A POLYNOMIAL VARIANT 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 polynomials with integer coefficients, which are not all constant, such that the product of any two of them is one greater than a square of a polynomial with integer coefficients.

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. The first D(1)-quadruple, the set {1, 3, 8, 120}, was found by Fermat. 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 [10]).

In the case n = −1, the conjecture is that there does not exist a D(−1)-quadruple (see [5]). It is known that some particular D(−1)-triples cannot be extended to D(−1)-quadruples (see [2], [6], [13], [14]). Let us mention that from [9, Theorem 4] it follows that there does not exist a D(−1)-33-tuple.

This 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}\} (see [4], [3]). In [8] 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 same property is equivalent to finding D(−1)-m-tuples.

∗This work was supported by the Austrian Science Foundation FWF, grant S8307-MAT.

2 ANDREJ DUJELLA AND CLEMENS FUCHS^{∗}

A polynomial variant of the above problems was first studied by Jones [11], [12], and it was for the case n = 1.

Definition 1. Let n be an integer. A set {a_{1}, a_{2}, …, a_{m}} of m polynomials with integer coefficients, which are not all constant, is called a polynomial D(n)-m-tuple if for all 1 ≤ 1 < j ≤ m the following holds: a_{i} · a_{j} + n = b_{ij}^{2}, where b_{ij} ∈ ℤ[x].

A natural question is how large such sets can be. Let us define

P_{n} = sup{|S| : S is a polynomial D(n)-tuple }.

¿From [9, Theorem 1] it follows that P_{n} ≤ 22 for all n ∈ ℤ. The above mentioned result about the existence of only finitely many D(1)-quintuples implies that P_{1} = 4.

In the present paper, we will prove that P_{−1} = 3. First of all, P_{−1} ≥ 3. More precisely, if a · b − 1 = r^{2}, then

{a, b, a + b + 2r}

is a polynomial D(−1)-triple. E.g.

{x^{2} + 1, x^{2} + 2x + 2, 4x^{2} + 4x + 5}

is a polynomial D(−1)-triple (see [2]). Therefore, we have to prove that P_{−1} < 4, and this is the statement of our main theorem.

Theorem 1. There does not exist a polynomial D(−1)-quadruple.

The proof of Theorem 1 is divided into several parts. In Section 2, we transform our problem into a system of polynomial Pellian equations, which leads to finding intersections of some binary recursive sequences. We obtain some useful information about initial terms of these sequences.

In Section 3, we show that there is no loss of generality in assuming that one element of our initial triple is equal to 1. This, together with results from Section 2, allow us to completely determine initial terms of corresponding sequences.

In Section 4, we prove Theorem 1 by showing that our sequences cannot have nontrivial common terms. This is done by comparing degrees and leading coefficients of corresponding polynomials.

2. Two sequences of polynomials

Let ℤ^{+}[x] denote the set of all polynomials with integer coefficients with positive leading coefficient. For a, b ∈ ℤ[x], a < b means that b − a ∈ ℤ^{+}[x]. The usual fundamental properties of inequality hold for this order. For a ∈ ℤ[x], we define |a| = a if a ≥ 0, and |a| = −a if a < 0.

If {a, b, c, d}, a < b < c < d is a polynomial D(−1)-quadruple, then d is non-constant. Assume now that a and b are constant polynomials. Considering leading coefficients of ad − 1 and bd − 1 we conclude that ab is a perfect square, contradicting the assertion that ab − 1 is also a perfect square. Therefore, we proved that in a polynomial D(−1)-quadruple there is at most one constant polynomial. It is also clear that all leading coefficients of the polynomials in a polynomial D(−1)-m-tuple have the same sign. This implies that there is no loss of generality in assuming that they are all positive, i.e. that all polynomials are in ℤ^{+}[x].

Let {a, b, c}, where 0 < a < b < c, be a polynomial D(−1)-triple and let r, s, t ∈ ℤ^{+}[x] be 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 d ∈ ℤ^{+}[x], d > c, is a polynomial such that {a, b, c, d} is a polynomial D(−1)-quadruple. We have

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

with u, y, z ∈ ℤ^{+}[x]. Eliminating d from (1) we obtain the following system of polynomial Pellian equations

az^{2} − cu^{2} = c − a, (2)

bz^{2} − cy^{2} = c − b. (3)

We will describe the sets of solutions of equations (2) and (3). We will follow the arguments in the classical case of Pellian equations in integers (cf. [7]).

Lemma 1. If (z, u) and (z, y), with u, y, z ∈ ℤ^{+}[x], are polynomial solutions of (2) and (3) respectively, then there exist z_{0}, u_{0} ∈ ℤ[x] and z_{1}, y_{1} ∈ ℤ[x] with

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

0 ≤ |u_{0}| < s, (4)

0 < z_{0} < c, (5)

0 ≤ |y_{1}| < t, (6)

0 < z_{1} < c, (7)

and there exist integers m, n ≥ 0 such that

z\sqrt{a} + u\sqrt{c} = (z_{0}\sqrt{a} + u_{0}\sqrt{c})(s + \sqrt{ac})^{2m}, (8)

z\sqrt{b} + y\sqrt{c} = (z_{1}\sqrt{b} + y_{1}\sqrt{c})(t + \sqrt{bc})^{2n}, (9) where this means that the coefficients of \sqrt{a}, \sqrt{b} and \sqrt{c} respectively on both sides are equal.

Proof. It is clear that it suffices to prove the statement of the lemma for equation (2). First observe that

(s + \sqrt{ac})^{2m} = (s^{2} + ac + 2s\sqrt{ac})^{m} = (2ac − 1 + 2s\sqrt{ac})^{m}

and by multiplying with the conjugate (s − \sqrt{ac})^{2m} we see that

(s + \sqrt{ac})^{2m}(s − \sqrt{ac})^{2m} = (s^{2} − ac)^{2m} = (−1)^{2m} = 1. (10)

Let now (z, u) be a solution of (2) in polynomials from ℤ^{+}[x]. Consider all pairs (z^{∗}, u^{∗}) of polynomials of the form

z^{∗}\sqrt{a} + u^{∗}\sqrt{c} = (z\sqrt{a} + u\sqrt{c})(s + \sqrt{ac})^{2m}, m ∈ ℤ.

By (10) it is clear that (z^{∗}, u^{∗}) satisfies (2).

We would like to show that z^{∗} > 0. We write (s + \sqrt{ac})^{2m} = A + B\sqrt{ac}, where A, B ∈ ℤ[x] satisfying A^{2} − acB^{2} = 1. Therefore we have

z^{∗}\sqrt{a} + u^{∗}\sqrt{c} = (z\sqrt{a} + u\sqrt{c})(A + B\sqrt{ac}) = (Az + cuB)\sqrt{a} + (Au + azB)\sqrt{c},

and this yields

z^{∗} = Az + cuB.

Now, if m ≥ 0 then we have A, B > 0 and thus z^{∗} > 0. On the other hand, if m < 0 we have A > 0, B < 0. If we assume that z^{∗} ≤ 0, we have Az ≤ −Bcu and both sides are > 0. Squaring yields A^{2}z^{2} ≤ B^{2}c^{2}z^{2}. Using the fact that A^{2} − acB^{2} = 1 we obtain z^{2}B^{2}ac + z^{2} ≤ B^{2}c^{2}u^{2} and therefore

z^{2} ≤ cB^{2}(cu^{2} − az^{2}) = cB^{2}(a − c) < 0,

a contradiction.

Among all pairs (z^{∗}, u^{∗}), we can now choose a pair with the property that z^{∗} is minimal, and we denote that pair by (z_{0}, u_{0}). Define polynomials z′ and u′ by

z′\sqrt{a} + u′\sqrt{c} = (z_{0}\sqrt{a} + u_{0}\sqrt{c})(2ac − 1 − 2sε\sqrt{ac}),

where ε = 1 if u_{0} > 0, and ε = −1 if u_{0} < 0. From the minimality of z_{0} we conclude that z′ = z_{0}(2ac − 1) − 2csu_{0}ε ≥ z_{0} and this leads to cs|u_{0}| ≤ z_{0}(ac − 1) and further to c|u_{0}| ≤ sz_{0}. Squaring this inequality we obtain

acz_{0}^{2} − c(c − a) = c^{2}u_{0}^{2} ≤ acz_{0}^{2} − z_{0}^{2}

and finally

z_{0}^{2} ≤ c(c − a) < c^{2},

which implies (5). Now we have

cu_{0}^{2} = az_{0}^{2} − c + a ≤ ac^{2} − a^{2}c − c + a < ac^{2} − c = cs^{2} (11) and therefore we obtain also (4). Hence, we have proved that there exists a solution (z_{0}, u_{0}) of (2), which satisfies (4) and (5), and an integer m ∈ ℤ such that

z\sqrt{a} + u\sqrt{c} = (z_{0}\sqrt{a} + u_{0}\sqrt{c})(s + \sqrt{ac})^{2m}.

It remains to show that m ≥ 0. Suppose that m < 0. Then, as above, we have (s + \sqrt{ac})^{2m} = A − B\sqrt{ac}, with A, B ∈ ℤ^{+}[x] satisfying A^{2} − acB^{2} = 1. We have u = Au_{0} − z_{0}Ba and from the condition u > 0 we obtain Au_{0} > z_{0}Ba and by squaring u_{0}^{2} > B^{2}a(c − a) ≥ ac − a^{2}, which by (11) implies

ac^{2} − a^{2}c ≤ cu_{0}^{2} ≤ ac^{2} − a^{2}c − c + a.

This implies −c + a ≥ 0, which is clearly a contradiction. □

The solutions z arising, for given (z_{0}, u_{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}, u_{0}) with the above properties and integer m ≥ 0, where

v_{0} = z_{0}, v_{1} = (2ac − 1)z_{0} + 2scu_{0}, v_{m+2} = (4ac − 2)v_{m+1} − v_{m}. (12)

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

w_{0} = z_{1}, w_{1} = (2bc − 1)z_{1} + 2tcy_{1}, w_{m+2} = (4bc − 2)w_{n+1} − w_{n}. (13)

Now the following congruence relations follow easily from (12) and (13) by induction.

Lemma 2. Let the sequences (v_{m}) and (w_{n}) be given by (12) and (13). Then we have

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

Proof. It suffices to prove the statement of the lemma for v_{m}. By looking on (12) we have

v_{0} = z_{0}, v_{1} ≡ −z_{0} (mod 2c).

Proceeding the induction step, we see using (12)

v_{m+2} ≡ −2(−1)^{m+1}z_{0} − (−1)^{m}z_{0} = (−1)^{m+2}z_{0} (mod 2c),

as stated. □

Now we can prove the following lemma, which says that a solution of v_{m} = w_{n} implies also a solution at the beginning of the sequences.

Lemma 3. If the equation v_{m} = w_{n} has a solution, then z_{0} = z_{1}.

Proof. Assume that v_{m} = w_{n} has a solution. By Lemma 2 we conclude

z_{0} ≡ ±z_{1} (mod 2c).

If we assume that z_{0} ≡ z_{1} (mod 2c), then we can conclude by using (4) and (7) from Lemma 1, namely

0 < z_{0} < c, 0 < z_{1} < c,

that z_{0} = z_{1} holds. If we assume that z_{0} ≡ −z_{1} (mod 2c), we have 2c|z_{0} +z_{1}, which contradicts the fact that z_{0} + z_{1} < 2c. This finishes the proof. □

3. Reduction to the case a = 1

In this section, we show that it suffices to prove that polynomial D(−1)- triples {a, b, c}, where a = 1, cannot be extended to a polynomial D(−1)- quadruple.

Lemma 4. Let {a, b, c, d} with 0 < a < b < c < d be a polynomial D(−1)- quadruple. Then there exists d_{0} ∈ ℤ^{+}[x] with d_{0} < c such that ad_{0} − 1, bd_{0} − 1, cd_{0} − 1 are perfect squares.

Proof. We are interested in sequences (v_{m}) (and (w_{n})) such that z^{2} = v_{m}^{2} = w_{n}^{2} = cd − 1, where d ∈ ℤ^{+}[x]. This implies that v_{m}^{2} ≡ −1 (mod c). By Lemma 2 this means

z_{0}^{2} ≡ −1 (mod c).

In this case we define

d_{0} = \frac{z_{0}^{2} + 1}{c} ∈ ℤ^{+}[x].

For this d_{0} we have

cd_{0} − 1 = z_{0}^{2}.

By Lemma 3 we find

bd_{0} − 1 = b\frac{z_{1}^{2} + 1}{c} − 1 = \frac{cy_{1}^{2} + c − b + b}{c} − 1 = y_{1}^{2}

and finally also

ad_{0} − 1 = a\frac{z_{0}^{2} + 1}{c} − 1 = \frac{1}{c}(az_{0}^{2} + a − c) = \frac{1}{c}cu_{0}^{2} = u_{0}^{2}

holds. Furthermore, we have

cd_{0} = z_{0}^{2} + 1 < c^{2},

which implies

d_{0} < c.
□

Assume now that {a, b, c, d} is a polynomial D(−1)-quadruple with minimal d. We may use Lemma 4 to construct d_{0}. From the minimality of d, it follows that {a, b, c, d_{0}} is not a polynomial D(−1)-quadruple and this means that d_{0} ∈ {a, b}. But this implies that d_{0}^{2} − 1 is a perfect square, which can only hold in the case when d_{0} = 1. Since b > a ≥ 1, we conclude that a = 1.

Remark 1. It follows that it suffices to consider polynomial D(−1)- quadruples, which contain the constant polynomial 1.

Now let {1, b, c} with 1 < b < c be a polynomial D(−1)-triple. By the previous discussion, we have d_{0} = 1. This implies that z_{0}^{2} + 1 = c and therefore we have z_{0} = ±s. Because of the fact that z_{0} > 0 we have z_{0} = s. In the same way, we can conclude that z_{1} = s. Now we have

cu_{0}^{2} = z_{0}^{2} − c + 1 = c − 1 − c + 1 = 0

and this yields u_{0} = 0. Finally we get

cy_{1}^{2} = bz_{1}^{2} − c + b = b(c − 1) − c + b = bc − c = cr^{2}

and therefore y_{1} = ±r. To sum up, it suffices to consider the following three sequences

v_{0} = s, v_{1} = (2c − 1)s, v_{m+2} = (4c − 2)v_{m+1} − v_{m}, (14)

and

w_{0} = s, w_{1} = (2bc − 1)s + 2tcr, w_{n+2} = (4bc − 2)w_{n+1} − w_{n}, (15)

w′_{0} = s, w′_{1} = (2bc − 1)s − 2tcr, w′_{n+2} = (4bc − 2)w′_{n+1} − w′_{n}. (16)

4. Proof of Theorem 1

Let {1, b, c} be a polynomial D(−1)-triple. Let us repeat the defining equations:

b − 1 = r^{2}, c − 1 = s^{2}, bc − 1 = t^{2}.

In what follows, we need the leading coefficients of b and c. We know that b and c are non-constant, and thus their leading coefficients are perfect squares. Let us give them names:

lc(b) = β^{2}, lc(c) = γ^{2},

where β and γ are positive integers. Let v_{m} and w_{n}, w′_{n} be the remaining sequences from the last section. To finish the proof, we have to show that no nontrivial solution is obtained from these sequences. The trivial solution is always v_{0} = w_{0} = s, which leads to d = 1, which does not yield the extension of our triple {1, b, c}. We divide the proof in three cases. The first one is handled in the following lemma.

Lemma 5. The equation v_{m} = w_{n} has no nontrivial solution.

Proof. First let us mention that deg v_{m} < deg v_{m+1}, m = 0, 1, 2, . . .. To be precise we have

(17) deg v_{m} = \frac{1}{2} deg c + m deg c, m ≥ 0.

This follows at once by induction using the recurring formula (14). The same is also true for the second sequence w_{n} with

(18) deg w_{n} = \frac{1}{2} deg c + n (deg b + deg c) , n ≥ 0.

Again, by induction, we can now read off the leading coefficient of v_{m}, which is

2^{2m−1}γ^{2m+1}, m ≥ 1.

We have lc(v_{0}) = γ, lc(v_{1}) = 2γ^{3} and using the recursive formula (14) we get

lc(v_{m+1}) = 4γ^{2}lc(v_{m}) = 4γ^{2}2^{2m−1}γ^{2m+1} = 2^{2(m+1)−1}γ^{2(m+1)+1}.

In the same way, we find the leading coefficient of w_{n}, which is

2^{2n}β^{2n}γ^{2n+1}.

First we have lc(w_{0}) = γ, lc(w_{1}) = 2β^{2}γ^{2}γ + 2βγγ^{2}β = 4β^{2}γ^{3}. By using the recursive formula for w_{n}, one finds

lc(w_{n+1}) = 4β^{2}γ^{2}lc(w_{n}) = 2^{2n+2}β^{2n+2}γ^{2n+3}.

If the equation v_{m} = w_{n} has a solution, we must have equal leading coefficients, which means

2^{2m−1}γ^{2m+1} = 2^{2n}β^{2n}γ^{2n+1}.

This implies

(\frac{2^{m−n}γ^{m−n}}{β^{n}} )^{2} = 2,

which yields

\sqrt{2} = \frac{2^{m−n}γ^{m−n}}{β^{n}} ∈ ℚ,

a contradiction. Thus v_{m} = w_{n} cannot hold and the proof is finished. □

To handle the equation v_{m} = w′_{n}, we have to distinguish whether deg b < deg c or deg b = deg c holds.

Lemma 6. Assume that deg b < deg c. Then the equation v_{m} = w′_{n} has no nontrivial solution.

Proof. First we calculate

w′_{1}w_{1} = (2bc − 1)^{2}s^{2} − 4t^{2}c^{2}r^{2} =
= −4b^{2}c^{2} + 4bc + c − 1 + 4bc^{3} − 4c^{2}.

Because of our assumption deg b < deg c, we obtain that the dominating summand is 4bc^{3}. Therefore we get

lc(w′_{1}w_{1}) = 4β^{2}γ^{6}

and

deg w′_{1}w_{1} = 3 deg c + deg b.

On the other hand, we already know that

lc(w_{1}) = 4β^{2}γ^{3}

and

deg w_{1} = deg b + \frac{3}{2} deg c.

Hence, we can conclude that

lc(w′_{1}) = γ^{3} and deg w′_{1} = \frac{3}{2} deg c.

Now by induction and by the recursion (16) we get that deg w′_{n} < deg w′_{n+1} and that the leading coefficient of w′_{n} is given by

2^{2n−2}β^{2n−2}γ^{2n+1}, n ≥ 1.

Namely we have lc(w′_{0}) = γ, lc(w′_{1}) = γ^{3} and using (16) we obtain

lc(w′_{n+1}) = 4β^{2}γ^{2}lc(w′_{n}) = 2^{2n}β^{2n}γ^{2n+3}.

Again, if v_{m} = w′_{n} has a solution, we can conclude by comparing the leading coefficients that

2^{2m−1}γ^{2m+1} = 2^{2n−2}β^{2n−2}γ^{2n+1}.

As before we get

\sqrt{2} = 2^{n−m}γ^{n−m}β^{n−1} ∈ ℚ,

which is a contradiction. This yields that in this case no solution can exist. □

Before we can prove the remaining part, we need the following useful gap principle for the elements of a polynomial D(−1)-m-tuple. The principle is a direct modification from the integer case (see [9, Lemma 3]). The analogous statement for polynomial D(1)-triples was proved by Jones in [12].

Lemma 7. Let {a, b, c} be a polynomial D(−1)-triple. Then there exist polynomials e, u, y, z ∈ ℤ[x] such that

ae + 1 = u^{2}, be + 1 = y^{2}, ce + 1 = z^{2}

and

c = a + b − e + 2(abe + ruy).

Proof. Define

e = −(a + b + c) + 2abc − 2rst.

Then

(ae + 1) − (at − rs)^{2} = −a(a + b + c) + 2a^{2}bc − 2arst + 1 −
−a^{2}(bc − 1) + 2arst − (ab − 1)(ac − 1) = 0.

Hence, we may take u = at − rs, and analogously y = bs − rt, z = cr − st. We have

abe + ruy = −ab(a + b + c) + 2a^{2}b^{2}c − 2abrst + abrst −
−a(ab − 1)(bc − 1) − b(ab − 1)(ac − 1) + rst(ab − 1) =
= abc − (a + b) − rst,

10 ANDREJ DUJELLA AND CLEMENS FUCHS^{∗}

and finally a+b−e+2(abe+ruy) = 2a+2b+c−2abc+2rst+2abc−2a−2b−2rst = c.

□

Using this lemma we can finish our proof. Lemma 8. Assume that deg b = deg c. Then the equation v_{m} = w′_{n} has no nontrivial solution.

Proof. First we conclude by Lemma 7 that there exist polynomials e, f, g, h such that e + 1 = f^{2}, be + 1 = g^{2}, ce + 1 = h^{2} (19) and

c = 1 + b − e + 2(be + rfg).

By looking at the proof of Lemma 7, we see that we have

e = −1 − b − c + 2bc − 2rst.

We want to show that e = 0. Let us assume e ≠ 0 and define

ē = −1 − b − c + 2bc + 2rst.

Then

(20) deg ē = deg b + deg c = 2 deg c = deg c^{2}. Let us calculate

eē = (2bc − 1 − b − c)^{2} − 4r^{2}s^{2}t^{2} =
= (2bc − 1 − b − c)^{2} − 4(b − 1)(c − 1)(bc − 1) =
= 1 + b^{2} + c^{2} − 2b − 2bc − 2c + 4.

This yields

deg e + deg ē = deg eē ≤ deg c^{2}.

Using (20), we can conclude

deg e ≤ 0.

But looking at (19) we see that

e + 1 = φ^{2} and e = ψ^{2}

must hold with φ, ψ ∈ ℤ. This is only possible if e = 0.

This implies now that f = 1, g = 1 and c = 1 + b + 2r. Next let us express all polynomials in terms of the polynomial r. We have

b = r^{2} + 1,

and therefore

c = r^{2} + 2r + 2.

Next we calculate s^{2} = c − 1 = b + 2r = r^{2} + 2r + 1 = (r + 1)^{2}, thus s = r + 1.

In the same way, we get via t^{2} = bc − 1 = (r^{2} + 1)(r^{2} + 2r + 2) − 1 = r^{4} + 2r^{3} + 3r^{2} + 2r + 1 = (r^{2} + r + 1)^{2}, that

t = r^{2} + r + 1.

This gives us

w′_{1} = (2bc − 1)s − 2tcr =
= (2r^{4} + 4r^{3} + 6r^{2} + 4r + 3)(r + 1) − 2(r^{3} + r^{2} + r)(r^{2} + 2r + 2) =
= 2r^{2} + 3r + 3.

¿From this we conclude that

deg w′_{1} = deg c

and by induction, using the recurring formula (16), we get

deg w′_{n} = deg c + 2(n − 1) deg c, n ≥ 1.

Let us assume that v_{m} = w′_{n} has a solution. Then by comparing the degree of v_{m}, which is given by (17), and the degree of w′_{n}, we get

\frac{1}{2} deg c + m deg c = deg c + 2(n − 1) deg c

and

\frac{1}{2} + m = 2n − 1,

a contradiction. Therefore v_{m} = w′_{n} cannot have a solution and the proof is finished. □

Now Theorem 1 follows directly from Lemma 5, Lemma 6 and Lemma 8.

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

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

[13] 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.

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

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 A-8010 Graz, Austria. e-mail: clemens.fuchs@tugraz.at