COMPLETE SOLUTION OF A PROBLEM OF DIOPHANTUS AND EULER

ANDREJ DUJELLA^{∗} AND CLEMENS FUCHS^{‡}

Abstract. In this paper, we prove that there does not exist a set of four positive integers with the property that the product of any two of its distinct elements plus their sum is a perfect square. This settles an old problem investigated by Diophantus and Euler.

1. Introduction

Let n be an integer. A set of m positive integers is called a Diophantine m-tuple with the property D(n) or simply D(n)-m-tuple, if the product of any two of them increased by n is a perfect square. For the case n = 1 this problem was first studied by Diophantus and he found a set of four positive rationals with the above property: \{\frac{1}{16}, \frac{33}{16}, \frac{17}{4}, \frac{105}{16}\}. The first D(1)-quadruple however, the set {1, 3, 8, 120}, was found by Fermat. Euler was able to add the fifth positive rational, \frac{777480}{8288641}, to the Fermat’s set (see [4] and [16]). Gibbs [15] found examples of sets of six positive rationals with the property of Diophantus. The folklore conjecture is that there does not exist a D(1)-quintuple. In 1969, Baker and Davenport [1] proved that the Fermat’s set cannot be extended to a D(1)-quintuple. Recently, the first author proved that there does not exist a D(1)-sextuple and there are only finitely many D(1)-quintuples (see [11]).

In the case n = −1, the conjecture is that there does not exist a D(−1)- quadruple (see [6]). It is known that some particular D(−1)-triples cannot be extended to D(−1)-quadruples, namely this was verified for the triples {1, 2, 5} (by Brown in [3], see also [23, 19, 22, 20]), {1, 5, 10} (by Mohanty and Ramasamy in [21]), {1, 2, 145}, {1, 2, 4901}, {1, 5, 65}, {1, 5, 20737}, {1, 10, 17}, {1, 26, 37} (by Kedlaya [19]) and {17, 26, 85} (again by Brown in [3]). Moreover, Brown proved that the following infinite families of D(−1)- triples cannot be extended to quadruples:

{x^{2} + 1, (x + 1)^{2} + 1, (2x + 1)^{2} + 4}, if x ≢ 0 (mod 4),
{2, 2x^{2} + 2x + 1, 2x^{2} + 6x + 5},  if x ≡ 1 (mod 4).

∗The first author was supported by the Ministry of Science, Education and Sports, Republic of Croatia, grant 0037110.

‡The second author was supported by the Austrian Science Foundation FWF, grant S8307-MAT.

The first author proved the conjecture in [7] for all triples of the form {1, 2, c}.

Let us mention that from [10, Theorem 4] it follows that there does not exist a D(−1)-33-tuple. This is the best known upper bound for this problem at present.

The n = −1 case is closely connected with an old problem of Diophantus and Euler. Namely, Diophantus studied the problem of finding numbers such that the product of any two increased by the sum of these two gives a square. He found two triples \{4, 9, 28\} and \{\frac{3}{10}, \frac{21}{5}, \frac{7}{10}\} satisfying this property. Euler found a quadruple \{\frac{5}{2}, \frac{9}{56}, \frac{9}{224}, \frac{65}{224}\} and asked if there is an integer solution of this problem (see [5], [4] and [16]). In [9] an infinite family of rational quintuples with the same property was given. Since

xy + x + y = (x + 1)(y + 1) − 1,

we see that the problem of finding integer m-tuples with the property that for any two distinct elements the product plus their sum is a perfect square is equivalent to finding D(−1)-m-tuples.

A polynomial variant of the above problems was first studied by Jones [17], [18], and it was for the case n = 1. Recently, polynomial variants were also studied by the authors. They were able to completely prove the polynomial variants of both conjectures for n = 1 (here even a stronger version of the quintuple conjecture was proved since the question of finding polynomial quintuples was answered with the result in [11]) in [13] and also for the case n = −1 in [12].

In this paper we completely solve the problem investigated by Diophantus and Euler, namely we prove the following result:

Theorem 1a. There does not exist a set of four positive integers with the property that the product of any two if its distinct elements plus their sum is a perfect square.

By the above correspondence this implies the following equivalent result on D(−1)-m-tuples.

Theorem 1b. There does not exist a D(−1)-quadruple {a, b, c, d} with 2 ≤ a < b < c < d.

In the proof of the above Theorem we first show that if {a, b, c, d} is a D(−1)-quadruple with minimal d, then {1, a, b, c} is also a D(−1)-quadruple. This idea was introduced for polynomials already in [12] when we proved the conjecture for polynomials with integral coefficients.

We have the following corollaries, which immediately follow from the above theorems.

Corollary 1. There does not exist a D(−1)-quintuple.

Moreover, we can state the following restrictions for D(−1)-quadruples:

Corollary 2. If {a, b, c, d} is a D(−1)-quadruple with 0 < a < b < c < d, then a = 1 and b ≥ 5.

It seems that the case a = 1 is more involved and much harder. It can be compared with the strong version of the quintuple conjecture for n = 1, which says that every D(1)-triple can be extended to a D(1)-quadruple in an essentially unique way.

We also want to remark that for general n, it was proven by the first author in [6] that if n ≢ 2 (mod 4) and n ∉ S = {−4, −3, −1, 3, 5, 12, 20}, then there exists at least one Diophantine quadruple with the property D(n). The conjecture is that for n ∈ S there does not exist a Diophantine quadruple with the property D(n). This paper gives support to this conjecture for the case n = −1.

The strategy of the paper follows the same line as the proofs of almost all other recent results on non-extendability of D(n)-m-tuples (especially we will follow the line of [11]). In Section 1 we reduce the problem of finding d which extends {a, b, c} to a D(−1)-quadruple to a system of simultaneous Pellian equations, which leads to the consideration of intersections of linear recurring sequences. Afterwards we apply congruence relations to show that these sequences cannot have intersections for small indices (in Section 2 we prove that the indices are easily related and in Section 3 we consider the small cases), which leads to a “gap principles”, which say that in such a triple there must be a certain gap between b and c (Section 3). Using this gap principle and a theorem about simultaneous approximations of square roots which are close to 1 due to Bennett, we obtain in Section 4 an upper bound for d. Comparing this upper bound with lower bounds obtained by congruence methods we show that such an extension does not exist, which is done on Section 5.

The new ideas in this paper are the following: we explore the assumption that the quadruple {a, b, c, d} is minimal and we consider the associated quadruple {1, a, b, c} first (observe that we assume that a ≥ 2), which exists as mentioned above (and this idea comes from the much easier polynomial case handled in our paper [12]). Here we show that we have an excellent “gap principle”, namely c > 40000b^{9}. Now this gap is good enough to apply Bennett’s theorem and it is also good enough to apply the congruence method introduced in [14]. Here we need to have control on the fundamental solutions of the Pellian equations under consideration. We get upper bounds for them, which are better than in general, again from the fact that {1, a, b, c} is also a D(−1)-quadruple.

2. Reduction to intersections of recursive sequences

Let {a, b, c}, where 2 ≤ a < b < c, be a D(−1)-triple and let r, s, t be positive integers defined by

ab − 1 = r^{2}, ac − 1 = s^{2}, bc − 1 = t^{2}.

In this paper, the symbols r, s, t will always have this meaning. Assume that there exists a positive integer d > c such that {a, b, c, d} is a D(−1)- quadruple. We have

(1)  ad − 1 = x^{2}, bd − 1 = y^{2}, cd − 1 = z^{2},

with integers x, y, z. Eliminating d from (1) we obtain the following system of Pellian equations

(2)  az^{2} − cx^{2} = c − a,
(3)  bz^{2} − cy^{2} = c − b.

We will describe the sets of solutions of equations (2) and (3) in the following lemma.

Lemma 1. If (z, x) and (z, y), with positive integers x, y, z, are solutions of (2) and (3) respectively, then there exist integers z_{0}, x_{0} and z_{1}, y_{1} with

(i) (z_{0}, x_{0}) and (z_{1}, y_{1}) are solutions of (2) and (3) respectively, (ii) the following inequalities are satisfied:

(4)  0 ≤ |x_{0}| < s,
(5)  0 < z_{0} < c,
(6)  0 ≤ |y_{1}| < t,
(7)  0 < z_{1} < c,

and there exist integers m, n ≥ 0 such that

(8)  z\sqrt{a} + x\sqrt{c} = (z_{0}\sqrt{a} + x_{0}\sqrt{c})(s + \sqrt{ac})^{2m},
(9)  z\sqrt{b} + y\sqrt{c} = (z_{1}\sqrt{b} + y_{1}\sqrt{c})(t + \sqrt{bc})^{2n}.

Proof. The proof runs along the same line as the proof of [8, Lemma 1]. A polynomial version of this lemma was proved in [12, Lemma 1]. □

The solutions z arising, for given (z_{0}, x_{0}), from formula (8) for varying m ≥ 0 form a binary recurrent sequence (v_{m})_{m≥0} whose initial terms are found by solving equation (8) for z when m = 0 and 1, and whose characteristic equation has the roots (s + \sqrt{ab})^{2} and (s − \sqrt{ab})^{2}. Therefore, we conclude that z = v_{m} for some (z_{0}, x_{0}) with the above properties and integer m ≥ 0, where

(10) v_{0} = z_{0}, v_{1} = (2ac − 1)z_{0} + 2scx_{0}, v_{m+2} = (4ac − 2)v_{m+1} − v_{m}.

In the same manner, from (9), we conclude that z = w_{n} for some (z_{1}, y_{1}) with the above properties and integer n ≥ 0, where

(11) w_{0} = z_{1}, w_{1} = (2bc − 1)z_{1} + 2tcy_{1}, w_{m+2} = (4bc − 2)w_{n+1} − w_{n}.

Now the following congruence relations follow easily from (10) and (11) by induction:

v_{m} ≡ (−1)^{m}z_{0} (mod 2c), w_{n} ≡ (−1)^{n}z_{1} (mod 2c).

From this relations it follows immediately (since z_{0} + z_{1} < 2c) that if the equation v_{m} = w_{n} has a solution, then we must also have

z_{0} = z_{1}.

Moreover, we can also conclude that m and n have the same parity, i.e.

(12)  m ≡ n (mod 2).

Furthermore, important relations are obtained by considering the sequences (v_{m}) and (w_{n}) modulo c^{2}. This method was first introduced by Pethő and the first author in [14].

Lemma 2. We have

v_{m} ≡ (−1)^{m}(z_{0} − 2acm^{2}z_{0} − 2csmx_{0}) (mod 8c^{2}),
w_{n} ≡ (−1)^{n}(z_{1} − 2bcn^{2}z_{1} − 2ctny_{1}) (mod 8c^{2}).

Proof. The proof follows immediately by induction from (10) and (11), respectively. □

From this lemma we can at once conclude that, if v_{m} = w_{n} then

(13)  am^{2}z_{0} + smx_{0} ≡ bn^{2}z_{1} + tny_{1} (mod 4c).

The next important step is to show that to a D(−1)-quadruple {a, b, c, d} with 2 ≤ a < b < c < d another D(−1)-quadruple can be associated. By assuming that {a, b, c, d} has minimal d under all such D(−1)-quadruples, we will conclude that {1, a, b, c} is also a D(−1)-quadruple.

First we need the following lemma, which is now easy to prove and which will play a key role in the rest of the proof. A polynomial version of this lemma was obtained in [12].

Lemma 3. Let {a, b, c, d} with 0 < a < b < c < d be a D(−1)-quadruple. Then there exists a positive integer d_{0} with d_{0} < c such that ad_{0} −1, bd_{0} −1, cd_{0} − 1 are perfect squares.

Proof. We look for a positive integer d such that z^{2} = v_{m}^{2} = w_{n}^{2} = cd − 1. Therefore, v_{m}^{2} ≡ −1 (mod c). Hence, z_{0}^{2} ≡ −1 (mod c). Now, we define

d_{0} = \frac{z_{0}^{2} + 1}{c}.

It is easy to check that d_{0} < c and that ad_{0} − 1 = y_{1}^{2}, bd_{0} − 1 = x_{0}^{2}, which implies that ad_{0} − 1, bd_{0} − 1, cd_{0} − 1 are perfect squares. □

Assume now that {a, b, c, d} with 2 ≤ a < b < c < d is a D(−1)- quadruple with minimal d among all such quadruples. We may use Lemma 4 to construct d_{0}. From the minimality of d, it follows that d_{0} = 1 must hold and therefore {1, a, b, c} is also a D(−1)-quadruple.

This has another important implication on the size of the fundamental solutions of (10) and (11) described in Lemma 1.

Lemma 4. Let the integers z_{0}, z_{1}, x_{0}, y_{1} be as in Lemma 1. Then we have

z_{0} = z_{1} = \sqrt{c − 1}, |x_{0}| = \sqrt{a − 1}, |y_{1}| = \sqrt{b − 1}.

Proof. Since we have

1 = d_{0} = \frac{z_{0}^{2} + 1}{c},

we immediately can conclude that

z_{0} = z_{1} = \sqrt{c − 1}.

Also, by x_{0}^{2} = ad_{0} − 1, y_{1}^{2} = ad_{0} − 1, we get |x_{0}| = \sqrt{a − 1}, |y_{1}| = \sqrt{b − 1}, as claimed in the lemma. □

From the fact that {1, a, b, c} is again a D(−1)-quadruple we can conclude that

a = α^{2} + 1, b = β^{2} + 1, c = γ^{2} + 1.

Moreover, to the D(−1)-triple {1, a, b} we can again associate a pair of Pellian equations as above, which give all extensions to quadruple, as for example c. Moreover, we can associate two linear recursive sequences (ṽ_{m}) and (w̃_{n}) to the triple, such that every solution ṽ_{m} = w̃_{n} gives us an extension of {1, a, b} to a D(−1)-quadruple. We have

(14)  ṽ_{0} = z̃_{0}, ṽ_{1} = (2b − 1)z̃_{0} + 2βbx̃_{0}, ṽ_{m+2} = (4b − 2)ṽ_{m+1} − ṽ_{m}.

and

(15) w̃_{0} = z̃_{1}, w̃_{1} = (2ab − 1)z̃_{1} + 2rbỹ_{1}, w̃_{m+2} = (4ab − 2)w̃_{n+1} − w̃_{n},

with

0 < z̃_{0} = z̃_{1} < b, |x̃_{0}| < β, |ỹ_{1}| < r.

We also know that we have γ = ṽ_{m} = w̃_{n} for certain values of m and n.

3. Relationships between m and n

In this section we will prove an unconditional relationship between m and n, which will be used several times later on.

Lemma 5. If v_{m} = w_{n}, then n ≤ m ≤ 2n.

Proof. We have the following estimates for v_{1}:

v_{1} = (2ac − 1)z_{0} + 2scx_{0} ≥ (2ac − 1)z_{0} − 2sc|x_{0}|
= \frac{(4a^{2}c^{2} − 4ac + 1)z_{0}^{2} − 4c^{2}(ac − 1)x_{0}^{2}}{(2ac − 1)z_{0} + 2sc|x_{0}|} ≥ \frac{4c(ac − 1)(az_{0}^{2} − cx_{0}^{2}) + z_{0}^{2}}{(2ac − 1)z_{0} + 2sc|x_{0}|}
> \frac{4c(ac − 1)(c − a) + z_{0}^{2}}{2(2ac − 1)z_{0}} > \frac{4c(ac − 1)(c − a)}{4ac(c − 1)} > c − a,

where we have used (2). On the other side we trivially have

v_{1} ≤ 4ac^{2}.

Hence,

(16)  (c − a)(4ac − 3)^{m−1} < v_{m} < 4ac^{2}(4ac − 2)^{m−1}, for m ≥ 1.

In the same manner, we obtain

(17)  (c − b)(4bc − 3)^{n−1} < w_{n} < 4bc^{2}(4bc − 2)^{n−1}, for n ≥ 1.

Now, by comparing the lower bound for v_{m} with the upper bound of w_{n} and vice versa we will obtain the bounds.

We have

(4ac − 2)^{n−1} < (c − b)(4bc − 3)^{n−1} < 4ac^{2}(4ac − 2)^{m−1} < (4ac − 2)^{m+1}

and therefore

m + 1 > n − 1,

which implies m > n − 2 or m ≥ n − 1. Since m, n have the same parity (see (12)), we conclude m ≥ n.

Moreover, we have

(c−a)(4s^{2})^{m−1} = (c−a)(4ac−4)^{m−1} < (c−a)(4ac−3)^{m−1} < 4bc^{2}(4bc−2)^{n−1}.

It is easy to see that we have

4bc − 2 < 4s^{4}

and

\frac{4bc^{2}}{c − a} < 8s^{4},

which is true because we have

4bc^{2} ≤ 4(c − 2)c^{2} < 8(c − 1)^{3} ≤ 8(c − a)(ac − 1)^{2} = 8(c − a)s^{4}.

We have used here, that the function f(x) = (c − x)(cx − 1)^{2} for integers 1 ≤ x ≤ c − 1 takes it minimum at x = 1. Thus, we have

(4s^{2})^{m−1} < 8s^{4}(4s^{4})^{n−1},

which implies

2(m − 1) < 4 + 4(n − 1).

Observe that for the exponent of 4, we have m − 1 < \frac{3}{2} + n − 1, which is always satisfied with the above inequality. Therefore, we conclude m < 2n + 1 or m ≤ 2n and therefore we get what we have claimed. □

The result from above is also true for D(−1)-triples {a, b, c} with 0 < a < b < c, where possibly a = 1.

Lemma 6. If ṽ_{m} = w̃_{n}, then n ≤ m ≤ 2n.

Proof. The proof of Lemma 5 does not use the fact that a ≥ 2 and therefore the same is true for the sequences ṽ_{m} and w̃_{n} which come from the triple {1, a, b}. □

4. Gap principles

In this section we will prove that for our D(−1)-triple {a, b, c}, we have a large gap between b and c of the form c > b^{9}. This will be essential for the proof of the theorems. We get this result by studying the associated D(−1)-quadruple given by {1, a, b, c}.

Before we do this, we need the following useful gap principle for the elements of an arbitrary D(−1)-triple {a_{1}, a_{2}, a_{3}}.

Lemma 7. If {a_{1}, a_{2}, a_{3}} is a D(−1)-triple and 0 < a_{1} < a_{2} < a_{3}, then a_{3} = a_{1} + a_{2} + 2\sqrt{a_{1}a_{2} − 1} or a_{3} > 3a_{1}a_{2} ≥ 3a_{2}.

Proof. It follows from [10, Lemma 3] that there exist integers e, f, g such that we have

(18)  a_{3} = a_{1} + a_{2} − e + 2(a_{1}a_{2}e + \sqrt{a_{1}a_{2} − 1}fg),

with a_{1}e + 1 = f^{2}, a_{2}e + 1 = g^{2}. Moreover, since we may take (see [10, Lemma 3])

f = a_{1}\sqrt{a_{1}a_{3} − 1} − \sqrt{a_{1}a_{2} − 1}\sqrt{a_{2}a_{3} − 1} and
g = a_{2}\sqrt{a_{2}a_{3} − 1} − \sqrt{a_{1}a_{2} − 1}\sqrt{a_{2}a_{3} − 1},

it is easy to check that f, g are positive.

We have two cases: if e = 0, then the above equation gives

a_{3} = a_{1} + a_{2} + 2\sqrt{a_{1}a_{2} − 1},

which is the first part of our assertion. Otherwise, we have e ≥ 1. If a_{1} = 1 we get e ≥ 3 and therefore (18) implies a_{3} ≥ 7a_{1}a_{2}. In the case a_{1} = 2, we have e ≥ 4 and therefore a_{3} ≥ 8a_{1}a_{2}. Now, if a_{1} ≥ 3, again by (18), we can estimate a_{3} by

a_{3} ≥ 2a_{1}a_{2} + 2\sqrt{a_{1}a_{2} − 1}\sqrt{a_{1}a_{2} − a_{1} − a_{2} + 1}
> 2a_{1}a_{2} + 2(a_{1}a_{2} − a_{1} − a_{2} + 1) > 3a_{1}a_{2},

where the last inequality is true because of a_{1}a_{2} > 2a_{1} + 2a_{2} − 2 or (a_{1} − 2)(a_{2} − 2) ≥ 2.

Altogether, we have

a_{3} > 3a_{1}a_{2} ≥ 3a_{2},

which was the statement of the lemma. □

We are intended to prove that the equation

(19)  ṽ_{m} = w̃_{n}

does not have a solution for small m, n (besides the trivial solution ṽ_{0} = w̃_{0} which leads to some extension d_{0} < b). We will show that if (19) has a solution, then

m > 4 or n > 4

holds. By the relation between m and n proved in the last section (Lemma 6), we have to consider the following cases:

\begin{matrix}n = 1, & m = 1, \\ n = 2, & m = 2 or 4, \\ n = 3, & m = 3, \\ n = 4, & m = 4. \end{matrix}

We will show this separately in the lemmas below.

Lemma 8. We have ṽ_{1} ≠ w̃_{1}.

Proof. From the definition of the sequences it follows that we have

(2b − 1)z̃_{0} + 2βbx̃_{0} = (2ab − 1)z̃_{1} + 2rbỹ_{1}.

We have z̃_{0} = z̃_{1}. Therefore, by dividing by 2b we get

(20)  (a − 1)z̃_{0} = βx̃_{0} − rỹ_{1}.

We remark that the system of Pellian equations associated to the triple {1, a, b} was given by

z̃_{0}^{2} − bx̃_{0}^{2} = b − 1,
az̃_{0}^{2} − bỹ_{1}^{2} = b − a

which implies

(a − 1)z̃_{0}^{2} − b(ỹ_{1}^{2} − x̃_{0}^{2}) = 1 − a.

Squaring (20) and using the last relation, we get

(a − 1)b(ỹ_{1}^{2} − x̃_{0}^{2}) − (a − 1)^{2} = (b − 1)x̃_{0}^{2} + (ab − 1)ỹ_{1}^{2} − 2βrx̃_{0}ỹ_{1}.

Hence,

(x̃_{0}r − ỹ_{1}β)^{2} = x̃_{0}^{2}r^{2} + ỹ_{1}^{2}β^{2} − 2βrx̃_{0}ỹ_{1} = −(a − 1)^{2},

which is a contradiction, since the left side is positive and the right hand side of this equation is negative. Therefore, the lemma follows. □

Up to now we have shown that n, m ≥ 2. Our strategy is now the following: we use Lemma 4 to prove that there exists a certain gap between a and b.

Using this gap principle and the congruence relations from Lemma 2, we obtain larger lower bounds for m, n. We will repeat this until we end up with n > 4 or m > 4. First, we have the following lemma:

Lemma 9. If {1, a_{2}, a_{3}, a_{4}} is a D(−1)-quadruple and 1 < a_{2} < a_{3} < a_{4}, then

a_{4} > 4a_{2}^{1.5}a_{3}^{2.5}.

Proof. As in Section 2, we can associate to the triple {1, a_{2}, a_{3}} two sequences (v_{m}) and (w_{n}) with the above properties such that a_{4} comes from some solution z = v_{m} = w_{n} for certain indices n, m ≥ 1 by the formula a_{3}a_{4} = z^{2} + 1. For every possible solution z = w_{n} = v_{m}, we can now conclude by Lemma 7 that

z ≥ w_{2} ≥ \{ \begin{matrix}\frac{a_{3}}{2} · 3a_{2}a_{3} ≥ \frac{3}{2}a_{2}a_{3}^{2} & if a_{3} > 2a_{2} \\ \sqrt{a_{3}} · 3a_{2}a_{3} ≥ 3a_{2}a_{3}\sqrt{a_{3}} & if a_{3} < 2a_{2} \end{matrix}

Because of the relation a_{3}a_{4} − 1 = z^{2}, we have

a_{4} ≥ \{ \begin{matrix}\frac{9}{4}a_{2}^{2}a_{3}^{3} & if a_{3} > 2a_{2}, \\ 9a_{2}^{2}a_{3}^{2} ≥ \frac{9}{2}a_{2}^{1.5}a_{3}^{2.5} & if a_{3} < 2a_{2}. \end{matrix}

This implies that

a_{4} > 4a_{2}^{1.5}a_{3}^{2.5}.

From this the conclusion follows. □

Using this improved gap principle, we now can prove the next step, which is in fact the hardest one.

Lemma 10. We have ṽ_{2} ≠ w̃_{2}.

Proof. First we can apply Lemma 4 to the triple {1, a, b} and we conclude that there exists d_{0} < b such that d_{0} − 1, ad_{0} − 1 and bd_{0} − 1 are all perfect squares. We have two cases depending on whether d_{0} = 1 or not.

\underline{Case d_{0} = 1:} We have

z̃_{0} = β, x̃_{0} = 0, ỹ_{1} = ±α.

It follows that we have just three sequences to consider, namely (ṽ_{m}) corresponding to x̃_{0} = 0 and two sequences (w̃_{n}) and (w̃′_{n}) corresponding to the case ỹ_{1} = α and −α, respectively.

We pause for a moment to show some properties of the sequences (ṽ_{m}) and (w̃_{n}), (w̃′_{n}). By the proof of Lemma 1 in [8] it follows that ṽ_{0} ≤ ṽ_{1}. By induction using (10) we immediately conclude that ṽ_{0} ≤ ṽ_{1} ≤ ṽ_{2} ≤ …. The same is true for (w̃_{n}), so w̃_{0} ≤ w̃_{1} ≤ w̃_{2} ≤ … and (w̃′_{n}), hence, w̃′_{0} ≤ w̃′_{1} ≤ w̃′_{n} ≤ . . .. Furthermore, we trivially have w̃′_{n} ≤ w̃_{n} for all n ≥ 0. Moreover, for ỹ_{1} = +α, we have

(21)  ṽ_{m} < w̃_{m}, for all m ≥ 1,

which follows again by induction.

From this discussion it follows that for ỹ_{1} = α we have ṽ_{2} ≠ w̃_{2}, and therefore we just have to consider ỹ_{1} = −α and therefore only the sequence (w̃′_{n}).

Next, we calculate an upper bound for (ṽ_{m}). We have

(22)  ṽ_{m} < (4b)^{m−1}2b\sqrt{b}, m ≥ 1,

which can be easily seen by induction. Moreover, we need a lower bound for (w̃′_{n}). We have w̃_{1} ≤ 4ab\sqrt{b}. Moreover,

w̃_{1}w̃′_{1} = −4a^{2}b^{2} + 4ab + b − 1 + 4ab^{3} − 4b^{2}
> 4ab^{3} − 4a^{2}b^{2} − 4b^{2} + 4ab ≥ \frac{7}{3}ab^{3}.

For the last inequality we have used that

4a^{2}b^{2} + 4b^{2} − 4ab ≤ \frac{5}{3}ab^{3},

which is true under the assumption that b > 3a, which we will assume for the moment. Combining the upper bound for w̃_{1} with the lower bound for w̃_{1}w̃′_{1} we get

w̃′_{1} > \frac{7}{12}b\sqrt{b}.

Proceeding with induction we end up with

(23)  w̃′_{n} > (3ab)^{n−1} \frac{7}{12}b\sqrt{b}, n ≥ 1.

Now by comparing the upper bound from (22) with the lower bound from (23) with (n, m) = (2, 2), we get

(3ab)\frac{7}{12}b\sqrt{b} > (4b)2b\sqrt{b},

which is true for all a ≥ 5. We remark that in [7] it was proven that the pair {1, 2} cannot be extended to a D(−1)-quadruple, therefore we may assume that a ≥ 5 holds. In the same way we can prove that

(24)  ṽ_{m} < w̃′_{m} for all m ≥ 1,

since we have

(3ab)^{m−1} \frac{7}{12}b\sqrt{b} > (4b)^{m−1}2b\sqrt{b},

which implies

a > \frac{4}{3} (\frac{24}{7})^{\frac{1}{m−1}} ,

which is again true since a ≥ 5.

It remains to consider b ≤ 3a. In this case we can conclude by the gap principle (Lemma 7) applied to the triple {1, a, b} that we have

b = a + 2\sqrt{a − 1} + 1,
a = b − 2\sqrt{b − 1} + 1,
β = \sqrt{b − 1},
r = b − β = b − \sqrt{b − 1}.

Now, if we assume that ṽ_{2} = w̃_{2}, then by (13) we get

4β ≡ 4(b − 2β + 1)β − 2(b − β)α (mod 4b).

Therefore,

4β ≡ −8β^{2} + 4β − 2βα ≡ 8 + 4β − 2βα (mod 4b).

Hence, we may conclude

2αβ ≡ 8 (mod 4b),

which by 2αβ = 2\sqrt{a − 1}\sqrt{b − 1} < 2b implies that in fact equality must hold, implying that 2αβ = 8 or αβ = 4, which is trivially a contradiction.

Altogether, the proof of the first case is finishes.

\underline{Case d_{0} ≠ 1:} Now, we look back to our quadruple \{1, a, d_{0}, b\}. Since d_{0} ≠ 1, we can apply our improved gap principle from Lemma 9 and we get

b > 4a^{2.5}d_{0}^{1.5}.

From this it follows that

b^{2.5} > 4a^{2.5}(bd_{0})^{1.5} > 4a^{2.5}z̃_{0}^{3},

or

az̃_{0} < \frac{b}{4^{\frac{2}{5}} z̃0^{\frac{1}{5}}} < \frac{3}{5}b.

Now, we use again (13) and conclude that

4z̃_{0} + 2βx̃_{0} ≡ 4az̃_{0} + 2rỹ_{1} (mod 4b).

Since 4az̃_{0} is the largest term both sides can be bounded by 6az̃_{0} < 4b, which follows by the above inequality, and therefore this inequality implies that we have in fact equality here. Thus,

2z̃_{0} + βx̃_{0} = 2az̃_{0} − r|ỹ_{1}|,

because for ỹ_{1} > 0 we would trivially have a contradiction. This can be reformulated as

2(a − 1)z̃_{0} + βx̃_{0} = −r|ỹ_{1}|.

Now, since βx̃_{0} < (a − 2)z̃_{0}, which can be seen by squaring and using (14), we receive a contradiction again.

This finally finishes the proof of the lemma. □

Lemma 11. We have ṽ_{4} ≠ w̃_{2}.

Proof. Again we use Lemma 4 to find a d_{0} such that {1, a, d_{0}, b} is a D(−1)- quadruple. As before we consider the two cases depending on whether d_{0} = 1 or not.

Assuming d_{0} = 1, we are intended to show that we have

ṽ_{4} > w̃_{2} ≥ w̃′_{2},

where the last inequality is trivially true. We have the following lower bound for ṽ_{4}, which follows from (14), namely

ṽ_{4} ≥ (3ab)^{3}b\sqrt{b} ≥ 27a^{3}b^{4},

where we have used that ṽ_{1} = (2b − 1)β ≥ b\sqrt{b}. On the other side we have by (15)

w̃_{2} ≤ (4ab − 2)w̃_{1} ≤ 16a^{2}b^{3},

since we have w̃_{1} = (2ab − 1)β + 2r^{3}b = 2ab\sqrt{b} + 2ab^{2} ≤ 4ab^{2}. Comparing these bounds we trivially get what we want.

Now, we assume that d_{0} ≠ 1. By using (13) and the same estimate as above, we derive

16z̃_{0} + 4βx̃_{0} = 4az̃_{0} + 2rỹ_{1} > 2az̃_{0}.

Hence,

2βx̃_{0} > (a − 8)z̃_{0},

which is trivially true if a ≤ 8, which means a = 5. We will handle this case separately. Otherwise, we can square the above inequality and get

4β^{2}x̃_{0}^{2} > (a − 8)^{2}z̃_{0}^{2} > (a − 8)^{2}bx̃_{0}^{2},

where we used for the last inequality the Pellian equation lying behind (14), which is z̃_{0}^{2} − bx̃_{0}^{2} = b − 1. Dividing through x̃_{0}^{2} we derive

4b − 4 > (a − 8)^{2}b,

which is trivially a contradiction unless 5 < a < 10. But in this range there are no numbers of the form α^{2} +1 = a. Thus, we are left with the case a = 5. In this special case we have

16z̃_{0} + 4βx̃_{0} = 20z̃_{0} + 2rỹ_{1},

or

2(z̃_{0} − βx̃_{0}) = −rỹ_{1}.

This equation implies, since (βx̃_{0})^{2} = bx̃_{0}^{2} − x̃_{0}^{2} < z̃_{0}^{2}, that ỹ_{1} < 0. Thus, we have

2βx̃_{0} − rỹ_{1} > 3βx̃_{0} > 2z̃_{0},

which is a contradiction. The last inequality follows from

(3βx̃_{0})^{2} = 9bx̃_{0}^{2} − 9x̃_{0}^{2} > 5bx̃_{0}^{2} > 4bx̃_{0}^{2} + 4b > 4z̃_{0}^{2},

where we used once again z̃_{0}^{2} − bx̃_{0}^{2} = b − 1. Therefore, the claim of the lemma is proved. □

Now, we do the remaining two cases, namely (m, n) = (3, 3) or (4, 4) in the following lemma.

Lemma 12. We have ṽ_{3} ≠ w̃_{3} and ṽ_{4} ≠ w̃_{4}.

Proof. We start as in the proof of the last lemma and consider the triple {1, a, d_{0}, b}. Again we have to consider the two cases d_{0} = 1 and d_{0} ≠ 1.

Let us assume d_{0} = 1 first. In Lemma 10 we have already proved that ṽ_{3} < w̃′_{3} ≤ w̃_{3} and ṽ_{4} < w̃′_{4} ≤ w̃_{4} unless b ≤ 3a. In the latter case we have b = a + 2\sqrt{a − 1} + 1. But now from (13) we get for the case (m, n) = (3, 3) that

9β ≡ 9β(−2β + 1) − 3α(b − β) (mod 4b),

which implies

0 ≡ 18 + 3β(β − 1) (mod b),

where we used that α = β − 1 which is easy to verify. Since β^{2} ≡ −1 (mod b), we get

3β ≡ 15 (mod b),

and therefore β = 5. This implies b = 26 and a = 26 − 2 · 5 + 1 = 17, thus we have the triple {1, 17, 26}. In this case we have z̃_{0} = 5, ỹ_{1} = −4, x̃_{0} = 0 and it is easy to check directly that for the completely concrete sequences (ṽ_{m}) and (w̃′_{n}), we have ṽ_{3} < w̃′_{3} ≤ w̃_{3}. In the same way we can handle the case (m, n) = (4, 4). We start with

16β ≡ 16aβ − 4αr

and conclude that β = 7 must hold. This gives b = 50 and a = 50−2·7+1 = 37, and therefore the triple {1, 37, 50}. Here we have z̃_{0} = 7, ỹ_{1} = −6, x̃_{0} = 0, and we trivially get ṽ_{4} < w̃′_{4} ≤ w̃_{4}. Therefore, the proof of the first case is finished.

Next, we assume that d_{0} ≠ 1. As in the proof of Lemma 10 we conclude by the improved gap principle (Lemma 9) that we have

b > 4a^{2.5}d_{0}^{1.5} > 8a^{2.5}.

From this we get by bd_{0} ≥ z̃_{0}^{2} that b^{3} > 4a^{2.5}b^{0.5}z̃_{0}^{3} or

z̃_{0} < \frac{2b}{3a^{\frac{5}{6}} b^{\frac{1}{6}}} = \frac{2b^{\frac{5}{6}}}{3a^{\frac{5}{6}}} .

First, we get for the case (m, n) = (3, 3) by (13) that

9z̃_{0} + 3βx̃_{0} ≡ 9az̃_{0} + 3rỹ_{1} (mod 4b).

Both sides of this equation can be bounded by 12az̃_{0}. We want to show that this is < 4b. But this follows easily from

3az̃_{0} < 2a^{\frac{1}{6}} b^{\frac{5}{6}} < b,

since b > 8a^{2.5} ≥ 8 · 5^{1.5}a > 89a > 2^{6}a. From this it follows that we have equality in the above congruence relation. But the left hand side is bounded from above by 12az̃_{0} and the right side is bounded from below by 6bz̃_{0}, which show that

9z̃_{0} + 3βx̃_{0} < 12az̃_{0} < 6bz̃_{0} < 9az̃_{0} + 3rỹ_{1}

and therefore this equation cannot hold. We now turn to the case (m, n) = (4, 4). Here (13) reads

16z̃_{0} + 4βx̃_{0} ≡ 16az̃_{0} + 4rỹ_{1} (mod 4b).

This time both sides are bounded by 20az̃_{0} and we want to show that this is < 4b. This follows from

5az̃_{0} < \frac{10}{3}a^{\frac{1}{6}} b^{\frac{5}{6}} < b,

where we used as before

b > 4a^{2.5}d_{0}^{1.5} = 4a^{1.5}d_{0}^{1.5}a > 4 · 5^{1.5} · 10^{1.5}a > 1414a > (\frac{10}{3})^{6} a.

Observe that we have used that one of a, d_{0} is ≥ 5 and the other one ≥ 10 since there is no number in between which is a perfect square plus one. As before, we conclude that we have equality in the above congruence equation, which leads to a contradiction since the left side is bounded from above by 20az̃_{0} and the ride side is bounded from below by 12bz̃_{0}. This concludes the proof of the lemma. □

As stated at the beginning of this section, we are now able to prove a much improved gap principle in the triple {a, b, c} with which we started at the beginning.

Lemma 13. Let {a, b, c, d} be a D(−1)-quadruple with 2 ≤ a < b < c < d and with minimal d under all such quadruples, then we have

c > 40000b^{9}.

Proof. We know that in fact {1, a, b, c} is a D(−1)-quadruple. From this it follows that bc − 1 = z^{2} for an integer z with z = ṽ_{m} = w̃_{n} for certain integers m, n and by the lemmas in this section we can conclude that m > 4 or n > 4.

If we assume that m > 4, then we have

z ≥ ṽ_{5} > (b − 1)(4b − 3)^{4} > 200b^{5},

and therefore

c = \frac{z^{2} + 1}{b} ≥ \frac{ṽ_{5} + 1}{b} > 40000b^{9}.

If we assume that n > 4, then we have:

z ≥ w̃_{5} > (b − a)(4ab − 3)^{4} > \frac{b}{2}(3ab)^{4} > 40a^{4}b^{5},

where we assumed that b > 2a which gives b − a > \frac{b}{2}. It follows that

c = \frac{z^{2} + 1}{b} ≥ \frac{w̃_{5}^{2} + 1}{b} > 1600a^{8}b^{9} > 10^{5}b^{9},

since a ≥ 2. Otherwise, we have b − a = 2\sqrt{b − 1} − 1 > \sqrt{b} by Lemma 7. In this case we get

z ≥ w̃_{5} > \sqrt{b} · 81a^{4}b^{4}

and we conclude

c > 6400a^{8}b^{8} > 32000a^{7}b^{9} > 10^{5}b^{9},

again since a ≥ 2.

Putting these results together, we have proven that

c > 40000b^{9},

which was the claim of our lemma. □

5. Lower bounds for m, n

The intention in this section is to prove that the gap principle from the last section implies that we can estimate m, n in terms of c from below. The idea is that if v_{m} = w_{n}, then we can study this equation modulo 8c^{2}, which leads to equation (13). But if a, b, m, n are small compared with c, then these congruences are equations, which are in contradiction to the equation v_{m} = w_{n}. This will imply the lower bounds for m, n. We have the following lemma:

Lemma 14. If v_{m} = w_{n}, then m ≥ n > c^{0.064}.

Proof. We start with the congruence (13) which is

am^{2}z_{0} + smx_{0} ≡ bn^{2}z_{0} + tny_{1} (mod 4c).

Let us assume that n ≤ c^{0.064}. By Lemma 5 we have m ≤ 2c^{0.064} < c^{0.309} since c ≥ 17. All terms in the above congruence are bounded by bm^{2}z_{0}. By our assumption, Lemma 4 and the gap principle c > 40000b^{9} from Lemma 13, we get that

bm^{2}z_{0} < \frac{4}{40000^{\frac{1}{9}}} c^{\frac{1}{9} +2·0.061+ \frac{1}{2}} < c.

Hence,

am^{2}z_{0} + smx_{0} = bn^{2}z_{0} + tny_{1}

or

z_{0}(am^{2} − bn^{2}) = tny_{1} − smx_{0}.

Squaring leads to

−(am^{2} − bn^{2})^{2} ≡ −n^{2}y_{1}^{2} − m^{2}x_{0}^{2} − 2stmnx_{0}y_{1} (mod c)

or

[(am^{2} − bn^{2})^{2} − n^{2}y_{1}^{2} − m^{2}x_{0}^{2}]^{2} ≡ 4m^{2}n^{2}x_{0}^{2}y_{1}^{2} (mod c).

The left side is bounded by

[(bm^{2})^{2}]^{2} = b^{4}m^{8} < 256b^{4}n^{8} < 256b^{4}c^{0.064} < c

since we have c > 40000b^{9}, which implies that the above inequality is satisfied for 256b^{4} < 176b^{4.392} < c^{0.488} which is true for b > 2.6. Also the right side is by Lemma 4 bounded by

4m^{2}n^{2}ab < 16n^{4}b^{2} < 16b^{2}c^{0.256} < c.

Thus, we may conclude

(am^{2} − bn^{2})^{2} = (ny_{1} ± mx_{0})^{2},

which gives

am^{2} − bn^{2} = ±ny_{1} ± mx_{0} or am^{2} − bn^{2} = ±ny_{1} ∓ mx_{0}.

Therefore, we have

am^{2}z_{0} − bn^{2}z_{0} = tny_{1} − smx_{0} = ±ny_{1}z_{0} ± mx_{0}z_{0}, or
am^{2}z_{0} − bn^{2}z_{0} = tny_{1} − smx_{0} = ±ny_{1}z_{0} ∓ mx_{0}z_{0},

which implies

mx_{0}(s ± z_{0}) = ny_{1}(t ∓ z_{0}) or mx_{0}(s ± z_{0}) = ny_{1}(t ± z_{0}).

Let us first consider the first equation. Squaring once again leads to

n^{2}y_{1}^{2}(bc − 1 ∓ 2tz_{0} + z_{0}^{2}) = m^{2}x_{0}^{2}(ac − 1 ± 2sz_{0} + z_{0}^{2}).

Taking this equation modulo c and multiplying by z_{0} gives

n^{2}y_{1}^{2}(−2z_{0} ± 2t) ≡ m^{2}x_{0}^{2}(−2z_{0} ∓ 2s) (mod c).

Both sides are bounded by

4m^{2}b^{2}\sqrt{bc} < 32c^{2·0.064+ \frac{2}{9} + \frac{1}{2} + \frac{1}{18}} < c.

Thus, we have equality, which means

m^{2}x_{0}^{2}(z_{0} ± s) = n^{2}y_{1}^{2}(z_{0} ∓ t).

But by squaring the starting equation, we also have

m^{2}x_{0}^{2}(s ± z_{0})^{2} = n^{2}y_{1}^{2}(t ∓ z_{0})^{2},

and therefore by dividing this equation by the previous one we conclude that either s + z_{0} = z_{0} − t or z_{0} − s = z_{0} + t, which gives a contradiction in both cases.

The second equation can be handled exactly in the same way as the first one, leading to

m^{2}x_{0}^{2}(z_{0} ± s) = n^{2}y_{1}^{2}(z_{0} ± t),
m^{2}x_{0}^{2}(z_{0} ± s)^{2} = n^{2}y_{1}^{2}(z_{0} ± t)^{2}

and by dividing one equation through the other we get z_{0} ±s = z_{0} ±t, which is again a contradiction.

Therefore, the proof of the lemma is finished. □

6. Application of a theorem of Bennett

Now we are ready to calculate an upper bound for d in the D(−1)- quadruple {a, b, c, d}. For this we apply the following important and very useful result of Bennett [2].

Lemma 15. If a_{i}, p_{i}, q and N are integers for 0 ≤ i ≤ 2, with a_{0} < a_{1} < a_{2}, a_{j} = 0 for some 0 ≤ j ≤ 2, q nonzero and N > M^{9}, where

M = max_{0≤i≤2}{|a_{i}|},

then we have

max_{0≤i≤2} \{|\sqrt{1 + \frac{a_{i}}{N}} − \frac{p_{i}}{q} |\} > (130Nγ)^{−1}q^{−λ}

where

λ = 1 + \frac{log(33Nγ)}{log (1.7N^{2} ∏_{0≤i<j≤2}(a_{i} − a_{j})^{−2})}

and

γ = \{ \begin{matrix}\frac{(a−2−a_{0})^{2}(a_{2}−a_{1})^{2}}{2a_{2}−a_{0}−a_{1}} & if a_{2} − a_{1} ≥ a_{1} − a_{0}, \\ \frac{(a_{2}−a_{0})^{2}(a_{1}−a_{0})^{2}}{a_{1}+a_{2}−2a_{0}} & if a_{2} − a_{1} < a_{1} − a_{0}. \end{matrix}

We apply Lemma 15 to the numbers

θ_{1} = \frac{s}{a}\sqrt{\frac{a}{c}} = \sqrt{1 − \frac{b}{abc}} and θ_{2} = \frac{t}{b}\sqrt{\frac{b}{c}} = \sqrt{1 − \frac{a}{abc}}.

So θ_{1} and θ_{2} are square roots of rationals which are close to 1. First, we show that every solution of our problem induce good approximations of these numbers.

Lemma 16. All positive integer solutions x, y, z of (2) and (3) satisfy

max \{|θ_{1} − \frac{sbx}{abz}| , |θ_{2} − \frac{tay}{abz} |\} < \frac{c}{a}z^{−2}.

Proof. This is a special case of Lemma 1 in [10]. □

By combining this lemma with the lower bound from the theorem of Bennett we obtain an upper bound for d.

Lemma 17. Let {a, b, c, d} be a D(−1)-quadruple with 2 ≤ a < b < c < d and with minimal d, then

d < c^{51.06}.

Proof. We apply Lemma 15 with a_{0} = −b, a_{1} = −a, a_{2} = 0, N = abc, M = b, q = abz, p_{1} = sbx, p_{2} = tay. Since abc > b^{9} by our gap principle (Lemma 13), the condition N > M^{9} is satisfied. Considering γ from Lemma 15 as a function of a we easily get

\frac{b^{3}}{6} ≤ γ < \frac{b^{3}}{2} .

Lemma 15 and Lemma 16 imply

\frac{c}{az^{2}} > (130abcγ)^{−1}(abz)^{λ_{1}−2},

where

λ_{1} = \frac{log(33abcγ)}{log (1.7c^{2}(b − a)^{2})}.

This gives

z^{λ_{1}} < 65a^{2}b^{6}c^{2}

and

log z < \frac{log(65a^{2}b^{6}c^{2}) log(1.7c^{2}(b − a)^{−2})}{log (\frac{1.7c}{33ab(b−a)^{2}γ} )}.

We have

65a^{2}b^{6}c^{2} < 65b^{8}c^{2} < \frac{65}{40000^{\frac{8}{9}}} c^{2+ \frac{8}{9}} < c^{2.889},

1.7c^{2}(b − a)^{−2} < c^{2},

\frac{1.7c}{33ab(b − a)^{2}γ} > \frac{1.7c}{33abb^{2} \frac{b^{3}}{2}} > 0.103cb^{−7} > 391c^{1− \frac{7}{9}} > c^{0.222}.

Therefore, we get

log z < \frac{2.889 log c · 2 log c}{0.222 log c} < 26.028 log c.

Hence,

(25)  z < c^{26.028}

and

d = \frac{z^{2} + 1}{c} < c^{51.06},

which was the claim of the lemma. □

In the special case that a and b are very close, we can prove a better upper bound.

Lemma 18. Let {a, b, c, d} be a D(−1)-quadruple with 2 ≤ a < b < c < d and with minimal d. If b < 2a, then

d < c^{33.72}.

Proof. The proof runs the same line as the proof of the last lemma. By the assumption that b < 2a, we conclude by Lemma 7 applied to the D(−1)- triple {1, a, b} that

a = b − 2\sqrt{b − 1} + 1.

In this case we can get a better lower bound for

\frac{1.7c}{33ab(b − a)^{2}γ} > \frac{1.7c}{33b^{2}(2\sqrt{b})^{2} \frac{b^{3}}{2}} > 0.0257cb^{−6} > 30c^{1− \frac{6}{9}} > c^{0.333},

where we have used that b − a = 2\sqrt{b − 1} − 1 ≤ 2\sqrt{b}. Now, we get

log z < \frac{2.889 log c · 2 log c}{0.333 log c} < 17.352 log c,

and therefore, z < c^{17.352} and

d < \frac{z^{2} + 1}{c} < c^{33.72}.

That was the claim we wanted to prove. □

7. Proof of the theorems

Now, we put everything together to prove our theorems.

Proof of Theorems 1a and 1b.

Let {a, b, c, d} be a D(−1)-quadruple with 2 ≤ a < b < c < d and with minimal d. We use the lower bound, which is given by (17), to get

c^{n−1} < w_{n} = z < c^{26.028},

where we have used (25) from the proof of Lemma 17. It follows that n < 27.028 and thus n ≤ 27. But now we can use Lemma 14 to get

c^{0.064} < n ≤ 27,

which implies the following upper bound for c, namely

c < 27^{\frac{1}{0.064}} < 27^{15.63} < 2.36 · 10^{22}.

By our improved gap principle (Lemma 13), which was c > 40000b^{9}, we therefore derive

b ≤ 94.

It is easy to give a list of all D(−1)-triples {1, a, b} with b ≤ 94. They are

{1, 2, 5}, {1, 5, 10}, {1, 10, 17}, {1, 17, 26}, {1, 26, 37},
{1, 37, 50}, {1, 50, 65}, {1, 65, 82}, {1, 5, 65}.

The first eight of these triples are {1, i^{2} +1, (i+1)^{2} +1} for i = 1, …, 9. The only one in the above list, which is not of this form, is the last one, namely {1, 5, 65}. As remarked in the introduction it was proved already earlier that {1, 2, 5}, {1, 5, 10}, {1, 5, 65}, {1, 10, 17} and {1, 26, 37} cannot be extended to D(−1)-quadruples. So, we are left with the cases

{1, i^{2} + 1, (i + 1)^{2} + 1} for i = 4, 6, 7, 8, 9.

In all these cases a and b are very close (in fact they are closest possible) to each other. Therefore, by using Lemma 18 in the same way as above, we get

c^{n−1} < w_{n} = z < c^{17.36},

which implies n ≤ 18. Using the lower bound for n, we get

c < 18^{15.63} < 4.17 · 10^{19}.

This implies b ≤ 46. Hence, the only remaining case is

{1, 17, 26}.

In this case we have to consider c’s with 17c−1 = s^{2}, 26c−1 = t^{2} and where c − 1 is also a square. Therefore, we look at the Pell equation given by

17t^{2} − 26s^{2} = 9,

where the fundamental solutions are bounded by 0 < t_{0} < 26, |s_{0}| < \sqrt{17 · 26 − 1} = 21 and therefore are s_{0} = ±4, t_{0} = 5. All the solutions t satisfy the following linear recurrences

t_{0} = 5, t_{1} = 47 or 8783, t_{n+2} = 1766t_{n+1} − t_{n}.

We are interested in those t = t_{n} for which t^{2} = 27c − 1 ≤ 26 · 4.17 · 10^{19}, this gives t < 4 · 10^{10}, and which have the additional property that

c − 1 = \frac{t^{2} + 1}{26} − 1

is a square. It is checked easily that such a t does not exist, showing that {1, 17, 26} cannot be extended to a quadruple {1, 17, 26, c} with c < 4.17 · 10^{19}.

Finally we get that there is no D(−1)-quadruple {a, b, c, d} with 2 ≤ a < b < c < d. This completes the proof of our main result. □

8. Acknowledgements

This paper was written during a visit of the second author at the University of Zagreb in the frame of a joint Austrian-Croatian project granted by the Croatian Ministry of Science, Education and Sports and the Austrian Exchange Service (Nr. 20/2004). Both authors are most grateful for this support. The second author wants to thank Andrej Dujella for his kind hospitality.

References

[1] A. Baker and H. Davenport, The equations 3x^{2} −2 = y^{2} and 8x^{2} −7 = z^{2}, Quart. J. Math. Oxford Ser. (2) 20 (1969), 129-137.

[2] M. A. Bennett, On the number of solutions of simultaneous Pell equations, J. Reine Angew. Math. 498 (1998), 173-199.

[3] E. Brown, Sets in which xy +k is always a square, Math. Comp. 45 (1985), 613-620.

[4] L. E. Dickson, “History of the Theory of Numbers” Vol. 2, Chelsea, New York, 1966, 518-519.

[5] Diophantus of Alexandria, Arithmetics and the Book of Polygonal Numbers, (I. G. Bashmakova, Ed.) (Nauka, 1974) (in Russian), 85-86, 215-217.

[6] A. Dujella, On the exceptional set in the problem of Diophantus and Davenport, Application of Fibonacci Numbers Vol. 7, (G. E. Bergum, A. N. Philippou, A. F. Horadam, eds.), Kluwer, Dordrecht, 1998, 69-76.

[7] A. Dujella, Complete solution of a family of simultaneous Pellian equations, Acta Math. Inform. Univ. Ostraviensis 6 (1998), 59-67.

[8] A. Dujella, An absolute upper bound for the size of Diophantine m-tuples, J. Number Theory 89 (2001), 126-150.

[9] A. Dujella, An extension of an old problem of Diophantus and Euler II, Fibonacci Quart. 40 (2002), 118-123.

[10] A. Dujella, On the size of Diophantine m-tuples, Math. Proc. Cambridge Philos. Soc. 132 (2002), 23-33.

[11] A. Dujella, There are only finitely many Diophantine quintuples, J. Reine Angew. Math. 566 (2004), 183-214.

[12] A. Dujella and C. Fuchs, A polynomial variant of a problem of Diophantus and Euler, Rocky Mount. J. Math. 33 (2003), 797-811.

[13] A. Dujella and C. Fuchs, Complete solution of a polynomial version of a problem of Diophantus, J. Number Theory, to appear (Preprint: http://www.math.hr/∼ duje/polquad4.dvi).

[14] A. Dujella and A. Pethö, Generalization of a theorem of Baker and Davenport, Quart. J. Math. Oxford Ser. (2) 49 (1998), 291-306.

[15] P. Gibbs, Some rational Diophantine sextuples, preprint, math.NT/9902081.

[16] T. L. Heath, “Diophantus of Alexandria: A study in the history of Greek Algebra”. With a supplement containing an account of Fermat’s theorems and problems connected with Diophantine analysis and some solutions of Diophantine problems by Euler (Cambridge, England, 1910), Powell’s Bookstore, Chicago; Martino Publishing, Mansfield Center, 2003, 162-164, 344-347.

[17] B. W. Jones, A variation of a problem of Davenport and Diophantus, Quart. J. Math. Oxford Ser.(2) 27 (1976), 349-353.

[18] B. W. Jones, A second variation of a problem of Davenport and Diophantus, Fibonacci Quart. 15 (1977), 323-330.

[19] K. S. Kedlaya, Solving constrained Pell equations, Math. Comp. 67 (1998), 833-842.

[20] O. Kihel, On the extendibility of the set {1, 2, 5}, Fibonacci Quart. 38 (2000), 464– 466.

[21] S. P. Mohanty and A. M. S. Ramasamy, The simultaneous Diophantine equations 5y^{2} − 20 = x^{2} and 2y^{2} + 1 = z^{2}, J. Number Theory 18 (1984), 356-359.

[22] P. G. Walsh, On two classes of simultaneous Pell equations with no solutions, Math. Comp. 68 (1999), 385-388.

[23] D. X. Zheng, On the systems of Diophantine equations y^{2} − 2x^{2} = 1, z^{2} − 5x^{2} = 4 and y^{2} − 5x^{2} = 4, z^{2} − 10x^{2} = 9, Sichuan Daxue Xuebao 24 (1987), 25-29.

Andrej Dujella Department of Mathematics University of Zagreb Bijenička cesta 30, 10000 Zagreb, Croatia e-mail: duje@math.hr

Clemens Fuchs Institut für Mathematik TU Graz Steyrergasse 30, 8010 Graz, Austria e-mail: clemens.fuchs@tugraz.at