Effective Solution of the D(−1)-quadruple Conjecture

Andrej Dujella

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

Alan Filipin

Faculty of Civil Engineering, University of Zagreb Fra Andrije Kačića-Miošića 26, 10000 Zagreb, Croatia E-mail: filipin@grad.hr

and

Clemens Fuchs^{∗}

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

Abstract

The D(−1)-quadruple conjecture states that there does not exist a set of four positive integers such that the product of any two distinct elements is one greater than a perfect square. An effective proof is given by showing that if {a, b, c, d} is such a set then max{a, b, c, d} < 10^{10^{23}}, leaving open a completely determined (but currently computationally infeasible large) set of exceptions to check.

1 Introduction and results

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.

∗Corresponding author.

A lot of work has been done on D(n)-m-tuples in the last decade, but in fact the problem is much older. It was first studied by Diophantus in the case n = 1. 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. Later Euler was able to add the fifth positive rational, \frac{777480}{8288641}, to the Fermat’s set (see [6], [7, pp. 103-104, 232], [8, pp. 141-145] and [27, pp. 177-181]). Recently, Gibbs [25] 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 [2] 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 [15], this refined previous results [22, 12]). The last result is effective, namely if {a, b, c, d, e} is a D(1)-quintuple, then d < 10^{2171} and e < 10^{10^{26}} (cf. [15, Corollary 4, p. 210]). This proves the D(1)-quintuple conjecture in an effective way, leaving open a completely determined finite set to check.

For general n, it is easy to see, by considering congruences modulo 4, that if n ≡ 2 (mod 4) then there does not exist a D(n)-quadruple (see [5, 26, 34]), while the first author proved in [9] 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).

The best known upper bounds for the size of sets with the property D(n) are logarithmic in |n| (see [14, 16]). Moreover, for n prime it was shown recently that the size is bounded by the absolute constant 2^{170} (cf. [21]).

Therefore, in the case n = −1 the conjecture is that there does not exist a D(−1)-quadruple. In analogy to above this is known as the D(−1)- quadruple conjecture and it appeared explicitly in [10] for the first time. 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 [5], see also [39, 30, 38, 31]), {1, 5, 10} (by Mohanty and Ramasamy in [33]), {1, 2, 145}, {1, 2, 4901}, {1, 5, 65}, {1, 5, 20737}, {1, 10, 17}, {1, 26, 37} (by Kedlaya [30]) and {17, 26, 85} (again by Brown in [5]). Moreover, Brown proved that the infinite families {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) of D(−1)-triples cannot be extended to quadruples. The first author proved the conjecture in [11] for all triples of the form {1, 2, c}.

Very recently, important progress was made by Dujella and Fuchs in [20]. They proved that there does not exist a D(−1)-quintuple. More precisely, it is proved that there does not exist a D(−1)-quadruple {a, b, c, d} with 2 ≤ a < b < c < d. This implies that we are left by considering the case a = 1, i.e. D(−1)-quadruples of the form {1, b, c, d}. In [20] it was remarked that the case a = 1 seems more involved and much harder and that 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.

In the meantime the non-extendibility of {1, b, c} was confirmed for b = 5 (in [1]), for b = 10 (by the second author [23]), and for b = 17, 26, 37, 50 (by Fujita in [24]).

We mention that a polynomial variant of the above problems was first studied by Jones [28], [29], and it was for the case n = 1. Recently, polynomial variants were also studied by Dujella and Fuchs. 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 [15]) in [19] and also for the case n = −1 in [18].

The aim of the present paper is to go further and to prove that in fact there are at most finitely many D(−1)-quadruples {1, b, c, d}. We will prove the following result:

Theorem 1. Let {1, b, c} be a D(−1)-triple. Then:

(i) If c > b^{9}, then there does not exist an extension to a D(−1)-quadruple {1, b, c, d} such that d > c.

(ii) If 11b^{6} ≤ c ≤ b^{9}, then there does not exist an extension to a D(−1)- quadruple.

Assume that {1, b, c, d} with 1 < b < c < d is an extension to a D(−1)- quadruple.

(iii) If b^{3} < c < 11b^{6}, then c < 10^{238}, d < 10^{10^{20}},

(iv) if b^{1.1} < c ≤ b^{3}, then c < 10^{492}, d < 10^{10^{23}},

(v) if 3b < c ≤ b^{1.1}, then c < 10^{95}, d < 10^{10^{21}},

(vi) if c = 1 + b + 2\sqrt{b − 1}, then c < 10^{76}, d < 10^{10^{21}}.

Note that, by [20, Lemma 7], if c > b and c ≠ 1 + b + 2\sqrt{b − 1} then it holds c > 3b. Hence, the theorem covers all possible D(−1)-triples {1, b, c} such that 1 < b < c.

To sum up, we get the following effective solution of the D(−1)- quadruple conjecture:

Theorem 2. There are only finitely many D(−1)-quadruples. Moreover, if {a, b, c, d} is a D(−1)-quadruple, then max{a, b, c, d} < 10^{10^{23}}.

As in the case for the D(1)-quintuple conjecture, this proves the D(−1)- quadruple conjecture in an effective way. Unfortunately, the remaining set is to large to be checked by the means of a computer programm at the moment.

This is also the first nontrivial result (for integers ≢ 2 (mod 4)) related to the following conjecture

Conjecture 1. If n is not a perfect square, then there exist only finitely many D(n)-quadruples.

Since, by [9, Remark 3], all elements of a D(−4)-quadruple are even, Theorem 2 implies that Conjecture 1 is valid for n = −1 and n = −4.

Recently, for the case n = 1, it was shown by the first author in [17] that the number of D(1)-quintuples can be bounded by 10^{1930}. By using similar arguments together with Theorem 1, we can also derive an upper bound for the number of D(−1)-quadruples.

Theorem 3. The number of D(−1)-quadruples is bounded by 10^{903}.

The case n = −1 deserves special attention because this case is closely connected with another old problem investigated by 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 [6], [7, pp. 85-86, 215-217], [8, pp. 101-104] and [27, pp. 162-164, 344-347]). In [13] an infinite family of rational quintuples with the same property was given. Since

xy + x + y = (x + 1)(y + 1) − 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. In fact the main result in [20] completely solved this problem, namely it was shown that there does not exist a set of four positive integers such that the product of any two distinct elements plus there sum is a perfect square.

By the equivalence (1), we get the following corollaries to the above theorems, which extend the problem solved in [20].

Corollary 1. There are at most finitely many sets {a^{2}, b^{2}, c^{2}}, 0 < a < b < c of three positive perfect squares such that the product of any two of its distinct elements plus their sum is a perfect square.

This can be reformulated as (the case of positive elements cannot occur at all by [20, Theorem 1a]):

Corollary 2. There are at most finitely many sets of four integers with the property that the product of any two of its distinct elements plus their sum is a perfect square.

Of course, both corollaries are effective and therefore settle the problem up to a finite amount of computations.

In the next section we will start by collecting important information about this problem (especially from [20]). The strategy of proof follows the same lines as the proofs of almost all other recent results on nonextendability of D(n)-m-tuples. First we reduce the problem of finding d which extends {1, 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. We know that the indices of these sequences have the same parity and they are very simply connected. Moreover, we have a very important congruence relation. E.g. this relation implies that the sequences cannot have intersections for small indices. Moreover, by the gap principle, which was the main improvement in the proof in [20], we get precise information on the initial terms of the recurring sequences.

If c ≥ 11b^{6}, in Section 3 we obtain (Proposition 1) by Bennett’s theorem on simultaneous approximations of square roots which are close to 1 (in a slightly refined version for our context by Fujita [24]) parts (i) and (ii) (the case of very large solutions) of our main theorem (Theorem 1).

In Section 4 we will compute some general upper bound for the indices of the recurring sequences in terms of c by using Baker’s theory of linear forms in logarithms of algebraic numbers (in fact we will use Matveev’s result [32]).

In Section 5 we will give the proof of part (iii)-(v) (large, medium size, and small solutions, respectively) of the theorem (see Propositions 2, 3, 4), where we have to get lower bounds for n in terms of some power of c. For medium size (part (iv) of the theorem, i.e. b^{1.1} < c < b^{4}) and small (part (v) of the theorem, i.e. 3b < c < b^{1.1}) solutions we first have to refine the congruence relations. In fact this is the most important new part of the proof.

Finally, we will give the proof of the case of very small solutions (c ≤ 3b), namely part (vi) of the theorem, in Section 6 (Proposition 5).

The proof of the quantitative result in Theorem 3 will be given in the final Section 7.

2 Preliminaries

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

b − 1 = r^{2}, c − 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 {1, b, c, d} is a D(−1)- quadruple. We have

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

with integers x, y, z. Eliminating d from (2) we obtain the following system of Pellian equations

z^{2} − cx^{2} = c − a,  (3)
bz^{2} − cy^{2} = c − b.  (4)

By combining these two equations we additionally have y^{2}−bx^{2} = b−1. We will describe the sets of solutions of equations (3) and (4) in the following lemma.

Lemma 1. If (z, x) and (z, y), with positive integers x, y, z, are solutions of (3) and (4) 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 (3) and (4) respectively,

(ii) the following inequalities are satisfied:

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

and there exist integers m, n ≥ 0 such that

z\sqrt{a} + x\sqrt{c} = (z_{0}\sqrt{a} + x_{0}\sqrt{c})(s + \sqrt{ac})^{2m},  (7)
z\sqrt{b} + y\sqrt{c} = (z_{1}\sqrt{b} + y_{1}\sqrt{c})(t + \sqrt{bc})^{2n}.  (8)

Proof. This is [20, Lemma 1] □

From (7) we conclude that z = v_{m}, for some (z_{0}, x_{0}) with the above properties and integer m ≥ 0, where

v_{0} = z_{0}, v_{1} = (2c − 1)z_{0} + 2scx_{0}, v_{m+2} = (4c − 2)v_{m+1} − v_{m}. (9)

Hence for varying m ≥ 0 the solutions z form a binary recurrent sequence (v_{m})_{m≥0} whose initial terms are found by solving equation (7) for z when m = 0 and 1, and whose characteristic equation has the roots (s + \sqrt{c})^{2} and (s − \sqrt{c})^{2}. In the same manner, from (8), we conclude that z = w_{n} for some (z_{1}, y_{1}) with the above properties and integer n ≥ 0, where

w_{0} = z_{1}, w_{1} = (2bc − 1)z_{1} + 2tcy_{1}, w_{m+2} = (4bc − 2)w_{n+1} − w_{n}. (10)

Our system of equation (3) and (4) is thus transformed to finitely many equations of the form z = v_{m} = w_{n}.

From (9) and (10) we get by induction

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

(see [20, Lemma 2]). So if the equation v_{m} = w_{n} has a solution, then we have z_{0} = z_{1}. Moreover, we have the following properties:

Lemma 2. If v_{m} = w_{n}, n ≠ 0, then

(i) m ≡ n ( mod 2),

(ii) n ≤ m ≤ 2n,

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

Proof. Part (i) and (iii) follow at once from what we just said. Statement (ii) is [20, Lemma 5]. Observe that we remarked there that this statement holds for arbitrary D(−1)-quadruples and not only in the special situation considered there. □

Moreover, we have the following bounds for v_{m}, w_{n}, respectively.

Lemma 3. We have

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

Proof. This is part of the proof of [20, Lemma 5] (cf. (3.1) in [20] and the equation just before that). □

We collect some useful gap principles for the elements of D(−1)-triples and quadruples.

Lemma 4. Let {a, b, c} be a D(−1)-triple and 0 < a < b < c, then c = a + b + 2r or c > 3ab ≥ 3b. Let {1, b, c, d} be a D(−1)-quadruple and 1 < b < c < d, then we have d > c^{9}.

Proof. The first part of the Lemma is [20, Lemma 7]. The second part is a consequence of [20, Lemma 13]. Observe that in fact there it is proved for the D(−1)-quadruple {1, a, b, c} with 1 < a < b < c. The assumption that {a, b, c, d} with 2 ≤ a < b < c < d and d minimal under all such quadruples was only needed to ensure that {1, a, b, c} is really a D(−1)-quadruple again (by the use of the key lemma [20, Lemma 3]). □

This has the following important implication on the fundamental solutions of (9) and (10) described in Lemma 1.

Lemma 5. Let the integers z_{0}, z_{1}, x_{0}, y_{1} be as in Lemma 1. If c ≤ b^{9}, then

z_{0} = z_{1} = s, x_{0} = 0, y_{1} = ±\sqrt{b − 1} = ±r.

Proof. Let us define d_{0} by

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

It follows easily that

d_{0} − 1 = x_{0}^{2}, bd_{0} − 1 = y_{1}^{2}, cd_{0} − 1 = z_{0}^{2}

and d_{0} < c (this construction was the key lemma in [20], in fact it was [20, Lemma 3]). If d_{0} > 1, then {1, b, d_{0}, c} is again a D(−1)-quadruple and by the second part of Lemma 4 it follows that c > b^{9}, a contradiction. Hence, d_{0} = 1, which implies our assertion (cf. also [20, Lemma 4] for this final conclusion). □

It is easy to show that the equation v_{m} = w_{n} cannot have solutions with small indices.

Lemma 6. We have v_{1} ≠ w_{1}, v_{2} ≠ w_{2}, v_{4} ≠ w_{2}.

Proof. See [20, Lemma 8, Lemma 10 and Lemma 11]. Observe again that the conclusion there is independent of the existence of a minimal extension which was assumed there. □

Now we have collected all information we need to prove our main theorem which will be done in the following sections, respectively.

3 Very large solutions

The proof will follow from a lower bound for m, n in terms of some power of c, together with an logarithmic upper bound obtained from Bennett’s theorem. The idea for obtaining the lower bound is that if v_{m} = w_{n}, then we can study the congruence equation from Lemma 2(iii). We will show that if b, m, n are small compared with c, these congruences are equations, which are in contradiction to the equation v_{m} = w_{n}.

Lemma 7. If v_{m} = w_{n}, n ≠ 0, 1 and c ≥ 11b^{6}, then n > c^{\frac{1}{6}} .

Proof. From Lemma 2(iii) we have m^{2}z_{0}+smx_{0} ≡ bn^{2}z_{0}+tny_{1} (mod 4c). Assume now that n ≤ c^{\frac{1}{6}} . It is easy to see that max\{|m^{2}z_{0}|, |smx_{0}|, |tny_{1}|\} < |bn^{2}z_{0}|. We first consider the case c > b^{9}. Let d_{0} be such that cd_{0} − 1 = z_{0}^{2}. If d_{0} > 1 then {1, b, d_{0}, c} is a D(−1)-quadruple and we get by the second part of Lemma 4 that c > (max{b, d_{0}})^{9} > b^{6}d_{0}^{3}. Then we have

|bn^{2}z_{0}| < b · c^{\frac{1}{3}} · (cd_{0})^{\frac{1}{2}} < c^{\frac{1}{3}} · c^{\frac{1}{2}} · (b^{6}d_{0}^{3})^{\frac{1}{6}} < c.

If d_{0} = 1 then we get as in the proof of Lemma 5 that z_{0} = s. Thus we immediately get

|bn^{2}z_{0}| < c^{\frac{1}{3}} · c^{\frac{1}{2}} · c^{\frac{1}{2}} < c.

Now we consider the case 11b^{6} ≤ c ≤ b^{9}. Here we can use Lemma 5 to get

|bn^{2}z_{0}| < c^{\frac{1}{6}} · c^{\frac{1}{3}} · c^{\frac{1}{2}} = c.

Therefore we can conclude that m^{2}z_{0} + smx_{0} = bn^{2}z_{0} + tny_{1}. We have

m^{2}z_{0} + smx_{0} < m^{2}\sqrt{cd_{0}} + m\sqrt{cd_{0}} < m(m + 1)\sqrt{\frac{c}{b}}\sqrt{|y_{1}|^{2} + 1}

and

bn^{2}z_{0}−tn|y_{1}| > \sqrt{bc}|y_{1}|n^{2}−tn|y_{1}| > \sqrt{bc}|y_{1}|n^{2}−\sqrt{bc}|y_{1}|n = \sqrt{bc}|y_{1}|n(n−1).

This implies

m(m + 1)\sqrt{|y_{1}|^{2} + 1} > b|y_{1}|n(n − 1)

and by the relation m ≤ 2n and b ≥ 65 we get

5 ≥ \frac{2n + 1}{n − 1} > \frac{65|y_{1}|}{2\sqrt{|y_{1}|^{2} + 1}} > \frac{65}{2\sqrt{2}}

and therefore we get a contradiction that proves the statement. □

Next we need an upper bound for n in terms of c. To obtain it we use the following slightly modified version of a special case of Bennett’s theorem [4] (it is also a modified version of Rickert’s theorem [35]), which was proved in [24].

Lemma 8. Let b and N be integers with b ≥ 5 and N ≥ 2.39b^{7}. Then the numbers

θ_{1} = \sqrt{1 + \frac{1 − b}{N}} and θ_{2} = \sqrt{1 + \frac{1}{N}}

satisfy

max \{|θ_{1} − \frac{p_{1}}{q} | , |θ_{2} − \frac{p_{2}}{q} |\} > [32.1\frac{b^{2}(b − 1)^{2}}{2b − 1} N]^{−1} q^{−1−λ}

for all integers p_{1}, p_{2}, q with q > 0, where

λ = \begin{matrix}log & \frac{16.1b^{2}(b − 1)^{2}N}{2b − 1} \\ log & \frac{3.37N^{2}}{b^{2}(b − 1)^{2}} \end{matrix} < 1.

Proof. This is Theorem 3.5 in [24]. □

We apply this theorem with the numbers N = t^{2} = bc − 1 ≥ 11b^{7} − 1 > 2.39b^{7}, p_{1} = bsx, p_{2}bz, q = ty, i.e. to

θ_{1} = \frac{s\sqrt{b}}{t}, θ_{2} = \frac{\sqrt{bc}}{t}.

First we show that the solutions of our problem induce good approximations to the roots θ_{1}, θ_{2}.

Lemma 9. All positive integers solutions x, y, z of (3) and (4) satisfy

max \{|θ_{1} − \frac{bsx}{ty}| , |θ_{2} − \frac{bz}{ty} |\} < \frac{b − 1}{y^{2}}.

Proof. Since f(x) = b(x − 1)/(bx − 1) is an increasing function we get

|θ_{1} − \frac{bsx}{ty}| = \frac{s\sqrt{b}}{t} |1 − \frac{x\sqrt{b}}{y}| = \frac{s\sqrt{b}}{t} |\frac{1 − bx^{2}}{y^{2}} | |1 + \frac{x\sqrt{b}}{y}|^{−1} < \frac{b − 1}{y^{2}}.

Moreover, since by/y > \sqrt{bc} we have

|θ_{2} − \frac{bz}{ty} | = \frac{1}{t} |\sqrt{bc} − \frac{bz}{y} | = \frac{b}{t} |c − \frac{bz^{2}}{y^{2}} | |\sqrt{bc} + \frac{bz}{y} |^{−1}
< \frac{b}{t} \frac{c − b}{y^{2}} \frac{1}{2\sqrt{bc}} < \frac{1}{2y^{2}} \frac{bc − 1}{\sqrt{bc(bc − 1)}} < \frac{b − 1}{y^{2}}.

This proves the lemma. □

Now we are ready to prove the first two parts of Theorem 1. Namely, we show the following proposition.

Proposition 1. Let {1, b, c} with 1 < b < c be a D(−1)-triple. If c ≥ 11b^{6}, then there does not exist an integer d > c such that {1, b, c, d} is a D(−1)- quadruple. If 11b^{6} ≤ c ≤ b^{9}, then the triple {1, b, c} cannot be extended to a D(−1)-quadruple.

Proof. Combining the upper bound of Lemma 9 with the lower bound from Lemma 8 we get

[32.1\frac{b^{2}(b − 1)^{2}}{2b − 1} t^{2}]^{−1} (ty)^{−1−λ} < \frac{b − 1}{y^{2}},

which implies

y^{1−λ} < 32.1\frac{b^{2}(b − 1)^{3}}{2b − 1} t^{3+λ} < \frac{32.1b^{2}(b − 1)^{3}(bc − 1)^{2}}{2b − 1} .

We have

\frac{1}{1 − λ} = \begin{matrix}log & \frac{3.37(bc − 1)^{2}}{b^{2}(b − 1)^{2}} \\ log & \frac{3.37(2b − 1)(bc − 1)}{16.1b^{4}(b − 1)^{4}} \end{matrix} ,

which leads to

log y < log \frac{3.37(bc − 1)^{2}}{b^{2}(b − 1)^{2}} log \frac{32.1b^{2}(b − 1)^{3}(bc − 1)^{2}}{2b − 1} ; log \frac{3.37(2b − 1)(bc − 1)}{16.1b^{4}(b − 1)^{4}} .

We have (observe that c ≥ 11b^{6} and b ≥ 65, c ≥ 82)

log \frac{3.37(bc − 1)^{2}}{b^{2}(b − 1)^{2}} < log \frac{3.37b^{2}c^{2}}{b^{2}(b − 1)^{2}} < 2 log \frac{1.9c}{b − 1} < 2 log 0.03c,

log \frac{32.1b^{2}(b − 1)^{3}(bc − 1)^{2}}{2b − 1} < log (16.18b^{6}c^{2}) < 3 log 1.14c,

log \frac{3.36(2b − 1)(bc − 1)}{16.1b^{4}(b − 1)^{4}} > log \frac{2.29(2b^{2}c − bc − 2b + 1)}{b^{2}c} > 1.51.

Consequently,

log y < 3.98 log 0.03c log 1.14c.

Moreover, from (4) and the first part of Lemma 3 we get

y > \sqrt{\frac{b}{c}}z = \sqrt{\frac{b}{c}}v_{m} > \sqrt{\frac{b}{c}}(c − 1)(4c − 3)^{m−1} > (4c − 3)^{m−1}.

Assume that n ≠ 0, 1. Then by Lemma 2 and Lemma 7 we have m ≥ n > c^{\frac{1}{6}} , which implies

(c^{\frac{1}{6}} − 1) log(4c − 3) < 3.98 log 0.03c log 1.14c.  (11)

If we define the function

f(c) = 3.98 log 0.03c log 1.14c − (c^{\frac{1}{6}} − 1) log(4c − 3),

then it is easy to check that this function is degreasing for c > 0.55 · 10^{12}. Since c ≥ 11b^{6} ≥ 11 · 65^{6} > 0.82 · 10^{12}, inequality (11) cannot by fulfilled. Hence, we conclude that n = 0 or n = 1. Now, by Lemma 6, we have that z = v_{0} = w_{0}. But then d = (z^{2} + 1)/c < c. Moreover, if 11b^{6} ≤ c ≤ b^{9}, then Lemma 5 implies that z = s and d = 1. This proves the proposition. □

4 Diophantine approximation result

All other parts of Theorem 1 will be obtained by applying the theory of linear form of logarithms of algebraic numbers instead of Bennett’s approximation result. In this way it will not be possible to exclude any solution, but we will get effective upper bounds. In this section we will prepare the approximation tool that will be used in the proofs in case of solutions that are not very large.

In fact we will use a result by Matveev [32], which we quote in a suitable simplified version.

Lemma 10. Let Λ be a linear form in logarithm of l multiplicatively independent totally real algebraic numbers α_{1}, …, α_{l} with rational integer coefficients b_{1}, …, b_{l} (b_{l} ≠ 0). Let h(α_{j}) denote the absolute logarithmic height of α_{j}, 1 ≤ j ≤ l. Define the number D, A_{j}, 1 ≤ j ≤ l, and B by D = [ℚ(α_{1}, …, α_{l}) : ℚ], A_{j} = max{Dh(α_{j}), | log α_{j}|},

B = max \{1, max \{\frac{|b_{j}|A_{j}}{A_{l}} : 1 ≤ j ≤ l\}\} .

Then

log Λ > −C(l)C_{0}W_{0}D^{2}Ω,  (12)

where

C(l) = \frac{8}{(l − 1)!}(l + 2)(2l + 3)(4e(l + 1))^{l+1}, W_{0} = log (1.5eBD log(eD)) ,
C_{0} = log (e^{4.4l+7}l^{5.5}D^{2} log(eD)) ,  Ω = A_{1} ⋯ A_{l}.

Proof. See [32, Theorem 2.1]. □

Solving the recurrences (3) and (4) we get

v_{m} = \frac{s}{2} [(s + \sqrt{c})^{2m} + (s − \sqrt{c})^{2m}] ,

w_{n} = \frac{s\sqrt{b} ± r\sqrt{c}}{2\sqrt{b}} [(t + \sqrt{bc})^{2n} + (t − \sqrt{bc})^{2n})] .

We now turn the equation v_{m} = w_{n} into an inequality for a linear form in three logarithms to which Lemma 10 will be applied.

Lemma 11. If v_{m} = w_{n}, n ≠ 0, then

0 < 2n log(t + \sqrt{bc}) − 2m log(s + \sqrt{c}) + log \frac{s\sqrt{b} ± r\sqrt{c}}{2\sqrt{b}} < (3.96bc)^{−n+1}.

Proof. Let

P = s(s + \sqrt{c})^{2m}, Q = \frac{s\sqrt{b} ± r\sqrt{c}}{\sqrt{b}}(t + \sqrt{bc})^{2n}.

Then v_{m} = w_{n} implies P + (c − 1)P^{−1} = Q + \frac{c−b}{b}Q^{−1}. Furthermore, we have

P > 4s(c − 1) = 4\sqrt{c − 1}(c − 1) > c − 1,

Q > \frac{s\sqrt{b} − r\sqrt{c}}{\sqrt{b}}(2bc − 1 + 2\sqrt{bc}) ≥ 2c ≥ 2.

We get

P − Q = \frac{c − b}{b}Q^{−1} − (c − 1)P^{−1} < (c − 1)Q^{−1} − (c − 1)P^{−1}
= (c − 1)(P − Q)P^{−1}Q^{−1},

and therefore we have Q > P. Moreover, P > Q − (c − 1)P^{−1} > Q − 1, which implies that \frac{Q−P}{Q} < Q^{−1} ≤ \frac{1}{2}. Hence,

0 < log \frac{Q}{P} = − log \frac{P}{Q} = − log (1 − \frac{Q − P}{Q} ) < \frac{Q − P}{Q} + (\frac{Q − P}{Q} )^{2} .

From this we conclude

0 < log \frac{Q}{P} < \frac{1}{Q} + \frac{1}{Q^{2}} < \frac{2}{Q} = \frac{2\sqrt{b}}{s\sqrt{b} ± \sqrt{c}r} (t + \sqrt{bc})^{−2n}

< \frac{2\sqrt{b}}{s\sqrt{b} − r\sqrt{c}}(3.96bc)^{−n} = \frac{2\sqrt{b}(s\sqrt{b} + r\sqrt{c})}{c − b}(3.96bc)^{−n}

< \frac{4b\sqrt{c}}{c − b}(3.96bc)^{−n} < (3.96bc)^{−n+1}.

This proves the lemma. □

We use this result to prove a principle to obtain an upper bound for n in all remaining cases.

Lemma 12. If v_{m} = w_{n}, then

\frac{n − 1}{log (\frac{160.36n log(2.02bc)}{log c})} < 0.33 · 10^{12} · log c · log(bc).

Proof. We have l = 3, D = 4 and therefore C(3) ≤ 0.65·10^{9} and C_{0} ≤ 29.89. Moreover, we have α_{1} = s + \sqrt{c}, α_{2} = t + \sqrt{bc}, and

α_{3} = \frac{s\sqrt{b} ± r\sqrt{c}}{s\sqrt{b}}.

The minimal polynomials of α_{1}, α_{2}, α_{3} are α_{1}^{2} + (4c − 2)α_{1} + 1 = 0, α_{2}^{2} + (4bc−2)α_{2} +1 = 0, and b(c−1)α_{3}^{2} −2b(c−1)α_{3} +c−b = 0. Therefore, A_{1} = 2 log(s + \sqrt{c}) < 2 log(2\sqrt{c}) < 1.08 log c, A_{2} = 2 log(t + \sqrt{bc}) < log(\sqrt{2}bc) < 1.05 log(bc), and

A_{3} ≤ 2 log bc\frac{s\sqrt{b} + r\sqrt{c}}{s\sqrt{b}} < 2 log(2.02bc) < 2 log(3.96bc).

We also have

A_{3} ≥ log b(c − 1) ≥ log(64.2c) > 1.95 log c.

Thus,

B ≤ \frac{4m log(2.02bc)}{1.95 log c} ≤ \frac{2.06m log(2.02bc)}{log c}.

By Lemma 11 and Lemma 10 we therefore get

n − 1 < 0.33 · 10^{12} · log (\frac{80.18m log(2.02bc)}{log c}) log c log(bc).

Finally, we have

\frac{n − 1}{log (\frac{160.36n log(2.02bc)}{log c})} < 0.33 · 10^{12} · log c · log(bc),

which is the statement. □

Our remaining goal is to calculate lower bound for n in terms of small powers of c as above. Together with our gap conditions between b and c, the lemma above will enable us to get effective upper bounds for c in every case. This will be done in the next sections.

5 Large, medium and small solutions

We now start with the remaining cases in Theorem 1. Observe that we always have c ≤ 11b^{6} and therefore by Lemma 5 we have z_{0} = z_{1} = s, x_{0} = 0, y_{1} = ±r.

We start with the case of large solutions, i.e. with b^{3} ≤ c < 11b^{6}. We have the following lower bound for n.

Lemma 13. If v_{m} = w_{n}, n ≠ 0, 1 and b^{3} ≤ c < 11b^{6}, then n > c^{\frac{1}{12}} .

Proof. We again start with the relation in Lemma 2(iii), which reads here as m^{2}s ≡ bn^{2}s ± tnr (mod 4c). Assume that n ≤ c^{\frac{1}{12}} . Since max\{m^{2}s, tnr\} < bn^{2}s, and

bn^{2}s < c^{\frac{1}{3}} · c^{\frac{1}{6}} · c^{\frac{1}{2}} = c,

we have equality in the congruence above, thus m^{2}s = bn^{2}s ± tnr. This implies 4n^{2}s > n(bsn − tr) and therefore

\frac{b\sqrt{c}}{2}n < (b − 4)sn < tr < b\sqrt{c},

which gives a contradiction. □

Combining the previous lemma with the result from Baker’s theory from Lemma 12, we get part (iii) of Theorem 1.

Proposition 2. Let {1, b, c, d} with 1 < b < c < d be a D(−1)-quadruple. If b^{3} ≤ c ≤ 11b^{6}, then c < 0.55 · 10^{237} and d < 10^{10^{20}}.

Proof. Since b^{3} ≤ c, we get log(2.02bc) < 1.50 log c and therefore it follows by Lemma 12 that

\frac{n − 1}{log(240.54n)} < 0.29 · 10^{13} · \frac{4}{3}(log c)^{2} < 0.56 · 10^{15}(log n)^{2},

where we have used Lemma 13. Hence, n < 0.20 · 10^{17} and therefore c < 0.55 · 10^{237}.

From Lemma 3 we get z = v_{m} < 4c^{2}(4c − 2)^{m−1} ≤ 4^{m}c^{m+1}. Consequently,

d ≤ \frac{z^{2} + 1}{c} ≤ \frac{4^{2m}c^{2m+2}}{c} = 16^{m}c^{2m+1}.

Thus, log_{10} d ≤ m log_{10} 16 + (2m + 1) log_{10} c ≤ 0.19 · 10^{20} and therefore d ≤ 10^{10^{20}}. □

Next we handle the case of medium size solutions, i.e. with b^{1.1} ≤ c < b^{3}, and therefore prove part (iv) of Theorem 1.

Lemma 14. Let v_{m} = w_{n}, n ≠ 0, 1, and b^{1.1} ≤ c < b^{3}. If we moreover assume that c > 10^{100}, then n ≥ c^{0.04}.

Proof. We will consider the congruence of Lemma 2(iii). As in the lemmata above, we have

s(m^{2} − bn^{2}) ≡ ±rtn (mod 4c).  (13)

Let us define integers A and B by

A = 2bc − 2rst − c,
B = 2bc + 2rst − c.

We have AB = c^{2} + 4b^{2}c − 4b − 4c + 4. Clearly, B < 4bc and A > \frac{4b^{2}c}{4bc} = b. On the other hand, A < \frac{c^{2}+4b^{2}c}{2bc} < c. This implies that 2rst > 2bc − 2c, B > 4bc − 3c and

A < \frac{c^{2} + 4b^{2}c}{4bc − 3c} < \frac{c}{3b} + \frac{4}{3}b < 0.334c^{0.75} + 1.334c^{0.91} < c^{0.92},

which shows that A is small compared with c. Observe that here we use that c > 10^{100}.

Multiplying both sides of (13) by 2s, we get

2(m^{2} − bn^{2}) ≡ ∓An (mod c).

Assume now that n ≠ 0 and n < c^{0.04}. Then An < c and |2(m^{2}−bn^{2})| ≤ 2bn^{2} < 2c^{0.91+0.08} < c. Hence, we have an equality:

2bn^{2} − 2m^{2} = An.  (14)

The relation (14) and the inequalities n ≤ m ≤ 2n (cf. Lemma 2(ii)) imply

2(b − 4)n ≤ A ≤ 2(b − 1)n.  (15)

Let us define β by

c = 4b^{2}(2n − 1) + β.

We claim that |β| < c^{0.56}.

Let us first estimate the quantity γ defined by γ = 4bA − c − 4b^{2}. From AB = c^{2} + 4b^{2}c − 4b − 4c + 4 and 4bc − 3c < B < 4bc, we obtain

γ > \frac{1}{4bc}(−16b^{2} − 16bc + 16b) > −8,

γ < \frac{3c + 12b^{2}}{3b} = \frac{c}{b} + 4b.

Therefore, |γ| < c^{0.92}. Furthermore, from (15) we have 4bA = 8b^{2}n + δ, where |δ| ≤ 32bn < c^{0.966}. Since β = δ−γ, we get by putting these estimates together that

|β| < c^{0.966} + c^{0.92} < 2c^{0.966} < c^{0.97}.

Hence,

c = 4b^{2}(2n − 1) + β, |β| < c^{0.97}.  (16)

Analyzing (16) we will improve this estimate for |β|. Namely, (16) clearly implies 4b^{2} < c < 16nb^{2}. It follows that |γ| < \frac{c}{b} + 4b < 20bn < 10n\sqrt{c} < c^{0.55}, and also |δ| < 32bn < c^{0.553}. Hence, |β| < c^{0.56}, as we claimed.

Now we are ready to finish our argument. By squaring (14), we get

4(bn^{2} − m^{2})^{2} ≡ 4(b − 1)n^{2} (mod c),

which implies

−n^{4}β ≡ (2n − 1)(4(b − 1)n^{2} − 4m^{4} + 8bm^{2}n^{2}) (mod c).  (17)

But the absolute values of both sides of (17) are less that c^{0.72}. Hence, we have the equality in (17), and this implies (assuming that n ≠ 0)

n^{2}c = (2n − 1)(A^{2} − 4(b − 1)).

Since A^{2} ≡ 4(b − 1) (mod c), we find that (2n − 1)|n^{2}, which is possible only if n = 0 or n = 1. We obtained a contradiction, and hence we proved that n ≥ c^{0.04} □

Combining this lower bound with the upper bound which follows from Lemma 12, we will again obtain that c is bounded by an absolute constant.

Proposition 3. Let {1, b, c, d} with 1 < b < c < d be a D(−1)-quadruple. If b^{1.1} ≤ c < b^{3}, then c < 0.30 · 10^{491} and d < 10^{10^{23}}.

Proof. Since b ≤ c^{0.91}, we get log(2.02bc) ≤ 1.93 log c and log(bc) ≤ 1.91 log c. Now, by Lemma 14 and Lemma 12 we get

\frac{n − 1}{log(309.50n)} < 0.40 · 10^{15} (log n)^{2} .

Hence, n < 0.42 · 10^{20} and thus c < 0.30 · 10^{491}.

As in the proof of Proposition 2, we get by Lemma 3 that log_{10} d ≤ 2n log_{10} 16 + (4n + 1) log_{10} c ≤ 0.82 · 10^{23} and therefore d ≤ 10^{10^{23}}. □

Finally, we turn to part (v) of Theorem 1. This is the case of small solutions, i.e. with 3b < c < b^{1.1}.

To get the lower bounds for m, n in terms of some small power of c in the case of small solutions, we will use the following very useful construction, which in fact is the essence of the gap principle in Lemma 4.

Lemma 15. Let {a, b, c} be a D(−1)-triple. Define

e = −(a + b + c) + 2abc − 2rst.

Then there exist integers u, v, w such that

ae + 1 = u^{2}, be + 1 = v^{2}, ce + 1 = w^{2}

and

c = a + b − e + 2(abe + ruv).

Proof. This lemma is a special case of Lemma 3 in [14]. □

Using this lemma we now prove:

Lemma 16. If v_{m} = w_{n}, n ≠ 0, 1, 2 and 3b < c < b^{1.1}, then n ≥ 0.25 · c^{0.2}.

Proof. With the same notation as in Lemma 14, again we have the congruence

2(m^{2} − bn^{2}) ≡ ∓An (mod c).  (18)

In this case we get

A < \frac{c^{2} + 4b^{2}c}{4bc − 3c} = b + \frac{c + 3b}{4b − 3} < b + \frac{2c}{3b}.

Hence, A = b + α, where 0 < α < c^{0.1}.

By Lemma 15, there exist integers e, u, v, w such that

e + 1 = u^{2}, be + 1 = v^{2}, ce + 1 = w^{2},

and

b = 1 + c − e + 2ce − 2suw.

Hence, (b + e − 1)^{2} ≡ 4(c − 1)(e + 1)(ce + 1) ≡ −4(e + 1) (mod c), and

b^{2} + 2b(e − 1) + e^{2} + 2e + 5 ≡ 0 (mod c).  (19)

Moreover, e is small compared with c. Indeed, from c = 1 + b − e + 2be + 2ruv > 2be, we find that e < \frac{c}{2b} < c^{0.1}.

Now (18) implies

2m^{2} ± αn ≡ b(2n^{2} ∓ n) (mod c).  (20)

From (20), we also get

(2m^{2} ± αn)^{2} ≡ b^{2}(2n^{2} ∓ n)^{2} (mod c)  (21)

and

2(e − 1)(2n^{2} ∓ n)(2m^{2} ± αn) ≡ 2(e − 1)b(2n^{2} ∓ n)^{2} (mod c). (22)

By summing (21) and (22), and taking into account (19), we obtain

(2n^{2} ∓ n)^{2}(e^{2} + 2e + 5) + (2m^{2} ± αn)^{2}  (23)
+2(e − 1)(2n^{2} ∓ n)(2m^{2} ± αn) ≡ 0 (mod c).

Assume now that n < 0.25 · n^{0.2}.

Since n ≥ 3, e ≥ 3, m ≤ 2n and c ≥ 82, we have the following estimates:

|(2n^{2} ∓ n)^{2}(e^{2} + 2e + 5)| ≤ 6n^{4} · 3e^{2} · \frac{18c^{0.8}}{256} · \frac{c^{2}}{4b^{2}} < 0.02c,

(2m^{2} ± αn)^{2} ≤ 8m^{4} + α^{2}n^{2} ≤ \frac{c^{0.8}}{2} + \frac{c^{0.6}}{8} < 0.23c,

and

|2(e − 1)(2n^{2} ∓ n)(2m^{2} ± αn)| ≤ 6en^{2}(2m^{2} + αn) ≤ \frac{3c^{0.5}}{8} \frac{c^{0.4}}{2} < 0.13c.

Therefore, we have the equality in (23), i.e.

(e^{2} + 2e + 5)X^{2} + 2(e − 1)(e − 1)XY + Y^{2} = 0,  (24)

where X = (2n^{2} ∓ n), Y = 2m^{2} ± αn. Since the discriminant of (24) is negative (it is equal to −16u^{2}), we conclude that by multiplying (24) with 4(e^{2} + 2e + 5) we get

(2(e^{2} + 2e + 5)X + 2(e − 1)^{2}Y )^{2} + 16u^{2}Y^{2} = 0.

This implies X = Y = 0, which in turn gives n = 0, a contradiction. Hence, we proved that n ≥ 0.25 · c^{0.2}. □

Again by using Lemma 12 we will obtain that c is bounded by an absolute constant.

Proposition 4. Let {1, b, c, d} with 1 < b < c < d be a D(−1)-quadruple. If 3b < c < b^{1.1}, then c < 0.66 · 10^{94} and d < 10^{10^{21}}.

Proof. Since b < \frac{c}{3}, we have log(2.02bc) ≤ 2 log(0.83c) < 2 log c and log(bc) ≤ 2 log \frac{c}{9}. By c ≤ 1024n^{5} (cf. Lemma 16) and Lemma 12 it follows that

\frac{n − 1}{log(320.72n)} < 0.66 · 10^{12} · log(1024n^{5}) log(113.78n^{5}).

Hence, n ≤ 0.15 · 10^{19} and c ≤ 0.66 · 10^{94}.

By Lemma 3 we get log_{10} d ≤ 2n log_{10} 16 + (4n + 1) log_{10} c ≤ 0.55 · 10^{21} and therefore d ≤ 10^{10^{21}}. □

It remains to prove the smallest possible case, i.e. when we have c < 3b. This will be done in the following section.

6 Very small solutions

Now we consider the case c = 1 + b + 2\sqrt{b − 1} of very small solution (part (vi) in Theorem 1). Observe that we are left to consider c ≤ 3b. By the gap principle (Lemma 4), this implies that we have indeed c = 1 + b + 2r. We can write everything parametrized in r. To avoid confusions we set k = r in this section. Then we have to consider the D(−1)-triple

{1, k^{2} + 1, (k + 1)^{2} + 1}.

Moreover, we have

s = k + 1, t = k^{2} + k + 1, z_{0} = z_{1} = k + 1, x_{0} = 0, y_{1} = ±k,

and k ≥ 8.

If v_{m} = w_{n}, n ≠ 0, 1 then the congruence from Lemma 2(iii) has the form

m^{2}(k + 1) ≡ n^{2}(k^{2} + 1)(k + 1) ± n(k^{2} + k + 1)k  (25)
≡ n^{2}(k + 3) ∓ n(k + 1)  (mod (k^{2} + 2k + 2)).

We start by proving an upper bound for k in terms of n.

Lemma 17. In this case, if v_{m} = w_{n}, n ≠ 0, 1, then k < 16n^{2}.

Proof. Assume that n^{2} ≤ \frac{1}{16}k. We have the following estimates

m^{2}(k + 1) ≤ 4n^{2}(k + 1) < k(k + 1) < k^{2} + 2k + 2,

|n^{2}(k + 3) ± n(k + 2)| < \frac{1}{4}k(k + 3) + \frac{1}{2}\sqrt{k}(k + 1) < k^{2} + 2k + 2.

Therefore we have an equality in (25), which gives

m^{2}(k + 1) = n^{2}(k + 3) ∓ n(k + 1).

This implies m^{2} ≡ 3n^{2} ∓ n (mod k). Since m^{2} ≤ 4n^{2} < k and |3n^{2} ∓ n| < \frac{3}{4}k + \frac{1}{4} \sqrt{k} < k, we have again an equality. This implies m^{2} = 3n^{2} ∓ n. Moreover, we now also have the equation m^{2} = n^{2} ∓ n. Combining these two facts, we get 2n^{2} = ±n and therefore n = 0, a contradiction. This proves the lemma. □

We conclude the proof by combining Lemma 17 with Lemma 12 as in all cases above, to get an effective upper bound for k and therefore for c and d.

Proposition 5. Let {1, k^{2} +1, (k +1)^{2} +1, d} be a D(−1)-quadruple. Then k < 0.14 · 10^{38} and therefore c = (k + 1)^{2} + 1 < 0.19 · 10^{75} and d < 10^{10^{21}}.

Proof. We have log(2.02bc) = log(2.02(k^{2} +1)(k^{2} +2k +1)) ≤ 4.46 log k and log(k^{2} + 2k + 2) > log k. By combining Lemma 17 and Lemma 12 we get

\frac{n − 1}{log(715.21n)} < 0.33 · 10^{12} log ((16n^{2} + 1)^{2} + 1)
· log (((16n^{2}) + 1)((16n^{2} + 1)^{2} + 1)) .

This implies n < 0.93 · 10^{18} and therefore k < 0.14 · 10^{38}.

Moreover, we get c = (k + 1)^{2} + 1 < 0.19 · 10^{75} and by Lemma 3 we have log_{10} d ≤ 2n log_{10} 16 + (4n + 1) log_{10} c ≤ 0.21 · 10^{21} and finally this gives d < 10^{10^{21}}. □

Altogether, by combining Propositions 1, 2, 3, 4 and 5 above, we get Theorem 1. Theorem 2, as well as Corollary 1 and 2 are immediate consequences.

7 Quantitative result

We start by collecting two lemmata which will be needed for the proof of Theorem 3. In the following, ω(c) will denote, as usual, the number of different prime divisors of an integer c.

Lemma 18. Let c ∈ ℕ. The number of solutions x ∈ ℤ of the quadratic congruence

x^{2} ≡ −1 (mod c)

is bounded by 2^{ω(c)}.

Proof. This statement can be found in [37, p. 94]. □

We remark that the congruence has no solution if 3 or 4 divides c. Moreover, if c is even, then the number of solutions is bounded by 2^{ω(c)−1}.

Lemma 19. Let c ∈ ℕ. Then log(c) > \frac{1}{2}ω(c) log ω(c).

Proof. This can be found in [17]. The proof follows from results in [36]. □

Now we are ready to prove the quantitative result.

Proof of Theorem 3.

Let {1, b, c, d} be a D(−1)-quadruple. By Theorem 1 it follows that c < 10^{492} and d < 10^{10^{23}}. Let us first consider the triple {1, b, c} with b < c < 10^{492}. We have b − 1 = r^{2}, c − 1 = s^{2} and therefore r < s < 10^{246}. It is clear that the number of such triples is bounded by 10^{492}.

Let us now assume that {1, b, c} is fixed. Then, by Lemma 1, the number d is an element of the union of finitely many binary recurrent sequences (arising from (9) for some z_{0} with z_{0}^{2} = cd−1 and 0 < z_{0} < c). The number of such sequences is less than or equal to the number of solutions of the congruence z_{0}^{2} ≡ −1 (mod c). Consequently, this number is bounded by 2^{ω(c)} by Lemma 18.

Let us assume that 2^{ω(c)} ≥ c^{0.787}. By Lemma 19 we have c > ω(c)^{\frac{ω(c)}{2}} and therefore

2^{ω(c)} ≥ c^{0.787} > ω(c)^{0.3935ω(c)}.

This implies ω(c) ≤ 5. Hence, c ≤ 81.76 and therefore a contradiction, since c ≥ 82. This gives

2^{ω(c)} < c^{0.787}

and it follows that the number of recurrences mentioned above is bounded by

2^{ω(c)} < c^{0.787} ≤ 10^{388},

because of the upper bound for c.

On the other side, the sequences grow exponentially, namely by Lemma 3 we have

(c − 1)(4c − 3)^{m−1} < v_{m} = z = \sqrt{cd − 1} < 10^{10^{23}}.

Since b ≥ 82 we conclude that

81(4 · 82 − 3)^{m−1} < 10^{10^{23}}

and therefore m < 10^{23}. Therefore the number of d’s which extend the set {1, b, c} to a D(−1)-quadruple is at most 10^{411}.

Altogether, the number of D(−1)-quadruples is bounded by

10^{492} · 10^{411} = 10^{903}

and therefore we have proved Theorem 3. □

8 Acknowledgements

This paper was partly written 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); the authors would like to thank them for this support. Moreover, the first author was supported by the Ministry of Science, Education and Sports, Republic of Croatia, grant 0037110 and the third author was supported by the Austrian Science Foundation FWF, grant J2407-N12.

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