 a 1 a 2 . From L ≥ 5, a 1 ≥ 4 and a 2 ≥ 2.7λ we get L 2 a 1 a 2 /λ ≥ 270, so that We replace this condition by the two conditions Φ > 0, Θ > 0, where Φ 0 ≥ Φ + Θ. The term Φ is the main one, Θ is a sum of residual terms. As indicated in [START_REF] Laurent | Formes linéaires en deux logarithmes et déterminants d'interpolation[END_REF], 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, a 1 , and a 2 .

max LR|Λ| 2b 2 , LS|Λ| 2b 1 ≤ (1 + a 2 √ t )L 2 |Λ| 2 ≤ L 2 a 1 a 2 4λ e -4L
As in [14, Lemme 8] This proves that Φ > 0 provided that tΩ -L √ t -W ≥ 0.

We have Θ ≥ h -log(tL 2 a 1 a 2 ) -D log L √ e/(2π) + ε(tLa 1 a 2 -1). To prove that Θ ≥ 0, rewrite (7.22) We conclude by proving that Θ 0 and Θ and an elementary numerical verification shows that Θ 1 is positive for K ≥ 55, which holds as we previously saw.

Remark. We have proved that, under the hypotheses of our result, we can choose ε = 0.0262. More generally the condition on ε is ε K -log K + log 16π 9 √ e + f (K) ≥ 0 for all K ≥ K 0 , where K 0 = t 0 La 1 a 2 .

We apply the Corollary of Theorem 2 of [18]: Let a 1 , a 2 , h, t be real positive numbers, and ρ a real number, e 3/2 ≤ ρ ≤ e 3 . Put λ = log ρ, χ = h/λ and suppose that χ ≥ χ 0 for some number χ 0 ≥ 0 and that x -1 .

(
Then the condition on h in Theorem 2 of [START_REF] Mignotte | A corollary to a theorem of Laurent-Mignotte-Nesterenko[END_REF] is

h ≥ D log b 1 a 2 + b 2 a 1 + log λ + f ( K 0 ) + 0.023.
Here we can take χ 0 = 3 and it is easy to check that our present hypotheses imply K 0 > 195. Since f (x) < 1.285 for x ≥ 195, we get the result.

We notice that c 1 in the statement of Proposition 7.1 is a decreasing function of χ, so for χ ≥ 1 + √ 3 we have

c 1 ≤ 1 2a 1 a 2 .
We also have (notice that the hypotheses of the above Proposition imply χ ≥ 3 and λ + h ≥ 9). Using these remarks and simplifying the expression of C 0 using v ≥ 16 we get a simpler estimate.

c 2 ≤ 1.11λ

Then ( 4 ) (x 1 2

 412 , y 1 , z 1 ) = (m, n(l, m), , y 2 , z 2 ) = α 2l + α -2l 2 , 2 n(l, m) 2 -1, 2 n(l, m)are two positive integral solutions to (2) with c = m 2 -1 and d = n(l, m) 2 -1.

x 2 - 2 - 1 ) 2l 2 ,

 2212 (m 2 -1)z 2 = y 2 -n(l, m) 2 -1 z 2 = 1 and the system of hyperbolic Diophantine equationsax 2 -(a -1)z 2 = b(l, a)y 2 -b(l, a) -1 z 2 = 1 have two solutions in positive integers, given by (x, y, z) = (m, n(l, m), 1), (x, y, z) = (m + √ m 2 -1 ) 2l + (m -√ m 2n(l, m) 2 -1, 2n(l, m) l,a) -3, 4b(l, a) -1 , respectively. Call (a 1 , b 1 , c 1 , d 1 ) an equivalent form of (a, b, c, d) if there are positive integers a 0 , b 0 , c 0 , d 0 such that (a 1 , b 1 , c 1 , d 1 ) = (a/a 2 0 , b/b 2 0 , c/c 2 0 , d/d 2 0 ). Conjecture 1 ([33, Conjecture 1.1]). If the simultaneous Diophantine equations (6) have at least two solutions in positive integers, then their coefficients are given by

Theorem 1 . 1 .

 11 If a > 1 and b > 1 are distinct positive integers, then the Diophantine equations[START_REF] Batut | User's guide to PARI/GP, version 2.1[END_REF] have at most two solutions with x, y, z > 0.

Corollary 3 . 4 .

 34 aq 2 < bq 2 1 .

Lemma 4 . 1 .

 41 If α, β and (b -1)/(a -1) are multiplicatively dependent, then the equations (8) have at most one positive solution.

  20.2 + log 3 5.5 16 log(4e) = 180 • 16 4 e 4 20.2 + 5.5 log 3 + log 16 + log(log 4 + 1) < 1.925 • 10 10 .

For j 2 = 7 this yields a < 1 . 8 • 55 1 + 4 .5 log a λ 2 <

 218142 99(1 + 1.366 log a) 2 , and therefore a < 22000, b < 6.4 • 10 14 . For j 2 = 11 one obtains a < 71(1 + 1.366 log a) 2 , a < 14000, b < 1.4 • 10 23 .

k 2 = 3 :

 23 a < 22000 for j 2 = 7, a < 15000 for j 2 = 11, b < 3.9 × 10 23 , k 2 = 3 : a < 105000 for j 2 = 15, a < 23000 for j 2 ≥ 19, b < 4 × 10 38 , k 2 = 5 : a < 8.9 × 10 15 for j 2 = 9, a < 3 × 10 10 for j 2 = 13, b < 3.4 × 10 32 , k 2 = 5 : a < 3.5 × 10 7 for j 2 = 17, a < 7.5 × 10 5 for j 2 ≥ 21, b < 6.8 × 10 31 , k 2 = 7 : a < 7.4 × 10 8 for j 2 = 11, a < 1.3 × 10 6 for j 2 = 15, b < 1.6 × 10 15 , k 2 = 7 : a < 40000 for j 2 = 19, a < 4300 for j 2 ≥ 23, b < 8.6 × 10 14 , k 2 = 9 : a < 1.7 × 10 6 for j 2 = 13, a < 30000 for j 2 ≥ 17, b < 4.1 × 10 9 , k 2 = 11 : a < 57000, b < 7.7 × 10 6 , k 2 = 13 : a < 7000, b < 2 × 10 5 , k 2 = 15 : a < 1800, b < 20000, k 2 = 17 : a < 600, b < 3000.

Lemma 5 . 1 .

 51 There is no pair of integers (a, b) with 1 < a < b ≤ 2000 for which the equations ax 2 -(a -1)z 2 = by 2 -(b -1)z 2 = 1 have three positive solutions.Proof. We shall argue by reductio ad absurdum. Suppose that there exists a pair (a, b) of integers such that 1 < a < b ≤ 2000 and the equations ax 2 -(a -1)z 2 = by 2 -(b -1)z 2 = 1 have three positive solutions. Keeping the notation used up to now, we have[START_REF] Siegel | Über einige Anwendungen diophantischer Approximationen[END_REF]

j 3 >

 3 τ j 2 β (k2-3)/2 , with τ = 3.93 for m = 2, τ = 4.87 for m = 3, and τ = 5.63 for m ≥ 4.

  2 consists of applying Matveev's theorem for the linear form in logarithms Λ := k 3 log β -j 3 log α -log √ b with the choice α 1 := β, α 2 := α, α 3 := √ b, b 1 := k 3 , b 2 := -j 3 , b 3 := -1. If these algebraic numbers are multiplicatively dependent, then the equations (28) have at most one common solution in positive integers (cf. Lemma 4.1). In the notation introduced in the statement of Theorem A we have D = 4, χ = 1, h(β) = 1 2 log β, h(α) = 1 2 log α, h( √ b) = log b, and therefore one may choose A 1 := 2 log β, A 2 := 2 log α, A 3 := 4 log b. Since one has k 3 log β > max{ j 3 log α, log √ b } (by inequality (31)) and k 3 ≥ 11 (by Lemma 6.1), the choice B := k 3 is admissible. As C 1 < 1.925 • 10 10 , the conclusion of Matveev's theorem is log Λ > -4.928 • 10 12 log α log β log b log(38.92 k 3 ). Comparison of this inequality with relation (33) written for the current Λ results in j 3 < 2.464 • 10 12 log β log b log(38.92 k 3 ). Recall that j 3 > k 3 and β > 2m √ b ≥ 4 √ b, so that (36) j 3 < 4.928 • 10 12 log β log β 2m log(38.92 j 3 ).

Φ 0 =

 0 K(L -1) log ρ -(D + 1) log N -D(K -1) log b -gL (Ra 1 + Sa 2 ) > 0, when b > 2λtL 2 . Notice that the condition b > 2λtL 2 implies h ≥ D log(2λ 2 tL 2 ) + f (K)

  as Θ = Θ 0 (D -1) + Θ 1 , whereΘ 0 = log(λb ) + f (K) -log L + log 2π √ e ,andΘ 1 = εK -log K -2 log L + log 2π √ e + log(λb ) + f (K).

Proposition 7 . 1 .Λ = b 2 log α 2 -b 1 log α 1 ,

 7121 Consider the linear form in two logarithms where b 1 and b 2 are positive integers. Suppose that α 1 and α 2 are multiplicatively independent. Put D = [Q(α 1 , α 2 ) : Q] / [R(α 1 , α 2 ) : R].

Corollary 7 . 2 . 2 , c 2

 7222 With the notation and hypotheses of the above proposition, we have the lower bound log |Λ| ≥ -(C 0 + c 1 + c 2 )(λ + h) 2 a 1 a 2 , where = 0.177 • (a 1 a 2 ) -9/10 .

  2.7λ 2 + 2 16.2λ 2 , if λ ≥ 1, because of our hypotheses on a 1 , a 2 , and h. Thus we always have

	It is easy to check that the previous inequalities imply
		√	t 0 ≤	1.48 λ	.
	Hence, t 0 < 2.2 λ -2 and we can always choose t satisfying
		4 9λ 2 ≤ t ≤	2.2 λ 2 .
	Then we have
	tLa 1 a 2 ≥	4 9λ 2
		W Ω	≤	7 8.1λ 2 .

  |Λ| ≥ -λtL 2 a 1 a 2 -λ(L -0.5) -log L 2 (1 + a 2 √t ) and Proposition 4.2 follows. Now we verify that condition (III) is indeed satisfied. We have to prove that

				2 a1a2/(9λ) ,
	because (4/9)λ -2 ≤ t ≤ 2.2λ -2 . Therefore we have max LR|Λ| 2b 2 , LS|Λ| 2b 1	< 10 -10 .
	Thus, which implies	|Λ | ≤ |Λ| × L 2 (1 + a 2	√ t ),
	log

  (7.17) and(7.18) we see that Φ 0 ≥ Φ + Θ, where tLa 1 a 2 < K ≤ 1 + tLa 1 a 2 , hence Φ La 1 a 2

	By > tΩ -L	√ t -W,
	where		Ω = 3(L -1)λ -3h, W =	3 4	L a 2	+	1 a 1	.
	we get b 1 a 2 + b 2 a 1 which follows from the condition (7.17) log b ≤ log + log λ -	log (2πK/ K -1 √ e )	+ f (K) ≤	h D	-	ε D	-	log (2πK/ K -1 √ e )	,
	Here we have
	gL(Ra 1 + Sa 2 ) ≤	1 4	-	KL 12RS	L(Ra 1 + Sa 2 ) =	L(Ra 1 + Sa 2 ) 4	-	KL 2 12	a 1 S	+	a 2 R	,
	which implies (7.18) gL(Ra 1 + Sa 2 ) ≤	L 4	(a 1 L + a 2 + 2a 1 a 2 L	√ t ) -	KL 6 √ t	≤	L 4	(a 1 L + a 2 ) +	√ tL 2 a 1 a 2 3	.
	Put (7.21)	Φ = K(L -1)λ -Kh -	√ tL 2 a 1 a 2 3	-	L(a 1 L + a 2 ) 4
	and
	(7.22)

h ≥ D log b + log λ + f (K) + ε. Θ = ε(K -1) + h -D log L √ e/(2π) -log(KL).

  1 are both positive. Since b > 2tλL 2 and t ≥ 4/(9λ 2 ), we have log(λb ) > log(2tλ 2 L 2 ) > log(8/9) + 2 log L,

	and this implies that
	Θ 0 > log(8L/9) + f (K) + log(2π/	√	e) > log	8L 9	+	3 2	+ log	3 4	+ log	2π √ e	> 0.
	This also implies that	Θ 1 ≥ εK -log K + log	8 9	+ log	2π √ e	+ f (K).
	Thus,
		Θ 1 ≥ 0.0262K -log K + log	16π 9 √ e		+ f (K)

  Putv = 4χ + 4 + 1/χ, A = max{a 1 , a 2 }. Then we have the lower boundlog |Λ| ≥ -(C 0 + c 1 + c 2 )(λ + h) 2 a 1 a 2 ,Proof. The only difference with Theorem 2 of[START_REF] Mignotte | A corollary to a theorem of Laurent-Mignotte-Nesterenko[END_REF] is the definition of the term h. Put

	38) (39) and a where C 0 = K 0 := h ≥ max 7.5, 3λ, D log 1 λ 3    2 + 1 2χ(χ + 1)   1 3 + a 2 b 1 c 1 = λ(1.5λ + 2h) (λ + h) 2 a 1 a 2 , c 2 = + b 2 a 1 1 9 + 4λ + log λ + 1.285 + 0.023 , 3v 1 a 1 + 1 a 2 + 32 √ 2(1 + χ) 3/2 3v 2 √ a 1 a 2 1.11λ log A(2λ + 2h) 2 (λ + h) 2 a 1 a 2 . 1 λ √ 2 + 2χ 0 3 + 2(1 + χ 0 ) 9 + 2λ 3 1 a 1 + 1 a 2 + 4λ √ 2 + χ 0 3 √ a 1 a 2 2 a 1 a 2      and f (x) = log 1 + √ x -1 √ x x -1 + log x 6x(x -1) + 3 2 + log 3 4 + log x x-1	2

i ≥ max 4, λ, ρ | log α i | -log |α i | + 2Dh(α i ) , (i = 1, 2), a 1 a 2 ≥ 100.

  log a 1 a 2 (λ + h) 2 (λ + h) 2 a 1 a 2 + h) 4/5 (a 1 a 2 ) 9/10 < 0.177 • (a 1 a 2 ) -9/10

				=	1.11λ (λ + h) √ a 1 a 2	9/5	log a 1 a 2 (λ + h) 2 (λ + h) √ a 1 a 2 1/5 ,
	whence
	c 2 ≤	11.1λ e (λ + h) √ a 1 a 2	9/5 =	e(1 + χ)(λ	11.1

The first author has been partially supported by the CEEX Program of the Romanian Ministry of Education and Research, contract CEx05-D11-11/2005.