DIOPHANTINE m-TUPLES FOR LINEAR POLYNOMIALS. II. EQUAL DEGREES

ANDREJ DUJELLA^{∗}, CLEMENS FUCHS^{‡}, AND GARY WALSH

Abstract. In this paper we prove the best possible upper bounds for the number of elements in a set of polynomials with integer coefficients all having the same degree, such that the product of any two of them plus a linear polynomial is a square of a polynomial with integer coefficients. Moreover, we prove that there does not exist a set of more then 12 polynomials with integer coefficients and with the property from above. This significantly improves a recent result of the first two authors with R. F. Tichy [10].

2000 Mathematics Subject Classification: 11D09, 11C08. Keywords: diophantine m-tuples, linear polynomials, Mason’s inequality, 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 [2] 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 [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 [5]). Moreover, the first and the second author proved that there does not exist a D(−1)-quintuple (see [9]).

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

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

∗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.

This paper was partly written during a visit of the first author at the Department of Mathematics, Graz University of Technology within a joint Austrian-Croatian project granted by the Croatian Ministry of Science, Education and Sport and the Austrian Exchange Service (Nr. 20/2004).

where |S| denotes the number of elements in the set S. The first author proved that

M_{n} ≤ 31 for |n| ≤ 400,
M_{n} < 15.476 log |n| for |n| > 400

(see [4, 6]).

A polynomial variant of the above problems was first studied by Jones [12], [13], 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].

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

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

From [4, 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. The first and the second author proved that P_{−1} = 3 (cf. [7]). Moreover, in [8] they proved that if {a, b, c, d} is a polynomial D(1)- quadruple, then

(a + b − c − d)^{2} = 4(ab + 1)(cd + 1),

which implies that every polynomial D(1)-triple can be extended to a polynomial D(1)-quadruple in an essentially unique way, which in turn gives P_{1} = 4 once again.

Another polynomial variant of the problem was considered by the first author and Luca [11]. They considered sets of polynomials with the property that the product of any two elements plus 1 is a perfect kth power and they proved sharp upper bounds for the size of such sets.

The first and second author together with Tichy [10] considered 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 },

and let us denote by L_{n} the number of polynomials of degree n in a polynomial D(ax + b)-m-tuple S. Trivially, L_{0} ≤ 1. We proved that

L_{1} ≤ 8, L_{2} ≤ 5, L_{n} ≤ 3 for all n ≥ 3,

(see [10, Propositions 1,2 and 3]) and

L ≤ 26

(see [10, Theorem 1]). Moreover, we proved that there are at most 15 polynomials of degree ≥ 4 in such a set S.

In this paper we will give sharp upper bounds for L_{n} for all n ≥ 1. Moreover, we will significantly improve the upper bound for L.

Theorem 1. There does not exist a set of five linear polynomials with integer coefficients and the property that the product of any two of then plus the linear polynomial n = ax + b with integers a ≠ 0 and b is a square in ℤ[x].

This solves the problem for linear polynomials completely, in view of the following example:

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

is a polynomial D(16x + 9)-quadruple (see [3]).

The idea of the proof of Theorem 1 is the following: first we show that we may assume that one of the polynomials is a multiple of x, then we reduce the defining equations, which is a quadratic polynomial in x that is a square and therefore has vanishing discriminant, to a system of Diophantine equations for the coefficients. In the above example, the question of extendability reduces to finding all integer solutions of

n^{2}(3m − 8) + m^{2}(3n − 8)^{2} − m^{2}n^{2}(36mn − 9(m + n) + 265) = 0,

which gives

(3mn − 8m − 8n + 8)(3mn − 8m − 8n − 8)(m − n + 1)(m − n − 1) = 0,

from which a contradiction can be derived.

The next theorem now deals with the case of quadratic polynomials.

Theorem 2. There does not exist a set of four quadratic polynomials with integer coefficients and the property that the product of any two of them plus the linear polynomial n = ax + b with integers a ≠ 0 and b is a square in ℤ[x].

Also this result is best possible since the set

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

is a polynomial D(4x+3)-triple. Let us note that this triple can be extended to the D(4x + 3)-quadruple

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

(see [3]).

Corollary 1. We have

L_{1} ≤ 4, L_{n} ≤ 3 for all n ≥ 2.

Moreover, all these bounds are sharp.

In order to show that the bound L_{n} ≤ 3 for n ≥ 3 is sharp, let us consider the following examples

{x^{2n} − x, x^{2n} + 2x^{n} − x + 1, 4x^{2n} + 4x^{n} − 4x + 1},
{x^{2n−1} − 1, x^{2n−1} + 2x^{n} + x − 1, 4x^{2n−1} + 4x^{n} + x − 4},

for n = 1, 2, 3, …, which are polynomial D(x)-triples consisting of three polynomials with the same degree.

Using the new information from Theorems 1 and 2 together with a closer look at the case of polynomials with “large” degrees, we can prove the following result:

Theorem 3.

L ≤ 12.

In analog to the classical integer case, we prove our result for “large” degree by using Mason inequality [14], which is the function field analog of Baker’s method for linear forms in logarithms of algebraic numbers, to solve a certain elliptic equation over a function field in one variable, which is done by following the original ideas of Siegel [16].

In Section 2, we will consider the cases of equal degrees and give proofs of Theorems 1 and 2, which immediately imply Corollary 3. In Section 3 we prove an upper bound for the degree of the largest element in a D(n)- quadruple by considering the corresponding elliptic equation over a function field. In the last section (Section 4), by combining this upper bound with a gap principle and Theorems 1 and 2, we give a proof of Theorem 3.

2. Sets with polynomials of equal degree

First, we will handle the case of linear polynomials and therefore give a proof of Theorem 1. Afterwards, we consider the case of quadratic polynomials and therefore prove Theorem 2. Corollary 1 is then an immediate consequence of these two theorems together with the remark after [10, Proposition 2] in the first part to this paper.

2.1. Linear polynomials and proof of Theorem 1.

Let {ax+b, cx+d, ex+f} be a polynomial D(ux+v)-triple. Then {a^{2}x+ ab, acx+ad, aex+af} is a polynomial D(a^{2}ux+a^{2}v)-triple. By substitution ax = y, it follows that {ay + ab, cy + ad, ey + af} is a D(auy + a^{2}v)-triple, and finally by substitution y + b = z, we conclude that

{az, cz + d′, ez + f′} is a polynomial D(auz + v′)-triple,

where d′ = ad − cb, f′ = af − eb, v′ = a^{2}v − abu.

We may assume that gcd(a, c, e) = 1, since otherwise we substitute z′ = z gcd(a, c, e). This implies that a, c and e are perfect squares:

a = A^{2}, c = C^{2}, e = E^{2},

where A, C, E are positive integers. Furthermore, by specializing z = 0, we see that v′ is also a perfect square: v′ = V^{2}. But we have

v′ = a^{2}v − abu = A^{4}v − A^{2}bu = V^{2}.

Hence, V = AW with W^{2} = A^{2}v − bu.

Now from

A^{2}z · (C^{2}z + d′) + (A^{2}uz + A^{2}W^{2}) = (ACz ± AW)^{2},

we find by comparing the coefficients of z that A^{2}d′ + A^{2}u = ±2A^{2}CW and therefore d′ = ±2CW − u. Analogously, f′ = ±2EW − u. Hence, we obtained the set {A^{2}z, C^{2}z±2CW −u, E^{2}z±2EW −u} which is a polynomial D(A^{2}uz + A^{2}W^{2})-triple. It means that

(C^{2}z ± 2CW − u) · (E^{2}z ± 2EW − u) + (A^{2}uz + A^{2}W^{2})

is a square of a linear polynomial and this implies that the discriminant of this quadratic polynomial is equal to 0. The discriminant can be factored into 4 factors:

(C−E−A)(C−E+A)(±2CEW −Cu−Eu+Au)(±2CEW −Cu−Eu−Au),

which can be easily checked.

Assume now that there exists a D(ux+v)-quintuple consisting of 5 linear polynomials. The above construction shows that in this case there exists a D(A^{2}uz + A^{2}W^{2})-quintuple with one element equal to A^{2}z and with all other elements of the form

m_{i}^{2}z + 2m_{i}W − u for i = 1, 2, 3, 4.

Observe that the m_{i} can be positive or negative corresponding to the sign of W. Let

m_{1} = min{m_{1}, m_{2}, m_{3}, m_{4}}.

Then one of the remaining m_{i}’s is equal to m_{1} + A and the other two come from the factors ±2CEW − Cu − Eu + Au, ±2CEW − Cu − Eu − Au. The condition ±2ECW − Cu − Eu = Au or −Au is equivalent to (±2CW − u)(±2EW − u) = u^{2} + 2AW u or u^{2} − 2AW u. Therefore, let us denote

p_{i} := 2m_{i}W − u, i = 1, 2, 3, 4; P := u^{2} − 2AW u, Q := u^{2} + 2AW u.

We may assume that m_{2} = m_{1} + A and that

p_{1}p_{3} = P, p_{1}p_{4} = Q.

We want to prove that m_{3} = m_{2} + A or m_{4} = m_{2} + A. Suppose that this is not true, then p_{2}p_{3} = Q, p_{2}p_{4} = P. We have

AW u = Q − P = p_{3}(p_{2} − p_{1}) = 2p_{3}AW
= p_{4}(p_{1} − p_{2}) = −2p_{4}AW.

Since p_{3}p_{4} cannot be equal to P or Q, we have |m_{3} − m_{4}| = A. But then

Q − P = p_{2}(p_{3} − p_{4}) = ±2p_{2}W A,

which implies that p_{2} = p_{3} or p_{2} = p_{4}, a contradiction. Hence, we may assume that m_{3} = m_{2} + A. Then, p_{2}p_{4} = P. Moreover, from |m_{3} − m_{4}| = A, we conclude that m_{4} = m_{3} + A.

Let us insert m_{2} = m_{1} +A, m_{3} = m_{1} +2A, m_{4} = m_{1} +3A into the relation p_{1}p_{3} = p_{2}p_{4}. We obtain

4W (m_{2}m_{4} − m_{1}m_{3}) = 2(m_{2} + m_{4} − m_{1} − m_{3})u

or

4W A(2m_{1} + 3A) = 4Au,

and finally

u = 2m_{1}W + 3AW.

From

4AW u = Q − P = 2p_{1}AW = −2p_{4}AW,

we find that 2u = p_{1} = −p_{4}. This implies that 2m_{1}W = 3u and 2m_{4}W = −u and we get

(1)  4u = 2(m_{1} − m_{4})W = −6AW.

Furthermore, since p_{1}p_{4} = Q, we get −4u^{2} = u^{2} + 2AW u and therefore

−5u = 2AW,

which is a contradiction to equation (1). This proves that there does not exist a polynomial D(ux + v)-quintuple consisting of linear polynomials. □

2.2. Quadratic polynomials and proof of Theorem 2. 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].

Let {a, b, c} be a polynomial D(n)-triple containing only quadratic polynomials and with linear n ∈ ℤ[x]. Assume that a < b < c. In our previous paper ([10, Proof of Proposition 3]), we have shown that for fixed a and b such that ab + n = r^{2}, there are at most three possibilities for c, namely c = a + b + 2r and two possible c’s which come from

c_{1,2} = a + b + \frac{e}{n} + \frac{2}{n^{2}} (abe ± ruv),

where u, v ∈ ℤ^{+}[x] and e ∈ ℤ satisfy ae + n^{2} = u^{2}, be + n^{2} = v^{2}.

Observe now that

c_{1} · c_{2} = a^{2} + b^{2} + \frac{e^{2}}{n^{2}} − 2ab − \frac{2ae}{n} − \frac{2bc}{n} − 4n,

which implies that c_{2} < b, a contradiction.

Now we assume that a polynomial D(n)-m-tuple S contains a, b, c, c_{1}. The same argument as above applied to the pair {b, c} implies that c_{1} = d_{0} = b + c + 2t or c_{1} = d_{1} or c_{1} = d_{2} with

d_{1,2} = b + c + \frac{f}{n} + \frac{2}{n^{2}} (abf ± tũṽ),

where bc + n = t^{2} and with certain ũ, ṽ ∈ ℤ^{+}[x] and f ∈ ℤ satisfying bf + n^{2} = ũ^{2}, bf + n^{2} = ṽ^{2}. As before we get d_{2} < c and therefore c_{1} ≠ d_{2}. Moreover, in the proof of [10, Proposition 3] it is shown that be + n^{2} = v^{2} and bf + n^{2} = ũ^{2} with e, f ∈ ℤ implies that e = f. Hence, d_{1} > c_{1}. The only remaining case is

c_{1} = d_{0} = b + a + b + 2r + 2b + 2r = a + 4b + 4r,

which means that we have to deal with the only possible polynomial D(n)- quadruple of the form

{a, b, a + b + 2r, a + 4b + 4r}

with ab + n = r^{2}.

The only remaining condition for this set to be a polynomial D(n)- quadruple is a · (a + 4b + 4r) + n = z^{2}, which implies

a^{2} + 4(r^{2} − n) + 4ar + n = z^{2}

or

(2)  (a + 2r − z)(a + 2r + z) = (a + 2r)^{2} − z^{2} = 3n.

This is a contradiction, since the left hand side of (2) has degree ≥ 2 and the right hand side has degree 1. Consequently, we have proved that there are at most 3 polynomials in the D(n)-m-tuple S all having degree two. □

3. A certain elliptic equation

In this section we will reduce the problem of finding all extensions of {a, b, c} to a polynomial D(n)-quadruple to finding all solutions in ℤ[x] of a certain elliptic equation in an algebraic function field in one variable over the algebraically closed field of constants ℂ.

Assume that the set {a, b, c, d} is a polynomial D(n)-quadruple. Let ab + n = r^{2}, ac + n = s^{2}, bc + n = t^{2} where r, s, t ∈ ℤ^{+}[x]. 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].

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

(3)  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]. Instead of applying a general theorem for hyperelliptic equations in function fields due to Mason (cf. [15, Theorem 6]), as we did in our previous paper ([10, Lemma 2]), we will follow Siegel’s original approach (cf. [16] and the method of proof of [15, Theorem 6]).

Therefore, let

L := ℂ(x, \sqrt{ab}, \sqrt{ac})

be a function field in one variable over the field of complex numbers. Let O denote the ring of elements of L integral over ℂ[x]. These elements have the property that ν(f) ≥ 0 for all finite valuations on L. Let us recall the definitions of the discrete valuations on the field ℂ(x) where x is transcendental over ℂ. For ξ ∈ ℂ define the valuation ν_{ξ} such that for Q ∈ ℂ(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 ∈ ℂ[x], we put deg Q := deg A − deg B; thus ν_{∞} := − deg is a discrete valuation on ℂ(x). These are all discrete valuations on ℂ(x). Now let L as above be a finite extension of ℂ(x). Each of the valuations ν_{ξ}, ν_{∞} can be extended in at most [L : ℂ(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 ∈ ℂ and infinite if it extends ν_{∞}.

All solutions of interest for us come from solutions of (3) in L, where X + nbc, X + nac, X + nab are squares. Observe that this follows from the relations

X + nbc = abcd + nbc = u^{2}bc,
X + nac = abcd + nac = v^{2}ac,
X + nab = abcd + nab = w^{2}ac

and the fact that \sqrt{ab}, \sqrt{ac} and therefore also

\sqrt{bc} = \frac{\sqrt{ab}\sqrt{ac}}{a}

are elements of L. We denote

ξ_{1}^{2} = u^{2}bc = X + nbc, ξ_{2}^{2} = v^{2}ac = X + nac, ξ_{3}^{2} = w^{2}ac = X + nab and we define β_{i}, β̂_{i}, i = 1, 2, 3 by β_{1} = ξ_{2} −ξ_{3}, β̂_{1} = ξ_{2} +ξ_{3} with β_{2}, β̂_{2}, β_{3}, β̂_{3} defined similarly by permutation of indices. All these elements are contained in the ring O. Then β_{1}β̂_{1} = na(b − c), β_{2}β̂_{2} = nb(c − a), β_{3}β̂_{3} = nc(a − b), and

(4)  β_{1} + β_{2} + β_{3} = 0.

This is Siegel’s classical identity. Moreover, we have

(5)  β_{1} + β̂_{2} − β̂_{3} = −β̂_{1} + β_{2} + β̂_{3} = β̂_{1} − β̂_{2} + β_{3} = 0.

We note that each of β_{1}, β_{2} and β_{3} divide the fixed element

μ = −n^{3}abc(b − a)(c − a)(c − b)

in O. Hence, if ν is any finite valuation on L with ν(μ) = 0, then we have ν(β_{i}) = 0, i = 1, 2, 3 and so ν(β̂_{i}) = 0, i = 1, 2, 3, also. Now we apply Mason’s Inequality to Siegel’s identity (4) to get an upper bound for the degree of the polynomials X and therefore also for the polynomials d.

We need the following generalization of the degree from ℂ[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 ∈ ℂ(x) the height H(f) is just the number of poles of f counted according to multiplicity. We note that if f lies in ℂ[x], then H(f) = d deg f. Moreover, we have

(6)  max{H(f + h), H(fh)} ≤ H(f) + H(h)

for any two elements f, h in L.

Now we state the following theorem on the solutions of two-dimensional unit equations over an algebraic function field, which is usually referred to as Mason’s inequality and which can be seen as an analog of Baker’s theorem in linear forms of logarithms of algebraic numbers. A proof of this theorem can be found in the monograph of Mason (cf. [15, Lemma 2]).

Theorem 4. (R. C. Mason) Let γ_{1}, γ_{2} and γ_{3} be non-zero elements of L with γ_{1} + γ_{2} + γ_{3} = 0, and such that ν(γ_{1}) = ν(γ_{2}) = ν(γ_{3}) for each valuation ν not in the finite set V. Then either γ_{1}/γ_{2} lies in ℂ, in which case H(γ_{1}/γ_{2}) = 0, or

H(γ_{1}/γ_{2}) ≤ |V| + 2g − 2,

where |V| denotes the number of elements of V and g the genus of L/ℂ(X).

Now we are ready to prove the following lemma:

Lemma 1. Let {a, b, c, d}, a < b < c < d be a polynomial D(n)-quadruple with n ∈ ℤ[x]. Then

deg d ≤ 7 deg a + 11 deg b + 15 deg c + 14 deg n − 4.

Proof. We denote by W the set of absolute values on L containing all infinite ones together with those finite absolute values ν for which ν(μ) > 0. For brevity we denote M = 2g − 2 + |W|.

First, we need an upper bound for the genus g of L/ℂ(x). We consider the two Kummer extensions F_{1} := ℂ(x, \sqrt{ab}) and F_{2} := ℂ(x, \sqrt{ac}) and calculate the genus g_{1} of F_{1}/ℂ(x) and g_{2} of F_{2}/ℂ(x), respectively. It follows from [17, Corollary III.7.4] (see also Example III.7.6 on page 113) that

g_{1} = \frac{deg a + deg b − 2}{2}, g_{2} = \frac{deg a + deg c − 2}{2},

since neither ab nor ac can have odd degree (ab + n and ac + n are squares of polynomials and therefore have even degree). Observe that the degree of the extensions F_{1}/ℂ(x) and F_{2}/ℂ(x) is two in both cases. Now we can use Castelnuovo’s inequality (cf. [17, Theorem III.10.3]) to get an upper bound for the genus g of L = F_{1}F_{2}. We have

g ≤ 2\frac{deg a + deg b − 2}{2} + 2\frac{deg a + deg c − 2}{2} + 1 = 2 deg a + deg b + deg c − 3.

Next, we need an upper bound for the cardinality of the set W. It can be obtained by considering the number of zeros and poles of μ = −n^{3}abc(b − a)(c−a)(c−b). The number of zeros is bounded by the degree of the polynomial μ which is 3 deg n + deg a + 2 deg b + 3 deg c. Each zero can be extended to an absolute value of L in at most [L : ℂ(x)] = 4 ways. Moreover, there exist at most 4 infinite absolute values on L. Therefore,

|W| ≤ 4(3 deg n + deg a + 2 deg b + 3 deg c) + 4.

Now Mason’s theorem (Theorem 4) applied to the equation β_{1} +β_{2} +β_{3} = 0 yields that

H(β_{2}/β_{3}) ≤ M.

Further, M also serves as an upper bound for each of H(β̂_{2}/β_{3}), H(β_{2}/β̂_{3}) and H(β̂_{2}/β̂_{3}) because of equations (5). However, it is easy to check that

\frac{2(2X − nbc − nab)}{nc(a − b)} = \frac{β̂_{2}}{β_{3}} \frac{β̂_{2}}{β̂_{3}} + \frac{β_{2}}{β_{3}} \frac{β_{2}}{β̂_{3}} .

Hence, we have

H (\frac{2X − nbc − nab}{nc(a − b)}) ≤ 4M

and therefore

H(X) ≤ H(nb(a + c)) + H(nc(a − b)) + 4M,

where we have used that the height of a sum or a product is bounded by the sum of the heights (see (6)). Finally, since X = abcd, we get

H(X) = 4(deg a + deg b + deg c + deg d),
H(nb(a + c)) = 4(deg n + deg b + deg c),
H(nc(a − b)) ≤ 4(deg n + deg c + deg b),

and therefore, by taking into account the bound for M which is

M ≤ 4 deg a + 2 deg b + 2 deg c − 8 + 12 deg n + 4 deg a + 8 deg b + 12 deg c + 4,

we obtain the following upper bound

deg d ≤ 14 deg n + 7 deg a + 11 deg b + 15 deg c − 4

as claimed in our lemma. □

4. Proof of Theorem 3

Let S = {a_{1}, a_{2}, …, a_{m}} be a polynomial D(ux + v)-m-tuple with some integers u ≠ 0 and v. We already know that m ≤ 26 from the main result in [10]. From the fact that the product of each two elements from S plus ux+v 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. From Theorem 1 we get that there are at most 4 linear polynomials in S. Theorem 2 implies that there are at most 3 quadratic polynomials in S. The number of polynomials of degree μ ≥ 3 is also at most 3 and there is at most one constant in S.

We may assume that there is a polynomial of degree ≥ 2 in S. Therefore, we will consider separate cases depending on whether all degrees are even or all degrees are odd.

We use the following gap principle, which was already proved in our previous paper (cf. [10, Lemma 3]).

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

deg d ≥ deg b + deg c − 2.

Combining these gaps between the degrees of the elements in S with the upper bound proved in Lemma 1 we will get a much smaller upper bound for m.

First, we consider the case that all degrees of the a_{i} in S are odd. Let us assume the worst case, namely that there is the smallest possible gap between the degrees of the elements in S according to Lemma 2. In this case the following sequence of degrees is possible:

(7)  1, 1, 1, 1, 3, 3, 3, 5, 7, 11, 17, 27, 43, 69, 111, 179, ….

More precisely, we get lower bounds for the degrees of the elements of S:

deg a_{1} ≥ 1,  deg a_{2} ≥ 1,  deg a_{3} ≥ 1,  deg a_{4} ≥ 1,
deg a_{5} ≥ 3,  deg a_{6}, ≥ 3,  deg a_{7} ≥ 3,  deg a_{8} ≥ 5,
deg a_{9} ≥ 7,  deg a_{10} ≥ 11, deg a_{11} ≥ 17,  deg a_{12} ≥ 27,
deg a_{13} ≥ 43, deg a_{14} ≥ 69, deg a_{15} ≥ 111, …

We obtained this in the following way: there are at most 4 linear polynomials in S. The next possible (odd) degree is 3 and there are at most 3 polynomials of degree 3 in S. But having three polynomials of degree 3 enables us to use the above gap principle (Lemma 2) and we get that the next degree is

deg a_{8} ≥ 3 + 3 − 2 = 4,

and since the smallest odd number ≥ 4 is 5 we get the lower bound as stated in the table, namely deg a_{8} ≥ 5. Proceeding in this way, we produce the numbers in (7).

Since the linear polynomials in S play a special role here we will divide cases depending on how many linear polynomials our set S contains. If we assume that deg a_{1} = deg a_{2} = deg a_{3} = deg a_{4} = 1, then we have

deg a_{1} = 1,  deg a_{2} = 1,  deg a_{3} = 1,
deg a_{4} = 1,  deg a_{5} = A,  deg a_{6} ≥ A,
deg a_{7} ≥ A,  deg a_{8} ≥ 2A − 1,  deg a_{9} ≥ 3A − 2,
deg a_{10} ≥ 5A − 4,  deg a_{11} ≥ 8A − 7,  deg a_{12} ≥ 13A − 12,
deg a_{13} ≥ 21A − 20, deg a_{14} ≥ 34A − 33, …

with A ≥ 3.

Let us assume that A > 3 first. We get by Lemma 1 applied to {a_{1}, a_{2}, a_{3}, a_{m}} that

deg a_{m} ≤ 7 + 11 + 15 + 14 − 4 = 43,

which gives a contradiction unless m ≤ 12.

Now, we consider the case that A = 3. We first show that the configuration of degrees 1, 1, 1, 1, 3, 3, 3 is not possible. Assume that {a, b, c, d} ⊆ S is a polynomial D(ux + v)-quadruple such that deg a = 1, deg b = deg c = deg d = 3. For the polynomials e and \overline{e} defined by

e = n(a + b + c) + 2abc − 2rst, (8) \overline{e} = n(a + b + c) + 2abc + 2rst, (9)

where ab+n = r^{2}, ac+n = s^{2}, bc+n = t^{2}, we have that ae+n^{2}, be+n^{2}, ce+n^{2} are perfect squares (cf. [10, Lemma 1] applied to {a, b, c}) and

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

(see ([10, equation (2)]). It is plain that deg \overline{e} = 7, deg(e\overline{e}) ≤ 8 and therefore deg e ≤ 1. Hence, e = 0 or deg e = 1. If e = 0, then c = a + b + 2r. Also the third polynomial of degree 3 has the form d = b + c + 2t by the proof of [10, Proposition 2]. Thus, d = a + 4b + 4r and together with ad + n = z^{2}, we get

3n = (a + 2r − z)(a + 2r + z),

a contradiction. Therefore, we may assume that deg e = 1. From be+n^{2} = y^{2}, we have y ± n = e · f with deg f = 1. This gives b = ef^{2} ∓ 2fn. Hence, f|b. We want to prove that there are at most 3 such f’s corresponding to the possible linear factors of b. Assume that we have two such f’s (say f and f′) which correspond to the same linear factor of b, i.e. f′ = α · f, α ≠ 1. From

b = ef^{2} ∓ 2fn = e′f^{′2} ∓ 2f′n,

we find

f(e′ · α^{2} − e) = ±2n(α ± 1).

Thus, n|f and n|b. From be+n^{2} = y^{2} we find that n|y and n^{2}|be. Hence, n|e. Now ce + n^{2} being a perfect square, implies that n|c and n^{2}|bc contradicting the relation bc + n = t^{2}. Therefore, there are at most 3 polynomials f with the above property and consequently, there are at most 3 possibilities for the polynomial e. Altogether, this means that for fixed polynomials b and c of degree 3, there are at most 3 possibilities for the linear polynomial a (each e induce two possible a’s, but as we have shown above only one of them is indeed a polynomial). Hence, we proved that the configuration 1, 1, 1, 1, 3, 3, 3 is not possible. The remaining case to consider is

deg a_{1} = 1,  deg a_{2} = 1,  deg a_{3} = 1,
deg a_{4} = 1,  deg a_{5} = A,  deg a_{6} ≥ A,
deg a_{7} ≥ 2A − 1,  deg a_{8} ≥ 3A − 2,  deg a_{9} ≥ 5A − 4,
deg a_{10} ≥ 8A − 7,  deg a_{11} ≥ 13A − 12, deg a_{12} ≥ 21A − 20,
deg a_{13} ≥ 34A − 33, …

with A = 3. But as above we get deg a_{13} ≤ 43, and therefore

34A − 33 ≤ deg a_{13} ≤ 43,

which is a contradiction to A = 3.

Similarly, we get upper bounds for m in the case that we have

deg a_{1} = deg a_{2} = deg a_{3} = 1 and
deg a_{4} = A ≥ 3, deg a_{5} ≥ A, deg a_{6} ≥ A, deg a_{7} ≥ 2A − 1, …,
deg a_{13} ≥ 34A − 33,

where we get deg a_{13} ≤ 43 as before and therefore m ≤ 12. In the case

deg a_{1} = deg a_{2} = 1, deg a_{3} = A ≥ 3,

deg a_{4} ≥ A, deg a_{5} ≥ A, deg a_{6} ≥ 2A − 1, …, deg a_{13} ≥ 55A − 54,

we get

deg a_{13} ≤ 7 + 11 + 15A + 14 − 4 = 15A + 28

and therefore m ≤ 12. Next, we consider the case

deg a_{1} = 1, deg a_{2} = A, deg a_{3} = B,
deg a_{4} ≥ B, deg a_{5} ≥ 2B − 1, …, deg a_{12} ≥ 55B − 54

with 3 ≤ A ≤ B, where we get

deg a_{12} ≤ 7 + 11A + 15B + 14 − 4 ≤ 26B + 17

and therefore m ≤ 11. Observe that we can apply the gap principle already to get a lower bound for deg a_{4}, since we have three elements with degree ≥ 3. Finally, we consider the case

deg a_{1} = A, deg a_{2} = B, deg a_{3} = C,
deg a_{4} ≥ C, deg a_{5} ≥ 2C − 1, …, deg a_{12} ≥ 55C − 54

with 3 ≤ A ≤ B ≤ C, where we get

deg a_{12} ≤ 7A + 11B + 15C + 14 − 4 ≤ 33C + 10

and therefore m ≤ 11. Altogether, we see that there are at most 12 polynomials in S all of them having odd degrees.

The case where all polynomials in S have even degree can be handled in essentially the same way. Here the degrees 0 (which appears at most once) and 2 play a special role.

Let us start by showing that it is not possible to have polynomials {a, b, c, d} ⊆ S with deg a = A, deg b = deg c = deg d = B and a < b < c < d, 2 ≤ A < B. By the proof of [10, Proposition 2] we have d = b + c + 2t, where bc + n = t^{2}. Consider the triple \{a, b, c\} and let e and \overline{e} be the polynomials defined by (8) and (9), which exist by [10, Lemma 1]. Since deg \overline{e} = A+2B, deg(e\overline{e}) ≤ 2B+2 (by (10)), it follows that deg e ≤ 2−A ≤ 0. Hence, e is a constant. But by the proof of [10, Proposition 3] (we used these arguments already above), there is at most one nonzero constant e such that ae + n^{2} is a perfect square. Therefore, one of the polynomials c and d corresponds to e = 0. We may assume that c = a + b + 2r. Then d = a + 4b + 4r, and the condition ad + n = z^{2} leads again to

3n = (a + 2r − z)(a + 2r + z),

a contradiction.

Now assume that deg a_{1} = 0, deg a_{2} = 2, deg a_{3} = 2. Then deg a_{4} ≥ 2, deg a_{5} ≥ 4 (since there are at most 3 elements of degree 2 in the set S by Theorem 2), deg a_{6} ≥ 4, (since by the arguments from above with one polynomial of degree ≥ 2 there are at most two polynomials with the same degree B > 2) deg a_{7} ≥ 6, deg a_{8} ≥ 8, deg a_{9} ≥ 12, deg a_{10} ≥ 18, deg a_{11} ≥ 28, deg a_{12} ≥ 44, deg a_{13} ≥ 70. On the other hand, we get the upper bound

deg a_{13} ≤ 0 + 22 + 30 + 14 − 4 = 62,

which is a contradiction. Therefore, we get m ≤ 12 in this case.

Assume now that

deg a_{1} = 0,  deg a_{2} = A,  deg a_{3} = B,
deg a_{4} ≥ B,  deg a_{5} ≥ 2B − 2,  deg a_{6} ≥ 3B − 4,
deg a_{7} ≥ 5B − 8,  deg a_{8} ≥ 8B − 14,  deg a_{9} ≥ 13B − 24,
deg a_{10} ≥ 21B − 30, deg a_{11} ≥ 34B − 54, deg a_{12} ≥ 55B − 84

with 2 ≤ A < B and where we have again used the gap principle (Lemma 2) several times. Applying Lemma 1 to the quadruple {a_{1}, a_{2}, a_{3}, a_{12}} we get

deg a_{12} ≤ 11A + 15B + 14 − 4 ≤ 26B + 10,

a contradiction. Hence, m ≤ 11 in this case.

Finally, we consider the case that

deg a_{1} = A, deg a_{2} = B, deg a_{3} = C,

where 2 ≤ A ≤ B ≤ C. If C ≥ 4, then we have

deg a_{4} ≥ C, deg a_{5} ≥ 2C − 2, …, deg a_{13} ≥ 89C − 140.

By Lemma 1 we obtain

deg a_{13} ≤ 7A + 11B + 15C + 14 − 4 ≤ 33C + 10,

which gives a contradiction. If A = B = C = 2, then we have deg a_{1} = 2, deg a_{2} = 2, deg a_{3} = 2, deg a_{4} ≥ 4, deg a_{5} ≥ 4, deg a_{6} ≥ 6, deg a_{7} ≥ 8, deg a_{8} ≥ 12, deg a_{9} ≥ 18, deg a_{10} ≥ 28, deg a_{11} ≥ 44, deg a_{12} ≥ 70, deg a_{13} ≥ 112 and

deg a_{13} ≤ 14 + 22 + 30 + 14 − 4 = 76,

which gives a contradiction, showing that m ≤ 12. Altogether, we have at most 12 polynomials in S all having even degrees. □

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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

Gary Walsh Department of Mathematics University of Ottawa 585 King Edward St. Ottawa, Ontario, K1N 6N5, Canada. e-mail: gwalsh@mathstat.uottawa.ca