DIOPHANTINE m-TUPLES FOR LINEAR POLYNOMIALS

ANDREJ DUJELLA, CLEMENS FUCHS^{∗} AND ROBERT F. TICHY^{∗}

Abstract. In this paper, we prove that there does not exist a set with more than 26 polynomials with integer coefficients, such that the product of any two of them plus a linear polynomial is a square of a polynomial with integer coefficients.

2000 Mathematics Subject Classification: 11D09. Keywords: diophantine m-tuples, Mason’s inequality, hyperelliptic equations, function fields.

1. Introduction

Let n be a nonzero integer. A set of m positive integers {a_{1}, a_{2}, …, a_{m}} 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.

Diophantus [4] found the first quadruple {1, 33, 68, 105} with the property D(256). 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 [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 [7]).

The natural question is how large such sets can be. We define

M_{n} = sup{|S| : S has the propertyD(n)},

where |S| denotes the number of elements in the set S. The first author proved that M_{n} is finite for all n ∈ ℤ\{0}. In his proof, he estimated the number of “large” (greater than |n|^{3}), “small” (between n^{2} and |n|^{3}) and “very small” (less than n^{2}) elements of a set with the property D(n), using a theorem of Bennett [3] on simultaneous approximations of algebraic numbers and a gap principle in the first, a weaker variant of the gap principle in the second and a large sieve method due to Gallagher [9] in the third case respectively (cf. [6]). Let us introduce the following notation:

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

A_{n} = sup{|S ∩ [|n|^{3}, ∞⟩| : S has the propertyD(n)},
B_{n} = sup{|S ∩ [n^{2}, |n|^{3}⟩| : S has the propertyD(n)},
C_{n} = sup{|S ∩ [1, n^{2}⟩| : S has the propertyD(n)}.

His result was (cf. [6, Theorems 1, 2, 3 and 4])

A_{n} ≤ 21,

B_{n} ≤ 0.65 log |n| + 2.24,

C_{n} ≤ \{ \begin{matrix}265.55 log |n|(log log |n|)^{2} + 9.01 log log |n| & for |n| > 400, \\ 5 & for |n| ≤ 400. \end{matrix}

Therefore

M_{n} ≤ 32 for |n| ≤ 400,
M_{n} < 267.81 log |n|(log log |n|)^{2} for |n| > 400.

A polynomial variant of the above problems was first studied by Jones [10], [11], and it was for the case n = 1.

Definition 1. Let n ∈ ℤ[x] and let {a_{1}, a_{2}, …, a_{m}} be a set of m nonzero polynomials with integer coefficients. We assume that there does not exist a polynomial p ∈ ℤ[x] such that a_{1}/p, …, a_{m}/p and n/p^{2} are integers. The set {a_{1}, a_{2}, …, a_{m}} is called a polynomial D(n)-m-tuple if for all 1 ≤ i < j ≤ m the following holds: a_{i} · a_{j} + n = b_{ij}^{2}, where b_{ij} ∈ ℤ[x].

Let us mention that the assumption that there does not exist a polynomial p such that a_{1}/p, …, a_{m}/p and n/p^{2} are integers means for constant n that not all elements a_{1}, …, a_{m} of a polynomial D(n)-m-tuple are allowed to be constant (compare with Definition 1 in [8]). For linear n the condition under consideration is trivially always satisfied.

In analog to above results, we are interested in the size of

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

From [6, 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. Recently, the first and the second author proved that P_{−1} = 3 (cf. [8]) by succesfully transferring the needed methods to the polynomial case.

The results of [6], by specialization, give a bound for P_{n} in terms of the degree and the maximum of the coefficients of n. We conjecture that there should exist a bound for P_{n}, which depends only on the degree of n. As we have seen, this is true for constant polynomials, and in the present paper we will prove this conjecture for linear polynomials.

We want to handle the case of linear polynomials, i.e. n = ax + b, with integers a ≠ 0 and b. Let us define

L = sup{|S| : S is a polynomial D(ax + b)-tuple for some a ≠ 0 and b }.

It is easy to prove that L ≥ 4. E.g. the set

{x, 16x + 8, 25x + 14, 36x + 20}

is a polynomial D(16x + 9)-quadruple and the set

{1, 9x^{2} + 8x + 1, 9x^{2} + 14x + 6, 36x^{2} + 44x + 13}

is a polynomial D(4x + 3)-quadruple (see [5]).

We intend to prove that L < ∞. More precisely, we want to find some good upper bound for L.

The idea is to estimate the number of polynomials in S with given degree and to consider separate cases whether the degree is “large” or “small”.

In analog to the classical case, we prove our result for “large” degree by using a theorem due to Mason [12] on the polynomial solutions of hyperelliptic equations over function fields in one variable and a gap principle. Let S be a polynomial D(ax + b)-m-tuple with integers a ≠ 0 and b. We prove

Proposition 1. There are at most 15 polynomials in S with degree ≥ 4.

We want to remark that a weaker result can be shown by applying the results from the classical integer case. From that, it is possible to show that there are at most 21 polynomials in S with degree ≥ 4.

We have to estimate the number of constant, linear, quadratic and cubic polynomials in S. We denote these numbers by L_{0}, L_{1}, L_{2}, L_{3} respectively and we will consider them separately. First of all it is trivial to see that we have

L_{0} ≤ 1.

By using the mentioned gap principle once more, we get

Proposition 2. There are at most three polynomials in S of degree 3. Therefore, we have

L_{3} ≤ 3.

Let us remark that in fact the proof gives us the following result: There is no polynomial D(ax+b)-quadruple which consists of polynomials all having the same degree μ ≥ 3. For the case n = 1 this was already proved by Jones in [11].

By more detailed analysis we get

Proposition 3. There are at most five polynomials in S of degree 2. Therefore, we have

L_{2} ≤ 5.

Proposition 4. There are at most eight linear polynomials in S. Therefore, we have

L_{1} ≤ 8.

Altogether, we can prove the following bound for the size of polynomial D(n)-m-tuples for linear polynomials n = ax + b.

Theorem 1.

L ≤ 26.

In Section 2, we will collect auxiliary results which are needed to prove our results. In Section 3, we handle Propositions 1 and 2 which are the cases of large degrees. In Section 4, we prove the results for the small degrees, i.e. Propositions 3 and 4 and therefore finally get Theorem 1.

2. Auxiliary results

Let K be an algebraically closed field with characteristic 0. Let us begin by recalling the definitions of the discrete valuations on the field K(x) where x is transcendental over K. For ξ ∈ K define the valuation ν_{ξ} such that for Q ∈ K(x) we have Q(x) = (x − ξ)^{ν_{ξ}(Q)}A(x)/B(x) where A, B are polynomials with A(ξ)B(ξ) ≠ 0. Further, for Q = A/B with A, B ∈ K[x], we put deg Q := deg A − deg B; thus ν_{∞} := − deg is a discrete valuation on K(x). These are all discrete valuations on K(x). Now let L be a finite extension of K(x). Each of the valuations ν_{ξ}, ν_{∞} can be extended in at most [L : K(x)] =: d ways to a discrete valuation on L and in this way one obtains all discrete valuations on L. A valuation on L is called finite if it extends ν_{ξ} for some ξ ∈ K and infinite if it extends ν_{∞}.

We need the following generalization of the degree from K[x] to L. We define the height of f ∈ L by

H(f) = − ∑_{ν} min{0, ν(f)}

where the sum is taken over all valuations on L; thus for f ∈ K(x) the height H(f) is just the number of poles of f counted according to multiplicity. We note that if f lies in K[x] then H(f) = d deg f. We also want to define the height of a polynomial with coefficients in L. In order to do this let us denote for any finite set S of elements of L

ν(S) = min_{s∈S}{H(s)} and H(S) = − ∑_{ν} min{0, ν(S)}

where the sum again runs over all valuations in L. If P ∈ L[T] and S is the set of its coefficients, then the quantities ν(P) and H(P) are defined to be ν(S) and H(S) respectively.

Let O denote the ring of elements of L integral over K[x]. These elements have the property that ν(f) ≥ 0 for all finite valutations on L.

Now we are able to state the following theorem on the solutions of a hyperelliptic equation over an algebraic function field. A proof of this theorem can be found in the monograph of Mason (cf. [13, Theorem 6]).

Theorem 2. (R. C. Mason) Let α_{1}, …, α_{n} ∈ O. All the solutions X, Y ∈ O of the hyperelliptic equation

Y^{2} = (X − α_{1})(X − α_{2}) ⋯ (X − α_{n}) (1)

satisfy

H(X) ≤ 26H + 8g + 4(r − 1);

here H denotes the height of the polynomial on the right hand side of (1), g denotes the genus of L/K and r denotes the number of infinite valuations on L.

Let us note that this bound varies only as a linear function of the height of the hyperelliptic equation, in contrast with the multiply exponential bounds for the classical case obtained by Baker [1]. This shows that the fundamental inequality due to Mason on which the proof of this theorem is based and which is the function field analog of Baker’s method of linear forms in logarithms is very sharp.

Before we can go to the proofs of the results, we need the following useful construction with the elements of a polynomial D(n)-triple, where n is a polynomial with integer coefficients. The construction is a direct modification from the integer case (see [6, Lemma 3]). The analogous statement for polynomial D(1)-triples was proved by Jones in [11] and we did already use it in the case n = −1 (cf. [8]).

Lemma 1. Let {a, b, c} be a polynomial D(n)-triple and let ab+n = r^{2}, ac+ n = s^{2}, bc + n = t^{2}. Then there exist polynomials e, u, v, w ∈ ℤ[x] such that

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

More precisely,

e = n(a + b + c) + 2abc − 2rst.

Furthermore, it holds:

c = a + b + \frac{e}{n} + \frac{2}{n^{2}} (abe + ruv),

where u = at − rs, v = bs − rt.

Proof. We have

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

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

abe + ruv = abn(a + b + c) + 2a^{2}b^{2}c − 2abrst + abrst −
−a(ab + n)(bc + n) − b(ab + n)(ac + n) + rst(ab + n) =
= −abcn − n^{2}(a + b) + rstn,

and finally

a+b+ \frac{e}{n} + \frac{2}{n^{2}} (abe+ruv) = 2a+2b+c+ \frac{2abc}{n} − \frac{2rst}{n} − \frac{2abc}{n} −2a−2b+ \frac{2rst}{n} = c.

□

If we also define

\overline{e} = n(a + b + c) + 2abc + 2rst, (2)

then easy computation shows that

e · \overline{e} = n^{2}(c − a − b − 2r)(c − a − b + 2r). (3)

This relation will be very useful in the proof of Proposition 2, 3 and 4.

We conclude this section with the following definition: 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.

Observe that it is clear that all leading coefficients of the nonconstant polynomials in a polynomial D(n)-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].

3. Elements with large degrees

Assume that the set {a, b, c, d} is a polynomial D(n)-quadruple with n ∈ ℤ[x]. Let ab + n = r^{2}, ac + n = s^{2}, bc + n = t^{2} where r, s, t ∈ ℤ^{+}[x]. In this paper, the symbols r, s, t will always have this meaning. Moreover, we have

ad + n = u^{2}, bd + n = v^{2}, cd + n = w^{2},

with u, v, w ∈ ℤ^{+}[x]. Multiplying these equations, we get the following elliptic equation

(uvw)^{2} = (ad + n)(bd + n)(cd + n),

where we search for polynomial solutions d ∈ ℤ[x]. We will apply Mason’s Theorem 2 to this equation.

Lemma 2. Let {a, b, c, d}, 0 < a < b < c < d be a polynomial D(n)- quadruple with n ∈ ℤ[x]. Then

deg d ≤ 51(deg a + deg b + deg c) + 78 deg n.

Proof. Let us denote X = abcd and Y = abcuvw. Then by multiplying the above equation with a^{2}b^{2}c^{2} we get

Y^{2} = (X + nbc)(X + nac)(X + nab).

The polynomial on the right hand side becomes

(X + nbc)(X + nac)(X + nab) =
= X^{3} + n(ab + bc + ac)X^{2} + n^{2}abc(a + b + c)X + n^{3}a^{2}b^{2}c^{2}

so this polynomial has coefficients and roots in ℤ[x]. Let S be the set of coefficients of this polynomial, i.e.

S = {1, n(ab + bc + ac), n^{2}abc(a + b + c), n^{3}a^{2}b^{2}c^{2}}.

Since the elements of S are polynomials, we get for each ξ ∈ ℂ that

ν_{ξ}(S) = min_{s∈S}{0, ν_{ξ}(s)} = 0.

Moreover, we have

ν_{∞}(S) = min_{s∈S}{0, ν_{∞}(s)) = min_{s∈S}{− deg s} = − max_{s∈S} deg s,

and by comparing the degrees of the elements of S we get

ν_{∞}(S) = −2(deg a + deg b + deg c) − 3 deg n.

Therefore,

H(S) = − ∑_{ν} min{0, ν(S)} = − ∑_{ξ∈ℂ}min{0, ν_{ξ}(S)} − min{0, ν_{∞}(S)} =

= − min{0, ν_{∞}(S)} = 2(deg a + deg b + deg c) + 3 deg n.

Thus, we have for the height H of the polynomial on the right hand side of our elliptic equation

H = 2(deg a + deg b + deg c) + 3 deg n.

By Mason’s Theorem 2 with L = ℂ(x) and O = ℂ[x] we therefore get

deg X ≤ 52(deg a + deg b + deg c) + 78 deg n,

where we have used that the genus of the rational function field ℂ(x) is zero (which can be found e.g. in [14], page 22) and that ℂ(x) has only one infinite valuation, namely ν_{∞}. But now by the definition of X = abcd we get

deg d ≤ 51(deg a + deg b + deg c) + 78 deg n

as claimed in our lemma. □

Observe that due to the sharpness of the fundamental inequality this bound is very good. Especially, it does not depend on a gap which has to appear between the elements of the quadruple as in the classical case (cf. [6, Lemma 2]).

We use Lemma 1 to prove the following gap principle. It is very similar to [6, Lemma 4] in the classical case for integers.

Lemma 3. If {a, b, c, d} is a polynomial D(n)-quadruple, where n ∈ ℤ[x] and 2n^{2} < a < b < c < d, then

n^{2}d > 2bc.

Proof. We apply Lemma 1 to the triple {a, c, d}. Let e be defined as in Lemma 1. Since ce+n^{2} is a perfect square, we have that ce+n^{2} ≥ 0. Assume that e ≤ −1, then

ce + n^{2} < −2n^{2} + n^{2} = −n^{2} < 0,

a contradiction. Therefore, we have e ≥ 0. Observe that, if n > 0 we have

a^{2} < ac + r^{2} = ac + ab + n ⟺ na^{2} < na(b + c) + n^{2} ⟺
a^{2}t^{2} = a^{2}(n + bc) = na^{2} + a^{2}bc < a^{2}bc + na(b + c) + n^{2} =
= (ab + n)(ac + n) = r^{2}s^{2} ⟺
at < rs ⟺ u < 0,

and

b^{2} < bc + r^{2} = bc + ab + n ⟺ nb^{2} < nb(a + c) + n^{2} ⟺
b^{2}s^{2} = b^{2}(ac + n) = ab^{2}c + b^{2}n < ab^{2}c + nb(a + c) + n^{2} =
= (ab + n)(bc + n) = r^{2}t^{2} ⟺
bs < rt ⟺ v < 0.

In the same way, one can show that n < 0 implies

u > 0 and v > 0.

If e = 0, then d = a + c + 2s. If e ≥ 1, then

n^{2}d = n^{2}(a + c) + en + 2(ace + suv) > 2ac.

Note that we need here that uv > 0 which follows from the comments just made.

Analogously, we apply Lemma 1 to the triple {b, c, d} and obtain either d = b + c + 2t or n^{2}d > 2bc. However, d = b + c + 2t is impossible since s^{2} = ac + n < bc + n = t^{2} and therefore s < t which implies a + c + 2s < b + c + 2t and

n^{2}(b + c + 2t) < n^{2} · 4c < 2ac,

which follows from

t^{2} = bc + n ≤ (c − 1)c + n = c^{2} + n − c < c^{2} + n − n^{2} < c^{2}

since c > 2n^{2} and consequently t < c.

Hence, we proved

n^{2}d > 2bc,

as claimed in our lemma. □

Proof of Proposition 1.

Assume that {a, b, c, a_{4}, a_{5}, …, a_{16}} is a polynomial D(n)-16-tuple and |n|^{3} ≤ a < b < c < a_{4} < a_{5} < … < a_{16}. We apply Lemma 2 to the quadruple {a, b, c, a_{16}} and obtain

a_{16} < (abc)^{51}n^{79} < c^{153}n^{79} < c^{180}, (4)

since |n|^{3} < c.

Lemma 3 implies n^{2}a_{4} > bc > |n|^{3}c and a_{4} > c|n|. Furthermore, n^{2}a_{5} > a_{4}c > c^{2}|n| and |n|a_{5} > c^{2}. In the same manner, Lemma 3 gives

n^{2}a_{6} > a_{5}a_{4} > c^{3}, |n|^{5}a_{7} > |n|^{3}a_{6}a_{5} > c^{5}, |n|^{9}a_{8} > c^{8},
n^{16}a_{9} > c^{13},  |n|^{27}a_{10} > c^{21},  |n|^{45}a_{11} > c^{34},
n^{74}a_{12} > c^{55},  |n|^{121}a_{13} > c^{89},  |n|^{197}a_{14} > c^{144},
|n|^{320}a_{15} > c^{233}, |n|^{519}a_{16} > c^{377},

which implies (since |n|^{3} < c) that

c^{173}a_{16} > c^{377}

and therefore

a_{16} > c^{204},

a contradiction to (4). □

Proof of Proposition 2.

Let S = {a, b, c} with a < b < c be a polynomial D(n)-triple with linear n ∈ ℤ[x] and let deg a = deg b = deg c = 3. Then by (2) we get deg \overline{e} = 9. But from (3) it follows that deg e\overline{e} ≤ 8. Thus we have a contradiction unless e = 0, i.e. c = a + b + 2r. Consequently, if we fix a and b, then c is unique, which implies that S cannot be extended to a polynomial D(n)-quadruple. Therefore,

L_{3} ≤ 3,

as claimed in our proposition. □

Observe that Proposition 2 follows directly from Lemma 3, but the above proof gives more information on triples of cubic polynomials.

4. Elements with small degrees

First we prove Proposition 3. Here the argument from the proof of Proposition 2 does not longer work. The polynomials e which are induced by a polynomial D(ax + b)-triple in Lemma 1 are constants. The proof uses the fact that u^{2} − n^{2} = (u − n)(u + n) is a complete factorization of the polynomial a up to the constant factor.

Proof of Proposition 3.

Let {a, b, c} with a < b < c be a polynomial D(n)-triple with linear n ∈ ℤ[x] and let deg a = deg b = deg c = 2. Then by (2) we get deg \overline{e} = 6. Now (3) implies that e is a constant. Assume that two distinct non-zero e’s exist. We call them e and f. From ae + n^{2} = u^{2} we see that a is a product of two linear polynomials:

a = α(x − a_{0})(x − a_{1}).

Let us assume that we have

u − n = ε_{1}(x − a_{0}), u + n = ε_{2}(x − a_{1}),

where ε_{1}ε_{2} = αe. It implies

2n = x(ε_{2} − ε_{1}) + ε_{1}a_{0} − ε_{2}a_{1}.

In the same manner, we can conclude from af + n^{2} = ũ^{2} that

ũ − n = φ_{1}(x − a_{0}), ũ + n = φ_{2}(x − a_{1}),

or

ũ − n = φ_{1}(x − a_{1}), ũ + n = φ_{2}(x − a_{0})

holds. Let us first consider that the first of this equations holds. Then we get

2n = x(φ_{2} − φ_{1}) + φ_{1}a_{0} − φ_{2}a_{1},

where φ_{1}φ_{2} = αf. Hence, ε_{2} − ε_{1} = φ_{2} − φ_{1}, ε_{1}a_{0} − ε_{2}a_{1} = φ_{1}a_{0} − φ_{2}a_{1}. Consequently, we have a_{0}(ε_{1} − φ_{1}) = a_{1}(ε_{2} − φ_{2}) = a_{1}(ε_{1} − φ_{1}). We have two possibilities: ε_{1} = φ_{1} or a_{0} = a_{1}.

Let first assume ε_{1} = φ_{1}. This implies also ε_{2} = φ_{2} and therefore e = f, a contradiction. Now we assume that a_{0} = a_{1} holds. Then x − a_{0}|n and together with ab + n = r^{2} this implies x − a_{0}|r. Therefore (x − a_{0})^{2}|n, and we obtained a contradiction since n is a linear polynomial.

Now let us consider the second case. So assume that we have

ũ − n = φ_{1}(x − a_{1}), ũ + n = φ_{2}(x − a_{0}),

where φ_{1}, φ_{2} are as above. It implies

2n = x(φ_{2} − φ_{1}) + φ_{1}a_{1} − φ_{2}a_{0}.

Hence, ε_{2} − ε_{1} = φ_{2} − φ_{1}, ε_{1}a_{0} − ε_{2}a_{1} = φ_{1}a_{1} − φ_{2}a_{0}. This yields, a_{0}(ε_{1} + φ_{2}) = a_{1}(φ_{1} + ε_{2}) = a_{0}(ε_{1} + φ_{2}). We have again two possibilities: ε_{1} = −φ_{2} which implies ε_{2} = −φ_{1} and therefore e = f, a contradiction, or a_{0} = a_{1}. But as above this yields a contradiction with the assumption that n is a linear polynomial.

Therefore, there is at most one such constant e. It follows that for fixed a and b, there are at most three c, namely c = a + b + 2r and two possible c(e) which come from

c = a + b + \frac{e}{n} + \frac{2}{n^{2}} (abe + 2ruv),

where u, v satisfy ae + n^{2} = u^{2}, be + n^{2} = v^{2}. This last equations fix u and v only up to the sign and therefore we get two possible c’s in this case. Consequently we get

L_{2} ≤ 5,

which was claimed in the proposition. □

As in the proof of Proposition 3, we will see that also the proof of Proposition 4 heavily depends on the fact that we are considering linear polynomials. Especially, we will use that (3) is the complete factorization of the product e\overline{e}.

Proof of Proposition 4.

Let S = {a, b, c} with a < b < c be polynomial D(n)-triple with linear n ∈ ℤ[x] and let deg a = deg b = deg c = 1. Then by (2) we get deg \overline{e} = 3. Now (3) implies that

deg e ≤ 1.

From ab + n = r^{2} it follows that at most one of the elements in S is divisible by n. Indeed, assume that a and b are divisible by n. Then n|r and n^{2}|n, a contradiction. Thus we may assume that a, b, c are not divisible by n. We have

e + \overline{e} = 2n(a + b + c) + 4abc. (5) \overline{e} − e = 4rst. (6)

If n|e, then (3) implies that n|\overline{e} and therefore, by (5), we get n|abc, a contradiction.

Therefore, e = δ · (c − a − b ± 2r), δ ∈ ℚ. Assume that δ ≠ 0. We have

\overline{e} = n^{2}(c − a − b ∓ 2r)\frac{1}{δ}.

This implies

\frac{e}{δ} − \frac{\overline{e}δ}{n^{2}} = ±4r

or

\frac{n^{2}e}{δ} − \overline{e}δ = ±4n^{2}r

or

\frac{n^{2}e}{δ} − eδ = 4r(δst ± n^{2}).

This can be written as

\frac{e}{δ}(n^{2} − δ^{2}) = 4r(δst ± n^{2}).

Hence, there are two possibilities: r|e or r|n ± δ.

If r|e, then by (6) we have r|\overline{e} which yields

r^{2}|n^{2}(c − a − b − 2r)(c − a − b + 2r).

If r|n, then from ab + n = r^{2} we conclude r|a or r|b. Both cases lead to a contradictions since this would imply n|a or n|b. Observe that r and n only differ by a constant factor since they are both linear. Thus, r|c − a − b, say c = a + b + r · ρ, ρ ∈ ℚ. But from this we get

ac + n = a^{2} + ab + arρ + n = a^{2} + r^{2} + ρar = s^{2}

or

(2r + ρa)^{2} − (ρ^{2} − 4)a^{2} = (2s)^{2}.

Now, if ρ = ±2 then c = a + b ± 2r. Observe that by considering leading coefficients it is clear that a + b − 2r < b so this case is impossible and it remains the possible case c = a + b + 2r. Otherwise, if ρ ≠ ±2, we have

(ρ^{2} − 4)a^{2} = (2r + ρa − 2s)(2r + ρa + 2s)

which implies a|r and a|s, and moreover, using ab + n = r^{2}, we get a|n or equivalently n|a, a contradiction.

Therefore, it remains the case r|n ± δ. It means that δ is unique. It is defined by n ≡ ∓δ (mod r). Let n ≡ δ_{0} (mod r). We have the following five possibilities for c, namely c = a + b + 2r and c(e), where e = (c − a − b + 2r)(−δ_{0}) or e = (c − a − b − 2r) · δ_{0}. Each of this two e’s induce at most two c’s as we have seen at the end of the proof of Proposition 3. Therefore, we have at most seven linear polynomials in S which are not divisible by n. We get

L_{1} ≤ 8,

and so the proof is finished. □

Now we are ready to prove our bound for L.

Proof of Theorem 1.

Let S be a polynomial D(ax + b)-m-tuple with some integers a ≠ 0 and b. From the fact that the product of each two elements from S plus ax + b is a square of a polynomial with integer coefficients, it follows that if the set S contains a polynomial with degree ≥ 2, then it contains either polynomials with even or polynomials with odd degree only. Together with the upper bound for the number of polynomials in S with degree ≥ 4, this implies that we have

|S| ≤ 11 + 15 = 26.

This proves our theorem. □

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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

Robert F. Tichy Institut für Mathematik TU Graz Steyrergasse 30 A-8010 Graz, Austria. e-mail: tichy@tugraz.at