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Diophantine equations
Mihai Cipu, Maurice Mignotte
To cite this version:
Mihai Cipu, Maurice Mignotte. On the number of solutions of simultaneous hyperbolic Diophantine
equations. 2006. �hal-00129725�
DIOPHANTINE EQUATIONS
MIHAI CIPU AND MAURICE MIGNOTTE
Abstract. We prove that, for positive integersa,b,candd withc6= d,a > 1,b > 1, the number of simultaneous solutions in positive integers toax2−cz2= 1,by2−dz2= 1 is at most two. This result is the best possible one. We prove a similar result for the equationsx2−ay2= 1, z2−bx2= 1.
1. Introduction
The number of solutions of the simultaneous Diophantine equations (1) ax2−cz2=δ1, by2−dz2=δ2
was a question of constant interest in the last century. It is known already since A. Thue [28] and C.L. Siegel [27] that the equations (1) have finitely many solutions whencδ26=dδ1. Their works do not provide estimations for the number of solutions. Using the methods developed by W.M.
Schmidt [25, 26], H.P. Schlickewei [24] proved that the number of integer solutions to
(2) x2−cz2= 1, y2−dz2= 1
is at most 4×8278.Masser and 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. Bennett [7], 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 anyc6=d. A recent result of Yuan [32] shows that there are at most finitely many exceptions to this conjecture. To be precise, if max{c, d} ≥1.4×1057, 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 [8]. (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. Forlandmintegers greater than 1, set α=m+√
m2−1 and
(3) n(l, m) =α2l−α−2l
4√
m2−1. Then
(4) (x1, y1, z1) = (m, n(l, m),1) and
(5) (x2, y2, z2) =
α2l+α−2l
2 ,2n(l, m)2−1,2n(l, m)
are two positive integral solutions to (2) withc=m2−1 andd=n(l, m)2−1.
The proofs in [32, 8, 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
Mathematics Subject Classification 2000Primary: 11D09, Secondary: 11D45, 11J20, 11J86 Key words: simultaneous Pell equations, linear forms in logarithms
The first author has been partially supported by the CEEX Program of the Romanian Ministry of Education and Research, contract CEx05-D11-11/2005.
1
Diophantine approximation is very instrumental in obtaining all solutions of instances of (1).
Since the ground-braking 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 [1] (see, also, [2]) that (2) have 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 [33, 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 Y2 =X3+ (c+d)X2+cdX. K. Ono [21] 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 [29], 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, [9, 13, 20]). Other papers, like [22, 31], to name but a few, deal in an ad hoc manner with systems (1) or (6) whose coefficients are subject to quite special restrictions. There are also conditional results, assuming the ABC conjecture [31].
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 (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 [33] put forward a conjecture aiming to describe when there are at least two solutions in positive integers to
(6) ax2−cz2= 1, by2−dz2= 1.
Yuan claims that the coefficients of systems with two solutions are obtained as follows.
For integersl >1,m >1, anda >1, put n(l, m) =(m+√
m2−1 )2l−(m−√
m2−1 )2l 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 thatn(l, m) andb(l, a) are positive integers and the simultaneous Pell equations x2−(m2−1)z2=y2− n(l, m)2−1
z2= 1 and the system of hyperbolic Diophantine equations
ax2−(a−1)z2=b(l, a)y2− b(l, a)−1 z2= 1 have two solutions in positive integers, given by
(x, y, z) = (m, n(l, m),1), (x, y, z) = (m+√
m2−1 )2l+ (m−√
m2−1 )2l
2 ,2n(l, m)2−1,2n(l, m)
!
and
(x, y, z) = (1,1,1), (x, y, z) =
(√ a+√
a−1 )l+ (√ a−√
a−1 )l 2√
a ,4b(l, a)−3,4b(l, a)−1
,
respectively. Call (a1, b1, c1, d1) an equivalent form of (a, b, c, d) if there are positive integersa0, b0,c0, d0 such that (a1, b1, c1, d1) = (a/a20, b/b20, c/c20, d/d20).
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
(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.
The first aim of this paper is to study the number of solutions for the pair of hyperbolic Dio- phantine equations (6) when a >1, b >1. Yuan [33] proved that ifc 6=dand max{a, b, c, d}>
1.16×1059then the system has at most two positive solutions. Here we prove that the same result holds regardless of the size of the coefficients.
Theorem 1.1. Ifa >1andb >1are distinct positive integers, then the Diophantine equations(6) have at most two solutions withx, 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 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 [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 intwologarithms. As a result of our deliberations, in Section 4 one obtains 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 equations (6). 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
(7) x2−ay2= 1, z2−bx2= 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. Letaandb 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 (7) 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 (7) has at most one solution in positive integers.
Our results illustrate the power of the method we developed for the study of numbers of so- lutions to simultaneous hyperbolic Diophantine equations. The approach is quite adequate for
the investigation of other families of equations (as already seen in [10]). We leave it on another occasion to present the details for the systems (1) not covered yet.
2. Properties of solutions
In this section we start to study the positive solutions of the equations (6). 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 (6), regardless of the number of solutions.
First we note that, when dealing with simultaneous Diophantine equations of the type (6), it is sufficient to consider only coefficientsc anddone less thanaandb, 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 anda >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) withx, y, z >1, one has x multiple of x0, y multiple of y0, andz multiple of z0.
b) The equations (6)have as many positive solutions as a system of the formAX2−(A−1)Z2= BY2−(B−1)Z2= 1.
Proof. Let α = u√a+v√c, resp. β = u0√
b+v0√
d, be the fundamental solution of the Pell equationax2−cz2= 1, resp. by2−dz2= 1. Put for oddj andk
Uj := αj−α−j
α−α−1 , Uk0 := βk−β−k
β−β−1 , Vj:= αj+α−j
α+α−1, Vk0:= βk+β−k β+β−1 .
Sincea≥2 andb≥2, there are odd integersj0≥1,k0≥1, such thatx0=u Vj0, y0=u0Vk00, z0= v Uj0 =v0Uk00. For any solution (x, y, z) of (6) one hasx =u Vj, y=u0Vk0, z =v Uj =v0Uk0 for certain odd integersj≥j0,k≥k0.
Suppose, by way of contradiction, thatz0 does not dividez. ThenUj0>1 and partsa) andb) of Lemma 2.2 yieldj0-j,k0-k. By Euclidean division one gets
j = 2qj0±r, k= 2tk0±s, 0< r < j0, 0< s < k0. From the identity stated in partc) of the next lemma one obtains
Uj∓Ur= 2Vqj0±rUqj0, Uk0 ∓Us0 = 2Vtk00±sUtk0 0. Having in view Lemma 2.2a), one gets
z=v Uj ≡ ±v Ur (mod z0) and z=v0Uk0 ≡ ±v0Us0 (mod z0).
Since 2v Ur < v Ur+1 ≤ v Uj0 = z0 and 2v0Us0 < v0Us+10 ≤ v0Uk00 = z0, one concludes that v Ur = v0Us0. Hence, (u Vr, u0Vs0, v Ur) is a positive solution of (6) whose third component is smaller thanz0. This contradicts the choice of (x0, y0, z0).
So it is true that z0 divides z. Then z=gz0 for a certain positive integerg and j0|j, k0|k.
Lemma 2.2a) implies x=ex0,y=f y0for suitable positive integerse, f. Hence, 1 =a x2−c z2=a e2x20−c g2z02=a x20e2−g2(a x20−1).
One concludes that (e, f, g) is a positive solution for the simultaneous Pell equations a x20X2−
(a x20−1)Z2=b y02Y2−(b y02−1)Z2= 1.
Lemma 2.2. a) If Uk 6= 1, thenUk |Un if and only ifk|n.
b) Ifk≥1, thenVk |Vn if and only ifn/k is an odd integer.
c) Ifn= 2t m±r, with 0≤r≤mandt≥0, then Un∓Ur= 2Vtm±rUtm.
Proof. The properties are well-known. They are proved in various places, for instance, in [23] and
[17].
So we will consider from now onc=a−1 and d=b−1 for some integersb > a≥2 and study the system of Diophantine equations
(8) ax2−(a−1)z2= 1, by2−(b−1)z2= 1.
Put
(9) α=√
a+√
a−1, β=√ b+√
b−1
and consider (x, y, z) a positive integer solution to (8). Then z appears in two linear recurrent sequences
(10) z=Uj=Uk0,
for some positive odd integersj andk, where
(11) Uj= αj−α−j
2√
a−1 and Uk0 =βk−β−k 2√
b−1 , j, k∈
N
. Note that(12) αj< βk < αj
rb−1 a−1. Indeed, fromα,α−1, β,β−1>0 it results
βk−αj= 2z √
b−1 +β−k−2z√
a−1−α−j
>2z √
b−1−√ a−1
−α−j ≥ 2z
α −α−j>0.
This in turn implies
βk
αj = 1−α−2j 1−β−2k
rb−1 a−1<
rb−1 a−1. From (10) one infers that the linear form in logarithms
(13) Λ = log
b−1 a−1
+jlog α2−klogβ2 is positive and bounded from above by
(14) −2 log (1−α−2j)<2α2−2j α2−1. Hence,
(15) log Λ<−2jlogα+ log 2α2
α2−1 <−2jlogα+ 0.882.
The observation stated in the next lemma allows one to obtain a companion inequality to (12).
Lemma 2.3.
1 + 1
2(a−1) − 1
2(b−1) − 3 16(a−1)2
β2
α2 < b−1 a−1 <
1 + 1
2(a−1) β2
α2. Proof. We shall use the elementary inequality 2√
t2+t <2t+ 1 valid for positivet.
The left inequality in our lemma is implied by (b−1)(16a2−24a+ 5)−8a2+ 16a−8
(2b−1)<8(b−1)2(a−1)(2a−1 + 2p
a2−a).
Expanding and squaring, one finds the equivalent form
0<(128a2−16a−4)b4+ (−256a3+ 224a2+ 80a+ 60)b3
+ (256a4−128a3−448a2+ 448a−245)b2+ (−512a4+ 1152a3−720a2−16a+ 150)b + 192a4−512a3+ 432a2−96a−25 =:f(b).
Routine computations yieldf00(b)>0,f0(b)> f0(a)>0,f(b)> f(a)>0 for a≥2.
The right inequality is equivalent to 2(b−1) 2a−1 + 2p
a2−a
<(2a−1) 2b−1 + 2p b2−b
,
which follows from 4(b−1)<2b−1 + 2√
b2−b.
From (12) and this lemma it follows an inequality which has an important role in the rest of the paper:
(16) β2k−2<
1 + 1
2(a−1)
α2j−2.
A direct consequence of it is thatj≥kif both j andk are greater than 1. Since the application t7→(tj−t−j)/(t−t−1) (j >1) is increasing fort >1, one concludes from equation (10) that, if k >1 andj >1, thenj > 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 (8). If z = Uj =Uk0, with j > k, then j and k are congruent modulo 4. If j ≡ ±3 (mod 8), then8 divides a(j−k). Ifj≡ ±1 (mod 8), then8divides b(j−k).
Proof. The valuesUj,Uk0 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 2.5. For any t≥2one has (4a−3)t< U2t+1<(4a−1)t. Consequently, bU2t+11/t c= 4a−2or4a−3.
Proof. The first inequality results fromU2t+1=α2t+α2t−1+. . .+α−2t because bothαandα−1 are positive, andα2>4a−3. The upper bound forUt+1is obtained by induction, sinceU3= 4a−1 andU2t+3= (4a−2)U2t+1−U2t−1<(4a−2)U2t+1 for any positive indext.
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 rather distant. We shall follow Yuan’s strategy from [33], paying considerable attention to the details of the reasoning. Consequently, fork2 ≥7, our gap principle is much stronger than that used by Yuan. In [33] it is proved that the quotient j3/j2 is bounded from below by an expression linear in β, whereas our bound is of the typeβ(k2−3)/2. Our result made possible explicit computations to be described in subsequent sections.
Suppose that the system (8) has three positive solutions (xi, yi, zi) (i= 1,2,3). Then zi=αji−α−ji
2√
a = βki−β−ki 2√
b
for odd integersji andki (i= 1,2,3) with 1 =k1< k2< k3 and 1 =j1< j2< j3.
We first note that either j2 dividesj3 and k2 divides k3 or j3 = 2qj2±1 and k3 = 2q1k2±1 for some positive integers qand q1. 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 (Vr, Vs0, Ur) is a positive solution of (8) whose third component is smaller than z2, it follows thatUr =z1 = 1, whencer= 1, and thereforeVs0= 1 =V10, so thats= 1.
As seen above, there exist integers q, q1 ≥2 and σ, σ1 ∈ { −1,0,1}such that j3 =qj2+σ, k3=q1k2+σ1and bothqσandq1σ1are even. As a preparation for the proof of our gap principle we establish first the next result.
Lemma 3.1. σ=σ1.
Proof. Ifσ = 0, then j2 divides j3, so thatz3 =Uj3 =Uk03 is multiple ofz2 =Uj2 =Uk02. Thus, Uk02 dividesUk03, whencek2 dividesk3. This meansσ1= 0, as asserted. By symmetry, one obtains thatσ = 0 wheneverσ1= 0.
Suppose now that bothσandσ1are non-zero. FromUj3−σ= 2Vqj2+σUqj2 (cf. Lemma 2.2c)) andUj2 |Uqj2 it resultsz3=Uj3 ≡σ (modz2). Similarly, Uk03 ≡σ1 (modz2). The claim follows
from|σ−σ1| ≤2<4a−3< z2 (cf. Lemma 2.5).
Another observation needed later on is that the relationship betweenq and q1 is the same as betweenj2and k2.
Lemma 3.2. q > q1.
Proof. Ifj3 is multiple ofj2, thenk3 is divisible byk2 and we have z3=Uqj2= (x2√
a+z2√
a−1 )q−(x2√
a−z2√ a−1 )q 2z2√
a−1
=X
t≥0
q 2t+ 1
x2√
aq−2t−1
z22t(a−1)t and
z3=Uq1k2 =X
t≥0
q1
2t+ 1
y2
√bq1−2t−1
z22t(b−1)t. The desired conclusion follows becausey2
√b > x2√aandb > a.
It remains to examine the casej3=qj2+σ andk3=q1k2+σ withσ6= 0. We shall implicitly use the well-known inequalities
t−t2
2 <log(1 +t)< t valid for positivet.
Let us consider the linear forms in logarithms Λ2= log
b−1 a−1
+j2logα2−k2logβ2,
Λ3= log b−1
a−1
+j3logα2−k3logβ2. By relations (12) and (16), we have
0<log b−1
a−1
−Λ2= 2(k2logβ−j2logα)<log b−1
a−1
,
log b−1
a−1
−Λ2<2 logβ
α+ 1
2(a−1). Replacing in Λ3 the coefficientsj3 andk3in terms ofj2andk2, one gets
2(q−q1)j2logα= Λ3+ (q1−1) log b−1
a−1
+ 2σlogβ
α−q1Λ2
(17)
≥Λ3−log b−1
a−1
−2 logβ α+q1
log
b−1 a−1
−Λ2
. From Lemma 2.3 one obtains
log b−1
a−1
<2 logβ
α+ 1
2a−2,
so that the rightmost term in the chain of inequalities (17) is greater than
(18) Λ3−4 logβ
α− 1
2a−2+q1
log
b−1 a−1
−Λ2
.
From the left inequality in Lemma 2.3 one gets for all (a, b)6= (2,3) log
b−1 a−1
>2 logβ α+ log
1 + 1
8(a−1)2
>2 logβ
α+ 1
9(a−1)2. Numerical verification shows that always one has
(19) log
b−1 a−1
>2 logβ
α+ 1
9(a−1)2. On the other hand, by (14) we have
(20) Λ2< 2α−12
α2−1 < 1 (4a−4)6. Therefore, the expression in (18) is bounded from below by
2(q1−2) logβ α+q1
1
9(a−1)2− 1 (4a−4)6
− 1 2a−2. Since
β α
2
> b
a ≥1 + 1 a, we have
2 logβ α> 1
a− 1 2a2 ≥ 3
4a. Thus
2(q−q1)j2logα >3(q1−2) 4a +q1
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 q1 ≥ 4, we have obtained the conclusion of the Lemma, unlessq=q1= 2. In this special situation, from equation (17), Lemma 2.3 and relation (14) it results
2
(4a−4)6 >2Λ2−Λ3≥log b−1
a−1
−2 logβ
α> 1 9(a−1)2, which is false fora >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. qj2logα < q1k2logβ.
Proof. By Lemma 2.3 and relations (13) – (14) we have (k3−1) logβ2−(j3−1) logα2= log
b−1 a−1
−Λ3−logβ2 α2
> 1
9(a−1)2 −Λ3> 1
9(a−1)2 − 1
(4a−4)6 >0, so that (j3−1) logα <(k3−1) logβ. Hence,
(qj2+σ−1) logα <(q1k2+σ−1) logβ, or
qj2logα < q1k2logβ+ (σ−1) log(β/α).
Sinceσ is at most 1 andβ > α, the result follows.
Corollary 3.4. aq2< bq21.
Proof. We argue by contradiction. Supposing thataq2≥bq21, it results rb
a ≤ q q1
< k2logβ
j2logα < logβ logα. It is straightforward to establish that the mappingt7→log(√
t+√
t−1 )/√
t, defined fort≥√ 2, is increasing fort≤t0'3.276 and decreasing fort≥t0. Thus, fora >2 ora= 2 andb >6, from the assumption aq2≥bq12 one gets the contradictiona > b. Fora= 2< b≤6 one sees that the inequalities (12) are not satisfied.
We can now prove the main results of this section.
Proposition 3.5. Assumek2= 3. Then for allβ (resp. forβ ≥1000) one has j3>1.7j2β2/3 (resp. j3>1.98j2β2/3)for j2= 7, j3>1.7j2β4/5 (resp. j3>1.90j2β4/5)for j2= 11, j3>1.7j2β6/7 (resp. j3>1.83j2β6/7)for j2≥15.
Proof. We first examine the situation whenk3= 3q1. Thenj3=qj2, for some odd integersq and q1. Fromax22≡by22≡1 (mod z22) and
z3≡q(ax22)(q−1)/2≡q1(by22)(q1−1)/2 (mod z22)
one getsq≡q1 (modz22). Sincez2=β2+ 1 +β−2> β2andq > q1(see Lemma 3.2), one obtains q > z22> β4, which yieldsj3> j2β4, a much stronger result than the desired one.
Suppose now thatk3= 3q1+σ, j3=qj2+σ, with σ∈ { −1,1}and evenq,q1. This time one has
z3≡aqx2z2+σ≡bq1y2z2+σ (mod z22), whence
bq1y2≡aqx2 (modz2).
Fromz2= 4b−1,j2≥7 andax22≡by22≡1 (modz22) one infers 4aq2≡4bq12≡q12 (mod z2).
Note thatq > q1 implies 4aq2> q21, so that
q2≥ 1 +Uj2
4a .
We shall give all the details for the rest of the proof in the casej2 = 7. The same idea, with obvious modifications, works for the other cases.
SinceU7= 64a3−80a2+ 24a−1, one hasq2≥16a2−20a+ 6. We shall determine a positive K such that 16a2−20a+ 6≥K2β4/3. It is sufficient to have
K6≤ (16a2−20a+ 6)3
4(2b−1)2 = 2(8a2−10a+ 3)3 (32a3−40a2+ 12a−1)2.
Denote byf(a) the rational function in the right side of the last relation, and note thatβ ≥1000 impliesa≥26. Asf is increasing fora≥2, it is sufficient to haveK6≤f(2) (resp. K6≤f(26)).
We verify that K= 0.883 (resp. K= 0.991) works. Indeed, then we haveq >0.883β2/3and j3≥2q j2−1>1.766j2β2/3−1>1.7j2β2/3.
The gap principle for higher values of k2 is much stronger due to the presence of a greater exponent ofβ.
Proposition 3.6. If k2≥5, thenj3>3.96j2β(k2−3)/2.
Proof. Recall thatj3=qj2+σandk3=q1k2+σfor certain integersq > q1≥2 andσ∈ { −1,0,1} such that qσ and q1σ 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 considerσ = 0. Then z3=Uqj2 =X
t≥0
q 2t+ 1
x2√
aq−2t−1
z22t(a−1)t ≡q x2√ aq−1
(mod z22).
One gets similarlyz3 ≡q1 y2
√bq1−1
(modz22), so that q x2√ aq−1
≡q1 y2
√bq1−1
(modz22).
Sinceax22≡by22≡1 (modz22), andqandq1are odd integers, this yieldsq≡q1 (modz22). Having in view Lemma 3.2, one concludes that q > z22 > β2(k2−1) and thereforej3 =q j2 > j2β2(k2−1), which is stronger than the desired conclusion.
Ifσ6= 0, then bothqandq1are even. From
α2j2 = 2z22(a−1) + 1 + 2x2z2
pa2−a it results
z3=Uj3 ≡aqx2z2+σ (mod 2z22).
With the same reasoning one obtains
z3=Uk03 ≡bq1y2z2+σ (mod 2z22).
Therefore
(21) bq1y2≡aqx2 (mod 2z2)
and
bq1
2 2
≡aq 2
2
(mod z2).
Asaq2< bq21 andq > q1 (see Corollary 3.4 and Lemma 3.2), one obtains bq2≥b(q1+ 2)2≥aq2+ 4z2+ 4bq1+ 4b
≥4z2+ 16a+ 12b >4 βk2−1+βk2−3+βk2−5 + 12b, whence
q2>4βk2−3 4−1 b
+ 12.
Sinceb≥60 for each system (8) with at least three positive solutions, one has j3> j2
r
4βk2−3 4−1 b
+ 12−1>3.96j2β(k2−3)/2.
4. 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 [16]; thus we quote his result. This theorem enables us to get a first bound on coefficientsaandbfor 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 numberγ.
Theorem A([16]). Letλ1, λ2,λ3be
Q
–linearly independent logarithms of non-zero algebraic numbers and let b1, b2, b3 be rational integers with b16= 0. Defineαj = exp(λj) for j = 1, 2, 3 andΛ =b1λ1+b2λ2+b3λ3. LetD be the degree of the number field
Q
(α1, α2, α3)overQ
. Putχ= [
R
(α1, α2, α3) :R
].LetA1, A2,A3 be positive real numbers, which satisfy Aj ≥max
Dh(αj),|λj|,0.16 (1≤j ≤3).
Assume that
B≥max
|bj|Aj/A1 : 1≤j ≤3 . Define also
C1= 5×165
6χ e3(7 + 2χ) 3e
2 χ
20.2 + log 35.5D2log(eD) . Then
log|Λ|>−C1D2A1A2A3 log 1.5 eDBlog(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 4.1. If α, β and(b−1)/(a−1) are multiplicatively dependent, then the equations (8) have at most one positive solution.
Proof. Suppose, by way of contradiction, that the system (8) has at least one solution given by relation (11) withj >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 α2 and β2 are powers ofγ. Therefore,
Λ = log b−1
a−1
+jlog α2−klogβ2= log b−1
a−1
−tlogγ for some positive integert. Asb−1−(a−1)γt is a nonzero algebraic integer,
b−1−(a−1)γt
b−1−(a−1)γ−t is a positive rational integer. Hence,
b−1−(a−1)γt≥ 3 4(b−1), b−1
(a−1)γt −1≥ 3 4(b−1)2, and
Λ> 2 3(b−1)2.
Comparison with the upper bound for Λ given in (14) results in (b−1)2>(1−α−2)α2j.
Using the obvious inequality β2 > 4(b−1), one gets β4 > 16(1−α−2)α2j. But we have seen (cf. (16))
(k−1) logβ2<(j−1) logα2+ 0.5.
Hence,
(j−1) logα2+ 0.5> jlogα2+ log 4(1−α−2) , wherefrom the contradiction logα2<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 ex- plained above, if z > 1, thenz is given by relation (10) for suitable odd integersj > k >1. We apply Matveev’s result for the linear form
Λ = log b−1
a−1
+ 2j3logα−2k3logβ, which we rewrite as
Λ = log b−1
a−1
+rlogα2−k3log(β/αh)2, where j3=hk3+r, 0≤r < k3. We haveD= 4 and
h b−1
a−1
= log(b−1), h(α2) = logα, h (β/αh)2
= logβ.
Therefore we can take
A1= 2D log(β/2), A2=D logα, A3=D logβ, and
B = max
1, rlogα
2 log(β/2), k3 logβ 2 log(β/2)
= k3 logβ 2 log(β/2). With these choices it follows
C1= 5×165
6 e3·9·3e
2 20.2 + log 35.516 log(4e)
= 180·164e4 20.2 + 5.5 log 3 + log 16 + log(log 4 + 1)
<1.925·1010.
Matveev’s theorem yields in our specific situation the inequality log Λ>−C1·16A1A2A3log 6eBlog(4e)
=−C1·16·128 logαlogβlog(β/2) log 6eBlog(4e)
>−3942.4·1010logαlogβlog(β/2) log
19.46k3logβ log(β/2)
.
For reasons that will be explained in Section 5 (see Lemma 5.1), we may suppose thatb≥2000.
Under this assumption we get
19.46k3logβ
log(β/2) <24j3. Therefore, by (15) we have
(22) j3<1.9712·1013logβlog(β/2) log 24j3 .
This relation together with Propositions 3.5–3.6 and the obvious inequalities
(23) 2√
b−1< β <2√ b
yield upper bound for integersbfor which the simultaneous Pell equations (6) can have (at least) three positive solutions:
k2= 3 : b <3.1×1051forj2= 7, b <1.1×1042forj2≥11, k2= 5 : b <3.1×1032, k2≥7 : b <1.6×1015.
The admissible range for bshrinks very fast with growing k2. However, fork2 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 intwologarithms, for which we use the main result of [14] (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 =b2logα2−b1logα1,
with b1 and b2 positive integers. Let K ≥3, L≥2, R1, R2, S1, S2 be positive integers and ρ a real number greater than1. Put R=R1+R2−1, S=S1+S2−1, N =KL,
g=1 4− N
12RS, c= (R−1)b2+ (S−1)b1 2
K−1
Y
k=1
k!
!−2/(K
2−K)
. Leta1, a2 be positive real numbers such that
ai≥ρ|logαi| −log|αi|+ 2Dh(αi), for i= 1,2. Suppose that:
(I) Card
αr1αs2; 0≤r < R1, 0≤s < S1 ≥L,
(II) Card
rb2+sb1; 0≤r < R2,0≤s < S2 >(K−1)L, and
(III) K(L−1) logρ−(D+ 1) logN− D(K−1) logc−gL(Ra1+Sa2) > 0.
Then,
|Λ0| ≥ρ−KL+(1/2), where Λ0= Λ·max
LSeLS|Λ|/(2b2) 2b2
, LReLR|Λ|/(2b1) 2b1
.
In the case when the numberα1is not a root of unity we shall deduce the following result from Theorem B, which is a variant of Th´eor`eme 2 of [14], close to Theorem 1.5 of [18].
Proposition 4.2. Consider the linear form
Λ =b2logα2−b1logα1,
whereb1 andb2 are positive integers. Suppose thatα1 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 ρa real number greater than 1. Putλ= logρ and suppose that
h≥max
1,1.5λ,D
logb1
a2
+b2
a1
+ logλ+f(K)
+ε
, ε= 0.0262, ai≥max
4,2.7λ, ρ|logαi| −log|αi|+ 2Dh(αi) , (i= 1,2), a1a2≥20λ2,
where
f(x) = log 1 +√
x−1√ x
x−1 + logx
6x(x−1)+3
2+ log3
4+logx−1x x−1 and
L= 2 +b2h/λc ≥5, K= 1 +btLa1a2c. Then we have the lower bound
log|Λ| ≥ −λtL2a1a2−maxn
λ(L−0.5) + log
L2(1 +√ t)a2
,Dlog 2o , provided thatt satisfies t≤2.2λ−2 and
tΩ−L√
t−W ≥0, with
Ω = 3(L−1)λ−3h, W = 3 4
L a2
+ 1 a1
.
We postpone the proof of the above result to the last section of the paper. Now we apply Proposition 4.2 to the systemax2−(a−1)z2=by2−(b−1)z2= 1 with three solutions, the second of which is obtained fork2= 3 andj2= 7 or 11.
In the present study we have two linear forms Λi= log
b−1 a−1
+jilogα2−kilogβ2, i= 2,3.
As a consequence of Lemma 2.3 we have log
1 + 1
8(a−1)2
+ 2 log β
α
<logb−1 a−1 <log
1 + 1 2a−2
+ 2 logβ α, so that (cf. the proof of Lemma 3.2)
1
9a2+ (j2−1) logα2−(k2−1) logβ2<Λ2<(j2−1) logα2−(k2−1) logβ2+1 a. Define the integerl by
k2−1
2 j3=j2−1
2 k3+l−j2−k2
2 . It is easily seen thatl is even.
We apply the estimate for the linear form in two logarithms Λ :=b2logα2−b1logα1, with
α1= βk2−1
αj2−1, α2= b−1
a−1
(k2−1)/2
α2l−j2+k2. HereD= 4 and b1=k3,b2= 1.
Using inequality (16), one obtains h(α1) = k2−1
2 logβ≤ j2−1
2 logα+ 1
4a <j2−1
4 log(4a).
We claim that
0<logα1≤ 1 4(a−1). Indeed, by Lemma 2.3 and relations (13) – (14) we have
2 logα1= (k2−1) logβ2−(j2−1) logα2= log b−1
a−1
−Λ2−logβ2 α2
> 1
9(a−1)2 −Λ2> 1
9(a−1)2 − 1
(4a−4)6 >0, and
2 logα1<log b−1
a−1
−logβ2 α2 <log
1 + 1
2(a−1)
< 1 2(a−1), as claimed.
The next immediate concern is to show thatlis non-negative. The equality (k2−1)Λ3= log (b−1)(k2−1)/2
(a−1)(k2−1)/2αj2−k2 +llogα2−k3logβk2−1 αj2−1 is equivalent to
logα2= (k2−1)Λ3+k3logα1.
Both terms in the right side are positive. Therefore, α2 >1. Further, Lemma 2.3 and inequal- ity (16) imply
(k2−1) 2 log
b−1 a−1
−logα(j2−k2)<(k2−1) logβ
α+ k2−1
4(a−1) −logα(j2−k2)
= (k2−1) logβ−(j2−1) logα+ k2−1
4(a−1) < k2
4(a−1),
whence it follows
0<logα2< llogα2+ k2
4(a−1).
Sincek2= 3 anda≥2, one concludes thatl is non-negative. In fact, it is positive. We prove this claim parreductio ad absurdum.
Whenl= 0, we have (k2−1)(j3−1) = (j2−1)(k3−1), and therefore Λ2−Λ3= (j2−j3) logα2−(k2−k3) logβ2= (k3−k2) logβk2−1
αj2−1. By relations (19) and (20), one obtains
2 logβk2−1 αj2−1 = log
b−1 a−1
−Λ2−log β2
α2
> 1
9(a−1)2 − 1
(4a−4)6 > 7 64(a−1)2, whence the contradiction
1
(4a−4)6 >Λ2>Λ2−Λ3> 7(k3−3) 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 2.4 we know that j3 ≡k3 (mod 4), so that l is multiple of 4 and our claim follows. Note that the inequalities l≥2 forj2= 7 andl≥4 forj2= 11 are equivalent to
2l−j2+k2≥0.
Now we proceed to bound from above the logarithm and the height ofα2forabigger than 1000.
Lemma 2.3 yields
logα2= log b−1
a−1
+ (2l−j2+k2) logα
<2 logβ
α+ 1
2(a−1)+ (2l−j2+k2) logα.
We also obtain
h(α2)< k2−1
2 log(b−1) +2l−j2+k2
4 log(4a)≤ l
2+ 3
log(4a) and
0<logα2≤(l+ 1) log(4a).
The hypothesis of Proposition 4.2 requires a1≥ ρ−1
4(a−1)+ 2(j2−1) log(4a), 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 ifρ <1 + 8(a−1) log(4a). Specializationρ= 27 yields
a2= (30l+ 50) log(4a), t= 2
3λ+ 2
3h+ 0.003 2
,
h= 4 log
k3
(30l+ 50) log(4a)+ 1 2j2log(4a)
+ 4 logλ+ 4f(K) + 0.0262.
Then our estimate for linear forms in two logarithms implies log|Λ3| ≥ −λ t L2a1a2−λ(L−0.5)−log
L2(1 +√ t)a2
(24)
>−0.21016λ
1 + h λ
2
a1a2−λ(L−0.5)−log
L2(1 +√ t)a2
>−0.7
1 +h λ
2
a1a2, whereas from (15) we know that
log|Λ3|<−j3logα2+ 0.9.
In order to exploit these bounds for log Λ3, we need to relate k3 toa andl. We show that for 2l6=j2−k2 anda >1000 one has
k3≥1.998a llogα2. From (12) and (16) it follows
α(j2−1)k3+2l−j2+k2 < β2k3, β4<
1 + 1
2(a−1)
α2(j2−1), whence
α2(2l−j2+k2)/k3 <1 + 1 2(a−1). Therefore,
(2l−j2+k2) logα2< k3log
1 + 1
2(a−1)
< k3
2(a−1).
Since 2(a−1)≥1.998aandl≥j2−k2(recall that 2l > j2−k2andlis multiple of 4 ifj2= 11), it results
k3>2(a−1)(2l−j2+k2) logα2≥1.998a llogα2.
Therefore k3≥1.997al log(4a) fora >1000. We conclude that eithera≤1000 or
(25) j3logα2<0.7
1 +h λ
2
a1a2+ 0.9<1.41
1 + h λ
2
j2(30l+ 50) log2(4a), whence
(26)
0.998(j2−1)alog(4a) + 1−4 l
logα2<1.41
1 + h λ
2
j2
30 +50
l
log2(4a).
Forj2= 7 this yields
a <1.8·55
1 + 4.5 loga λ
2
<99(1 + 1.366 loga)2, and thereforea <22000,b <6.4·1014. Forj2= 11 one obtains
a <71(1 + 1.366 loga)2, a <14000, b <1.4·1023.
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 tight upper bounds for the values ofaandbfor which the equations (8) have at least three solutions in positive integers. The bounds foraare derived by using Proposition 2.5.
Lemma 4.3. Suppose that equations (8) have three solutions given by (10)for j1= 1< j2< j3
andk1= 1< k2< k3. Then the following statements hold.
k2= 3 : a <22000 forj2= 7, a <15000 forj2= 11, b <3.9×1023, k2= 3 : a <105000 for j2= 15, a <23000 forj2≥19, b <4×1038, k2= 5 : a <8.9×1015 forj2= 9, a <3×1010 forj2= 13, b <3.4×1032, k2= 5 : a <3.5×107 forj2= 17, a <7.5×105 forj2≥21, b <6.8×1031, k2= 7 : a <7.4×108 forj2= 11, a <1.3×106 forj2= 15, b <1.6×1015, k2= 7 : a <40000 forj2= 19, a <4300 forj2≥23, b <8.6×1014,
k2= 9 : a <1.7×106 forj2= 13, a <30000 forj2≥17, b <4.1×109, k2= 11 : a <57000, b <7.7×106, k2= 13 : a <7000, b <2×105, k2= 15 : a <1800, b <20000, k2= 17 : a <600, b <3000.
5. In the search of a third solution
Since the epoch-making paper of Baker and Davenport [4], techniques of computational Dio- phantine 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 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 onj3. Using the suitable gap principle (either Proposition 3.5 or Proposition 3.6), one bounds from above b.
Fromαj3 < βk3 it follows thatais below a certain small value. Then one performs a search for a third solution. However, when k2= 3 andj2≤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 Λ3as a linear form in two logarithms. Proposition 4.2 yields a tight upper bound ona, which forcesb to remain confined to a domain defined by the inequality (16). 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 ifbis very small.
Lemma 5.1. There is no pair of integers (a, b) with 1< a < b ≤2000 for which the equations ax2−(a−1)z2=by2−(b−1)z2= 1 have three positive solutions.
Proof. We shall argue byreductio ad absurdum. Suppose that there exists a pair (a, b) of integers such that 1 < a < b≤2000 and the equationsax2−(a−1)z2 =by2−(b−1)z2= 1 have three positive solutions. Keeping the notation used up to now, we have
(27) Λ3= log
b−1 a−1
+j3logα2−k3logβ2<−2j3logα+ 0.882.
A variant of Davenport’s lemma due to Peth˝o [12] proved itself most suitable in our context. We record it for handy reference.
Lemma C([12]). LetA, B, θ, µ be positive real numbers andM 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
