ON A PARAMETRIC FAMILY OF THUE INEQUALITIES OVER FUNCTION FIELDS

CLEMENS FUCHS^{∗} AND BORKA JADRIJEVIĆ^{†}

Abstract. In this paper we completely solve a family of Thue inequalities defined over the field of functions ℂ(T), namely

deg (X^{4} − 4cX^{3}Y + (6c + 2)X^{2}Y^{2} + 4cXY^{3} + Y^{4}) ≤ deg c, where the solutions x, y come from the ring ℂ [T] and the parameter c ∈ ℂ [T] is some non-constant polynomial.

2000 Mathematics Subject Classification: 11D59 (Primary), 11D75, 11A55, 11B37, 11J68 (Secondary). Keywords: Thue equations, function fields, continued fraction algorithm for polynomials, simultaneous Pellian equations

1. Introduction

Let F ∈ R [X, Y] be binary irreducible form of degree d ≥ 3 over some ring R. An equation of the form

F (X, Y) = m, m ∈ R^{×}

is called a Thue equation, due to Thue [22] who proved, in the case R = ℤ, that such an equation has finitely many solutions. In the last decade, starting with the result of Thomas in [21], several families (at the moment up to degree 8; see [9] and the references mentioned therein) of Thue equations have been considered, where the coefficients of the form F_{c}(X, Y) depend on an integral parameter c in R = ℤ or the integral closure of ℤ in an imaginary quadratic number field.

It is due to Gill [8] and later Schmidt [19] and Mason [14, 15] that single Thue equations can also be solved effectively over function fields. Recently, Lettl gave some new insights in this subject (cf. [11, 12]). However, no family of Thue equations over function fields was considered by these authors. Very recently, some families of Thue equations with parameter c ∈ ℂ[T] were studied by Ziegler and the first author. These were the first such results over function fields. In [6] all solutions (x, y) ∈ ℂ[T] × ℂ[T] of the family

X^{3} − (T + ξ + 1)X^{2}Y + (T + ξ)XY^{2} + Y^{3} = −ξT^{2} + ξT + 1

Date: May 29, 2006.

∗The first author was supported by the Austrian Science Foundation FWF, grants J2407-N12 and NFN S9611.

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

for all ξ ∈ ℂ were found, namely {ζ(T, 1) : ζ^{3} = 1} with some additional solutions for ξ = −1, 0, 1 (see [6]). Moreover, in [7], all solutions of the function field analogue of Thomas’ family [21]

X^{3} − (c − 1)X^{2}Y − (c + 2)XY^{2} − Y^{3} = ξ,

where c ∈ ℂ[T] is a non-constant polynomial and ξ ∈ ℂ^{×} := ℂ\{0}, were calculated. We have that all solutions (x, y) ∈ ℂ[T] × ℂ[T] are contained in {(ζ, 0), (−ζ, ζ), (0, −ζ) : ζ^{3} = ξ}. In the proof we essentially followed the ideas of Mason [14].

Apart from families of Thue equations also Thue inequalities

|F (X, Y)| ≤ m

over R = ℤ, where studied previously, but up to now only few parametrized inequalities of small degree have been solved (see [17, 13, 25, 26, 27]). We mention a result of Dujella and the second author (see [4]), who proved that the only primitive (i.e. with coprime components) solutions of the Thue inequality

|X^{4} − 4cX^{3}Y + (6c + 2)X^{2}Y^{2} + 4cXY^{3} + Y^{4}| < 6c + 4

where c ≥ 4 is an integer, are (x, y) ∈ {(±1, 0), (0, ±1), (1, ±1), (−1, ±1), (±1, ∓2), (±2, ±1)} (see also [3]). In [10] the second author jointly with Ziegler, generalized this result by considering the analogue family of equations over imaginary quadratic number fields.

In the present paper, we will consider a function field analogue of the family of Thue inequalities considered above, namely

(1)  deg (X^{4} − 4cX^{3}Y + (6c + 2)X^{2}Y^{2} + 4cXY^{3} + Y^{4}) ≤ deg c,

where solutions x, y come from the ring ℂ [T] and the parameter c ∈ ℂ [T] is some non-constant polynomial.

First observe that the problem is trivial for solutions (x, y) ∈ ℂ[T] × ℂ[T] with deg x = deg y = 0, since all (x, y) ∈ ℂ^{2} give solutions to the Thue inequality since in this case x^{4} − 4cx^{3}y + (6c + 2)x^{2}y^{2} + 4cxy^{3} + y^{4} is a polynomial of degree ≤ deg c. Therefore we restrict to solutions (x, y) ∈ (ℂ[T] × ℂ[T])\ℂ^{2}. Our main result is now the following theorem.

Theorem 1. The set of solutions L ⊆ (ℂ[T] × ℂ[T])\ℂ^{2} of the family of Thue inequalities

deg (X^{4} − 4cX^{3}Y + (6c + 2)X^{2}Y^{2} + 4cXY^{3} + Y^{4}) ≤ deg c,

where c ∈ ℂ[T] is a non-constant polynomial, is given by

L = {(a, 0), (0, a), (2a, a), (a, −2a) : a ∈ ℂ[T] with 0 < 4 deg a ≤ deg c}.

Clearly, the Thue inequality (1) can be rewritten as

(2) X^{4} −4cX^{3}Y +(6c+2)X^{2}Y^{2} +4cXY^{3} +Y^{4} = m, where deg m ≤ deg c.

We collect the solutions of (1) from Theorem 1 according to the corresponding right-hand side m of the equation (2) in the following table (where as above a ∈ ℂ[T] with 0 < 4 deg a ≤ deg c):

\begin{matrix}m & (x, y) ∈ L \\ a^{4} & (a, 0), (0, a) \\ 25a^{4} & (2a, a), (a, −2a) \end{matrix}

The novelty of this result is threefold. Firstly it is the first parametrized family of Thue inequalities over function fields solved up to now. Secondly if we look at the corresponding families of Thue equations, which are solved with this family, then it is the family of the largest degree considered yet (up to now only cubic families were considered, whereas we have a quartic family). Last but not least the method of proof is completely different from the way the results in [6, 7] are obtained as we will describe below.

In order to attack the inequality we will use a reduction method, which origins in the method of Tzanakis [24], and show that inequality (6) corresponds to systems of the form

(3)  (2c + 1)U^{2} − 2cV^{2} = m ,
(4)  (c − 2)U^{2} − cZ^{2} = −2m .

where deg m ≤ deg c. This will be done in Section 2, where we will also give some information about the splitting field of the form associated with the Thue equation.

In the proof we will also use that there are lots of analogues between ℂ [T] , ℂ (T) , ℂ ( ; (T^{−1}) ; ) and ℤ, ℚ, ℝ, respectively. Especially, one can consider continued fraction expansion of an element in ℂ ( ; (T^{−1}) ; ) . Namely, we have - as in classical theory - that each element in ℂ ( ; (T^{−1}) ; ) has a unique continued fraction expansion and an expansion of an element is finite if and only if the element is in ℂ (T) . We will give an overview about the theory of continued fractions for algebraic functions and prove some auxiliary results in Section 3.

In Section 4 we will use the theory of continued fractions in order to determine all values of m with deg m ≤ deg c, for which system (3) and (4) has a solution. In particular, a characterization in terms of continued fractions of α ∈ ℂ ( ; (T^{−1}) ; ) of all rational functions P/Q satisfying the inequality

|α − \frac{P}{Q}| < \frac{1}{|Q|^{2}}

(see Proposition 1 in Section 3) is used.

After that (in Section 5) we apply the congruence method, which was first used in this context in [5] (see also [2]), to show that system (3) and (4) has only trivial primitive (i.e. with coprime components) solutions (U, V, Z) = (±ζ, ±ζ, ±ζ) for some ζ ∈ ℂ^{×} (cf. Corollary 1).

Finally, the proof of the main theorem will be completed in Section 6.

2. Preliminaries and reduction to a system of inequalities

We consider the field of rational functions ℂ (T) with the following valuation:

v (\frac{P}{Q}) = deg P − deg Q, P, Q ∈ ℂ [T] , Q ≠ 0,

where we put v (0) = −∞. This valuation can be extended to a valuation on the field of formal Laurent series ℂ ( ; (T^{−1}) ; ) by

v (∑_{i=n}^{∞} a_{i}T^{−i}) = −n, a_{n} ≠ 0.

With

(5)  |α| = b^{v(α)},

where b ≥ 2 is an integer and α ∈ ℂ ( ; (T^{−1}) ; ) , we get a non-archimedian absolute value on ℂ ( ; (T^{−1}) ; ) . In this notation our inequality (1) has the form

(6)  |X^{4} − 4cX^{3}Y + (6c + 2)X^{2}Y^{2} + 4cXY^{3} + Y^{4}| ≤ |c| .

First we state some easy to prove facts about the underlying algebraic structure of the form F (X, Y) = X^{4} −4cX^{3}Y +(6c+2)X^{2}Y^{2} +4cXY^{3} +Y^{4}. We consider

f(X) = F (X, 1) = X^{4} − 4cX^{3} + (6c + 2)X^{2} + 4cX + 1.

The zeros of f(X) = 0 are given by

α_{1} = c + \sqrt{c(c − 2)} + \sqrt{c(c + \frac{1}{2})} + \sqrt{(c − 2)(c + \frac{1}{2})},
α_{2} = c − \sqrt{c(c − 2)} + \sqrt{c(c + \frac{1}{2})} − \sqrt{(c − 2)(c + \frac{1}{2})},
α_{3} = c + \sqrt{c(c − 2)} − \sqrt{c(c + \frac{1}{2})} − \sqrt{(c − 2)(c + \frac{1}{2})},
α_{4} = c − \sqrt{c(c − 2)} − \sqrt{c(c + \frac{1}{2})} + \sqrt{(c − 2)(c + \frac{1}{2})}.

Therefore, the splitting field of f is

ℂ (T, \sqrt{c(c − 2)}, \sqrt{c(c + \frac{1}{2})}) = ℂ(T)(α_{1}) =: L.

The field extension L/ℂ(T) is a Galois extension and the Galois group is isomorphic to C_{2} × C_{2}. Moreover, by using the Hurwitz formula [23, Theorem III.4.12, page 88] the genus g_{L} of L/ℂ(T) is bounded by g_{L} ≤ \frac{9}{2} deg c − 3, since all places in L that are ramified over finite places v_{a} in the rational function field satisfy 2c^{3}(a)−3c^{2}(a)−2c(a) = 0. From what we have collected up to now it is rather easy to prove, by using the theorem of Mason mentioned above (cf. [15, Chapter 1, Lemma 2]), that every solution (x, y) ∈ ℂ[T]×ℂ[T] of our inequality (6) satisfies max\{deg x, deg y\} ≤ \frac{41}{4} deg c − \frac{3}{4}.

As mentioned in the introduction we will use another method of proof to obtain our result. We consider Thue equation (2), that was

X^{4} − 4cX^{3}Y + (6c + 2)X^{2}Y^{2} + 4cXY^{3} + Y^{4} = m,

where solutions x, y come from the ring ℂ [T], parameter c is some nonconstant polynomial and m ∈ ℂ [T] where |m| ≤ |c|. We want to show how the solutions of the family of Thue equations (2) over the function field ℂ(T) corresponds to the solutions of the system (3) and (4) of Pellian equations over the same function field.

It is clear that every solution (x, y) ∈ ℂ [T] × ℂ [T] of Thue equation (2) yields a solution of the system

(2c + 1)U^{2} − 2cV^{2} = m ,
(c − 2)U^{2} − cZ^{2} = −2m .

where

(7) U = x^{2} + y^{2},  V = x^{2} + xy − y^{2},  Z = −x^{2} + 4xy + y^{2},

because to show this we just have to insert U, V, Z in the system of Pellian equations and we see that both equations reduce to the original Thue equation. In fact this can be done by the same computations as in [4], since these computations are still valid in ℂ [T]. Therefore, L (by the transformation (7)) is contained in the set of solutions of (3) and (4). Conversely, if U, V, Z is a solution of the system of Pellian equations we have to find all solutions of the system (7), to get all possible corresponding solutions of the Thue equation.

We mention that this relationship was also used by Dujella and Jadrijević in [4] and that there it was found by using the method of Tzanakis [24].

3. Continued fractions

In this section we will recall the theory of continued fraction expansion for algebraic functions in ℂ(T). First of all, we will start with definitions and some notation. Everything here holds also over an arbitrary field instead of the field of complex numbers ℂ. For details see [16] and [20].

We define rational functions in polynomials A_{0}, A_{1}, ... ∈ ℂ [T] by

[A_{0}] = A_{0}, [A_{0}, A_{1}] = A_{0} + \frac{1}{A_{1}} ,
[A_{0}, A_{1}, ..., A_{n}] = [A_{0}, A_{1}, ..., A_{n−2}, A_{n−1} + 1/A_{n}] , for n ≥ 2

Then

[A_{0}, A_{1}, ..., A_{n}] = [A_{0}, A_{1}, ..., A_{m−1}, [A_{m}, ..., A_{n}]] .

Setting P_{−2} = 0, Q_{−2} = 1, P_{−1} = 1, Q_{−1} = 0,

P_{n} = A_{n}P_{n−1} + P_{n−2}, Q_{n} = A_{n}Q_{n−1} + Q_{n−2}, n ≥ 0,

we observe that P_{n}, Q_{n} are polynomials, that

(8)  Q_{n}P_{n−1} − P_{n}Q_{n−1} = (−1)^{n} , n ≥ −1,

and

(9)  [A_{0}, A_{1}, ..., A_{n}] = \frac{P_{n}}{Q_{n}} .

An element of ℂ (T) may be uniquely expressed as [A_{0}, A_{1}, ..., A_{n}] = P_{n}/Q_{n}, where each A_{i} ∈ ℂ [T] and where deg A_{i} ≥ 1 for i ≥ 1. By (8), the polynomials P_{n}, Q_{n} ∈ ℂ [T] are relatively prime, i.e. (P_{n}, Q_{n}) = 1. Note that the pair P_{n}, Q_{n} is determined by (9) and is relatively prime only up to a common factor in ℂ^{×}, i.e. ζP_{n}, ζQ_{n} with ζ ∈ ℂ^{×} have the same properties. When A_{0}, A_{1}, ... ∈ ℂ [T] are given with deg A_{i} ≥ 1 for i ≥ 1, then [A_{0}, A_{1}, ..., A_{n}] as n → ∞ converges with respect | · | is the norm associated to the nonarchimedean absolute value on ℂ ( ; (T^{−1}) ; ) (as given by (5)), to an element of ℂ ( ; (T^{−1}) ; ) which will be denoted by

[A_{0}, A_{1}, A_{2}, …] = A_{0} + \frac{1}{A_{1} + \frac{1}{A_{2} + …}} .

Then we have that: Every element α ∈ ℂ ( ; (T^{−1}) ; ) \ℂ (T) can be uniquely expressed as such an infinite continued fraction. Moreover, precisely the elements from ℂ(T) have a finite expansion.

Writing α = [A_{0}, A_{1}, ...] , we call \frac{P_{n}}{Q_{n}} as given by (9) the nth convergent, A_{n} the nth partial quotient and α_{n} = [A_{n}, A_{n+1}, …] the nth complete quotient. We have

α = [A_{0}, A_{1}, ...A_{n−1}, α_{n}] = \frac{α_{n}P_{n−1} + P_{n−2}}{α_{n}Q_{n−1} + Q_{n−2}} , n ≥ 1.

Legendre’s theorem is also valid for this continued fraction expansion of elements of ℂ((T^{−1})), namely we have:

Proposition 1. Let α be an element in ℂ ( ; (T^{−1}) ; ) . If P, Q ∈ ℂ [T] are nonzero polynomials, (P, Q) = 1 satisfying the inequality

|α − \frac{P}{Q}| < \frac{1}{|Q|^{2}} ,

then there exists n ≥ 0 such that

\frac{P}{Q} = \frac{P_{n}}{Q_{n}} ,

where \frac{P_{n}}{Q_{n}} denotes the nth convergent of α.

Proof. This result can be found in [16]. □

For each α ∈ ℂ ( ; (T^{−1}) ; ) we write ⌊α⌋ for the polynomial part of α and {α} = α − ⌊α⌋ for the fractional part of α, i.e. for

α = ∑_{i=n}^{∞} a_{i}T^{−i} with a_{i} ∈ ℂ,

the polynomial part and the fractional part are

⌊α⌋ = ∑_{i=n}^{0} a_{i}T^{−i}, {α} = ∑_{i=1}^{∞} a_{i}T^{−i},

respectively. The above algorithms for obtaining the partial quotients can be reformulated in A_{0} = ⌊α⌋, B_{0} = {α} and A_{n} = ⌊B_{n−1}^{−1}⌋, B_{n} = {B_{n−1}^{−1}} for all n ≥ 1.

An infinite continued fraction [A_{0}, A_{1}, ...] is called periodic if there are integers k ≥ 0, m ≥ 1 such that A_{n+m} = A_{n} for all n ≥ k. Thus a periodic continued fraction can be written in the form

[A_{0}, A_{1}, .., A_{k−1}, \overline{A_{k}, A_{k+1}, ..., A_{k+m−1}}] ,

where the bar over A_{k}, A_{k+1}, ..., A_{k+m−1} indicates that this block of polynomials is repeated indefinitely.

In the classical theory of continued fraction expansions exactly the quadratic irrationals have a periodic expansion. This is not true for the continued fraction algorithm over ℂ(T). Nevertheless, if we have a periodic expansion then we can calculate it in the usual way, which we will show below.

Let α ∈ ℂ ( ; (T^{−1}) ; ) be an algebraic element of degree 2 over ℂ (T) , then α is said to be quadratic irrational. In this case α has the form

α = \frac{A + Δ}{B},

with A, B ∈ ℂ [T] , B ≠ 0 and Δ^{2} = D ∈ ℂ [T] and B|(D −A^{2}). A quadratic irrational α must be a root of the equation

(10)  X^{2} − Tr_{ℂ(T)/ℂ} (α) X + N_{ℂ(T)/ℂ} (α) = 0.

The other root of (10) is the algebraic conjugate of α, namely α′ = \frac{A−Δ}{B}. Let us define Δ = ±\sqrt{D}, where we choose the sign such that

|\frac{A − \sqrt{D}}{B} | ≤ |\frac{A + \sqrt{D}}{B} | .

Lagrange’s Theorem is not true in the form that every quadratic irrational over ℂ (T) has periodic continued fraction or pseudo-periodic continued fraction (the definition can be found in [19]). But if the quadratic irrational α = ^{\frac{A+\sqrt{D}}{B}} has a periodic expansion, then this expansion can be obtained using the following algorithm which we will obtain similarly as in the classical theory of continued fraction expansion (see [18, Chapter 7.7]). Let S_{0} = A, T_{0} = B and

(11) α_{n} = \frac{S_{n} + \sqrt{D}}{T_{n}} , A_{n} = ⌊α_{n}⌋ , S_{n+1} = A_{n}T_{n} − S_{n}, T_{n+1} = \frac{D − S_{n+1}^{2}}{T_{n}}

for n ≥ 0.

Lemma 1. Let (α_{n}), (A_{n}), (S_{n}), (T_{n}) be the sequences defined by (11), then α = [A_{0}, A_{1}, …] is the continued fraction expansion of α.

Proof. We have

α_{i} − A_{i} = \frac{S_{i} + \sqrt{D} − A_{i}T_{i}}{T_{i}} = \frac{\sqrt{D} − S_{i+1}}{T_{i}} = \frac{T_{i+1}}{\sqrt{D} + S_{i+1}} = \frac{1}{α_{i+1}}

which verifies that α = [A_{0}, A_{1}, …]. Now we use induction to prove that S_{i} and T_{i} are polynomials such that T_{i} ≠ 0 and T_{i}|(D − S_{i}^{2}). This holds for i = 0. If it is true at the ith stage, we observe that S_{i+1} = A_{i}T_{i} − S_{i} is a polynomial. Then the equation

T_{i+1} = \frac{D − S_{i+1}^{2}}{T_{i}} = \frac{D − S_{i}^{2}}{T_{i}} + 2A_{i}S_{i} − A_{i}^{2}T_{i}

establishes that T_{i+1} is also a polynomial. Moreover T_{i+1} cannot be zero, since if it were, we would have D = S_{i+1}^{2}, whereas D is not a perfect square.

Finally, we have T_{i} = \frac{D−S_{i+1}^{2}}{T_{i+1}} , so that T_{i+1}|(D − S_{i+1}^{2}). □

From this algorithm it follows that if there are integers j and k (j < k) such that (S_{j}, T_{j}) = (S_{k}, T_{k}), then α_{j+i} = α_{k+i} for all i ≥ 0. Hence, α = \frac{A+\sqrt{D}}{B} has a periodic expansion and that expansion is of the form

α = [A_{0}, …, A_{j−1}, \overline{A_{j}, …, A_{k−1}}].

Moreover, the converse is also true: if the expansion is periodic then there are integers j < k such that (S_{j}, T_{j}) = (S_{k}, T_{k}).

In the next lemma we will see how the convergents of the continued fraction expansion of a quadratic irrational α can be used to describe the solutions of a certain Pellian equation. This will be useful in describing all possible right-hand sides of the equation (2).

Lemma 2. Let AB be a polynomial which is not a perfect square in ℂ [T], and let P_{n}/Q_{n} denotes the nth convergent of continued fraction expansion of \sqrt{AB^{−1}}. Let the sequences (S_{n}) and (T_{n}) be defined by (11) for the quadratic irrational \sqrt{AB}/B. Then

(12)  AQ_{n}^{2} − BP_{n}^{2} = (−1)^{n}T_{n+1}

for n ≥ 0.

Proof. Since \sqrt{AB} ∉ ℂ (T), from

\frac{\sqrt{AB}}{B} = \frac{(S_{n+1} + \sqrt{AB})P_{n} + T_{n+1}P_{n−1}}{(S_{n+1} + \sqrt{AB})Q_{n} + T_{n+1}Q_{n−1}}

it follows that

S_{n+1}Q_{n} + T_{n+1}Q_{n−1} = BP_{n},
S_{n+1}P_{n} + T_{n+1}P_{n−1} = AQ_{n}.

By combining these two equalities with (8), we find (12). □

In the next section we will apply this lemma in order to reduce the Thue inequality to finitely many Diophantine equations.

4. From an inequality to a system of equations

In this section, we will consider the connections between solutions of the equations (3), (4) and continued fraction expansion of the corresponding quadratic irrationals in ℂ ( ; (T^{−1}) ; ) .

Applying the algorithm for quadratic irrationals from above to

θ_{1} := \sqrt{\frac{2c + 1}{2c}} = \frac{\sqrt{2c(2c + 1)}}{2c} and θ_{2} := \sqrt{\frac{c}{c − 2}} = \frac{\sqrt{c(c − 2)}}{c − 2}

we find the following continued fraction expansions:

Lemma 3. The continued fraction expansion of the quadratic irrationals θ_{1}, θ_{2} are given by

θ_{1} = [1, 4c + \frac{1}{2}, \overline{−16c − 4, 4c + 1}] and θ_{2} = [1, c − \frac{3}{2}, \overline{−4c + 4, c − 1}] .

Proof. We prove the statement by applying the above algorithm. For the quadratic irrational θ_{1} we have S_{0} = 0, T_{0} = 2c and therefore

α_{0} = \frac{\sqrt{2c(2c + 1)}}{2c} = 1 + \frac{1}{c} + O (\frac{1}{c^{2}} ) .

Thus we get A_{0} = 1. Next we get S_{1} = 1·2c−0 = 2c and T_{1} = \frac{2c(2c+1)−(2c)^{2}}{2c} = 1. It follows that

α_{1} = \frac{2c + \sqrt{2c(2c + 1)}}{1} = 4c + \frac{1}{2} − \frac{1}{16c} + O (\frac{1}{c^{2}} ) ,

and therefore A_{1} = 4c + \frac{1}{2}. Next we get S_{2} = 2c + \frac{1}{2}, T_{2} = −\frac{1}{4} and

α_{2} = \frac{2c + \frac{1}{2} + \sqrt{2c(2c + 1)}}{−\frac{1}{4}} = −16c − 4 + \frac{1}{4c} + O (\frac{1}{c^{2}} ) ,

thus A_{2} = −16c − 4. Now S_{3} = 2c + \frac{1}{2}, T_{3} = 1 and

α_{3} = \frac{2c + \frac{1}{2} + \sqrt{2c(2c + 1)}}{1} = 4c + 1 − \frac{1}{16c} + O (\frac{1}{c^{2}} ) ,

which implies A_{3} = 4c+1. Since in the next step we get S_{4} = 2c+ \frac{1}{2}, T_{4} = −\frac{1}{4}, which is the same pair as in step 2, we conclude that we have the periodic expansion as claimed in the statement.

For the second quadratic irrational θ_{2} we get in the same way with (S_{0}, T_{0}) = (0, c − 2), (S_{1}, T_{1}) = (c − 2, 2), (S_{2}, T_{2}) = (c − 1, −\frac{1}{2}), (S_{3}, T_{3}) = (c − 1, 2), (S_{4}, T_{4}) = (c − 1, −\frac{1}{2}), the expansion and therefore the proof is finished. □

Assume that (U, V, Z) is a solution of the system (3) and (4). We will show that \frac{V}{U} is a good approximation of θ_{1} and \frac{U}{Z} is a good rational approximation of θ_{2} if the absolute value of this fractions is less than 1. Otherwise the situation is handled in the following lemma.

Lemma 4. Let (U, V, Z) be a solution of (3), (4), respectively. If deg U > deg V or deg Z > deg U, then all solutions of (7) have deg x = deg y = 0.

Proof. Indeed, from deg V < deg U, it follows that deg (x^{2} + xy − y^{2}) < deg (x^{2} + y^{2}). This implies that x, y are polynomials of the same degree deg x = deg y = k. Let ξ_{x}, ξ_{y} ∈ ℂ^{×} be the leading coefficients of x, y, respectively. Since the leading coefficient of x^{2} + xy − y^{2} has to vanish, it follows that ξ_{x} = \frac{−1±\sqrt{5}}{2}ξ_{y}. Now, equation (2) implies k = 0 in the following way: Assume that k > 0. Observe that, since c is not constant and deg m ≤ deg c, we have deg(4cx^{3}y) = deg(6cx^{2}y^{2}) = deg(4cxy^{3}) = deg c + 4k > max{deg m, deg x^{4}, deg y^{4}, deg(2x^{2}y^{2})}. Therefore, by comparing the leading coefficients in (2), we get that

(13)  −4ξ_{x}^{2} + 6ξ_{x}ξ_{y} + 4ξ_{y}^{2} = 0,

which is in contradiction to the above relation between ξ_{x} and ξ_{y}. This contradiction shows that k = 0.

Now, we discuss the case deg U < deg Z. We have deg (x^{2} + y^{2}) < deg (−x^{2} + 4xy + y^{2}), which implies that x, y are polynomials of the same degree deg x = deg y = k. As above, let ξ_{x}, ξ_{y} ∈ ℂ^{×} denote the leading coefficients. Then ξ_{x} = ±iξ_{y}, and by using equation (13) (which was a consequence of (2) in the case k > 0), it follows in the same way as above that k = 0. □

So, we are left to consider the cases where deg U ≤ deg V and deg Z ≤ deg U and from now on will assume that this is the case. We proceed by showing that in this case we have good approximations of θ_{1} and θ_{2}, respectively.

Lemma 5. Let (U, V, Z) be a solution of (3), (4), respectively. If deg U ≤ deg V, then

|\sqrt{\frac{2c + 1}{2c}} − \frac{V}{U} | = |θ_{1} − \frac{V}{U} | < \frac{1}{|U|^{2}} .

If deg Z ≤ deg U, then

|\sqrt{\frac{c}{c − 2}} − \frac{U}{Z} | = |θ_{2} − \frac{U}{Z} | < \frac{1}{|Z|^{2}} .

Proof. The assumption deg U ≤ deg V is equivalent to |\frac{V}{U} | ≥ 1. Moreover, observe that |θ_{1}| = 1. Therefore, we find

|\sqrt{\frac{2c + 1}{2c}} − \frac{V}{U} | = |\frac{2c + 1}{2c} − \frac{V^{2}}{U^{2}} | · |\sqrt{\frac{2c + 1}{2c}} + \frac{V}{U} |^{−1}

≤ \frac{|m|}{|c| |U|^{2}} (min \{|\sqrt{\frac{2c + 1}{2c}}| , |\frac{V}{U} |\})^{−1} = \frac{|m|}{|c| |U|^{2}} ≤ |U|^{−2} .

As above from deg Z ≤ deg U it follows |\frac{U}{Z} | ≥ 1, |θ_{2}| = 1, and consequently we get

|\sqrt{\frac{c}{c − 2}} − \frac{U}{Z} | = |\frac{c}{c − 2} − \frac{U^{2}}{Z^{2}} | · |\sqrt{\frac{c}{c − 2}} + \frac{U}{Z} |^{−1}

≤ \frac{|m|}{|c| |Z|^{2}} (min \{|\sqrt{\frac{c}{c − 2}}| , |\frac{U}{Z} |\})^{−1} = \frac{|m|}{|c| |Z|^{2}} ≤ |Z|^{−2} ,

which was the claim of the lemma. □

We would like to apply Proposition 1 in order to determine all values of m ∈ ℂ [T], deg m ≤ deg c for which one of the equations

(2c + 1)U^{2} − 2cV^{2} = m
(c + 2)U^{2} − cZ^{2} = −2m

has a solution. According to Proposition 1 and Lemma 5, for all solutions (U, V, Z) the quotients \frac{V}{U} , \frac{U}{Z} are equal to a convergent of the continued fraction expansion of θ_{1}, θ_{2}, respectively. For the determination of the corresponding m’s, we use Lemma 2.

Since the period of the continued fraction expansions of θ_{1}, θ_{2} are equal to 2 in both cases, according to Lemma 2, we have to consider only the convergents for n = 0, n = 1, n = 2. By checking all possibilities, it is now easy to prove the following result.

Proposition 2. Let m ∈ ℂ[T], m ≠ 0 be a polynomial such that deg m ≤ deg c. Then:

(i) If equation (3) has a solution in relatively prime polynomials U and V in ℂ[T] with deg U ≤ deg V, then m = ζ^{2}, ζ ∈ ℂ^{×} and

(U, V) ∈ {ζ(Q_{2n}, P_{2n}), 2ζ(Q_{2n+1}, P_{2n+1}) : n ≥ 0},

where P_{n}/Q_{n} is a convergent of θ_{1}.

(ii) If equation (4) has a solution in relatively prime polynomials U and Z in ℂ[T] with deg Z ≤ deg U, then m = ξ^{2}, ξ ∈ ℂ^{×} and

(U, Z) ∈ {ξ(P′_{2n}, Q′_{2n}), 2ξ(P′_{2n+1}, Q′_{2n+1}) : n ≥ 0},

where P′_{n}/Q′_{n} is a convergent of θ_{2}.

Proof. Observe that from the proof of Lemma 3 it follows that T_{2n+1} = 1 for n = 0, 1, … and T_{2n} = −\frac{1}{4} for all n = 1, 2, …, therefore we get the conclusion by Lemma 2 in case (i). We just have to observe that the convergents are only fixed up to a constant ζ ∈ ℂ, thus we get the set stated in the lemma.

In the second case we observe that by Lemma 3 we have T_{2n+1} = 2 for n = 0, 1, … and T_{2n} = −\frac{1}{2} for n = 1, 2, …, therefore we get the second assertion (again by using Lemma 2). □

It follows that we only have solutions if the right-hand side of (2) is constant. Moreover, we mention that for m = 0 there does not exist a solution, since we assume that c is non-constant.

Now we will discuss the solvability in relatively prime polynomials U, V and Z of the system of equations (3) and (4), where m ∈ ℂ^{×}.

We will need the recursive relations for the convergents with odd and even subscripts. First we do it for the convergents of θ_{1}. From Q_{0} = 1, Q_{1} = 4c+ \frac{1}{2} and

Q_{2n} = (−16c − 4) Q_{2n−1} + Q_{2n−2}, n ≥ 1,
Q_{2n+1} = (4c + 1) Q_{2n} + Q_{2n−1}, n ≥ 1,

it follows easily that

Q_{2n} = −2 (32c^{2} + 16c + 1) Q_{2n−2} − Q_{2n−4}, n ≥ 2,
Q_{2n+1} = −2 (32c^{2} + 16c + 1) Q_{2n−1} − Q_{2n−3}, n ≥ 2,

and the analogous relations are valid for P_{2n} and P_{2n+1}. Now we derive the recursive relations for the convergents of θ_{2}. From P′_{0} = 1, P′_{1} = c − \frac{1}{2} and

P′_{2n} = (−4c + 4) P′_{2n−1} + P′_{2n−2}, n ≥ 1,
P′_{2n+1} = (c − 1) P′_{2n} + P′_{2n−1}, n ≥ 1,

it follows

P′_{2n} = −2 (2c^{2} − 4c + 1) P′_{2n−2} − P′_{2n−4}, n ≥ 2,
P′_{2n+1} = −2 (2c^{2} − 4c + 1) P′_{2n−1} − P′_{2n−3}, n ≥ 2,

respectively, and the analogous relations are valid for Q′_{2n} and Q′_{2n+1}.

Using the proposition and the recurrences for the convergents we now get the following lemma, which reduces the problem to solve 8 systems of equations which are given by the intersections of the above linear recurrences.

Lemma 6. Let (U, V, Z) ∈ ℂ[T] × ℂ[T] × ℂ[T] be solution of the system of Pellian equations (3) and (4). Then there exist non-negative integers m and n such that U = ζv_{m} or U = 2ζv′_{m} with m = ζ^{2}, ζ ∈ ℂ^{×} is the solution of one of the following equations

v_{m} = ±w_{n} or v_{m} = ±2w′_{n} or 2v′_{m} = ±w_{n} or v′_{m} = ±w′_{n},

where the sequence (v_{m}) is given by

v_{0} = 1, v_{1} = − (64c^{2} + 24c + 1) , (14)

v_{m+2} = −2 (32c^{2} + 16c + 1) v_{m+1} − v_{m}, for m ≥ 0,

the sequence (w_{n}) by

w_{0} = 1, w_{1} = −4c^{2} + 6c − 1, (15)

w_{n+2} = −2 (2c^{2} − 4c + 1) w_{n+1} − w_{n}, for n ≥ 0,

the sequence (v′_{m}) by

v′_{0} = 4c + \frac{1}{2}, v′_{1} = −\frac{1}{2} (512c^{3} + 320c^{2} + 48c + 1) , (16)

v′_{m+2} = −2 (32c^{2} + 16c + 1) v′_{m+1} − v′_{m}, for m ≥ 0,

and finally the sequence (w′_{n}) is given by

w′_{0} = c − \frac{1}{2}, w′_{1} = −\frac{1}{2} (8c^{3} − 20c^{2} + 12c − 1) , (17)

w′_{n+2} = −2 (2c^{2} − 4c + 1) w′_{n+1} − w′_{n}, for n ≥ 0.

Proof. If the system of equations (3) and (4) has solutions then, by comparing the right hand sides of the equations (3) and (4), it follows that ζ = ±ξ (where ζ, ξ are the constants in the general form of Proposition 2). Now, we just have to set v_{m} = Q_{2m}, v′_{m} = Q_{2m+1}, w_{n} = P′_{2n} and w′_{n} = P′_{2n+1}. Then the assertion about U follows from Proposition 2. The recurrences follow from what we have said above. □

Therefore, in order to prove Theorem 1, it suffices to show that v_{m} = w_{n} implies m = n = 0 and the equations v_{m} = −w_{n}, v′_{m} = ±w′_{n}, v_{m} = ±2w′_{n}, 2v′_{m} = ±w_{n} have no solutions.

For this purpose we will consider the sequences modulo certain multiples of c. This was done earlier in various contexts of systems of Pellian equations (e.g. in [1, 5, 3, 4]).

5. Congruence relations

The aim of this section is to find all possible values of U by the so called congruence method introduced first in [5].

Lemma 7. Let the sequences (v_{m}), (w_{n}), (v′_{m}) and (w′_{n}) be defined by (14), (15), (16) and (17). Then for all m, n ≥ 0 we have

(18)  v_{m} ≡ (−1)^{m} [8m(2m + 1)c + 1] (mod 64c^{2}),
(19)  w_{n} ≡ (−1)^{n−1}[2n(2n + 1)c − 1] (mod 4c^{2}).
(20)  v′_{m} ≡ (−1)^{m} [4 (m + 1) (2m + 1)c + \frac{1}{2}] (mod 64c^{2})
(21)  w′_{n} ≡ (−1)^{n} [(n + 1) (2n + 1)c − \frac{1}{2}] (mod 4c^{2})

Proof. All relations are obviously true for m, n ∈ {0, 1}. Assume that (18) is valid for m − 2 and m − 1. Then

v_{m} = (−2) (32c^{2} + 16c + 1) v_{m−1} − v_{m−2}
≡ (−2) (16c + 1) [(−1)^{m−1} (8 (m − 1) (2m − 1)c + 1)]

− [(−1)^{m−2} (8 (m − 2) (2m − 3)c + 1)]

≡ (−1)^{m} [32c + 2 · 8 (m − 1) (2m − 1)c + 2]
−(−1)^{m} [8 (m − 2) (2m − 3)c + 1]
= (−1)^{m} [8m(2m + 1)c + 1] (mod 64c^{2}).

Thus the first congruence follows by induction. Similarly, one can show the other relations for w_{n}, v′_{m}, and w′_{n} by using the recurrence relations (15), (16), and (17), respectively. □

By using the congruences from the last lemma we can show that all solutions arising from the equations v_{m} = ±w_{n}, v′_{m} = ±w′_{n}, v_{m} = ±2w′_{n}, 2v′_{m} = ±w_{n} are just the trivial ones.

Proposition 3. If v_{m} = w_{n} then n = m = 0. The equations v_{m} = −w_{n}, v′_{m} = ±w′_{n}, v_{m} = ±2w′_{n}, 2v′_{m} = ±w_{n} have no solutions in non-negative integers n, m.

Proof. Suppose that m and n are non-negative integers such that v_{m} = w_{n}. Then, of course, v_{m} ≡ w_{n} (mod 4c^{2}). By Lemma 7, we have (−1)^{n} ≡ (−1)^{m} (mod 2c) and therefore n and m are both even or both odd. Furthermore, Lemma 7 implies

8m(2m + 1)c + 1 ≡ 1 − 2n(2n + 1)c (mod 4c^{2})

and

(22)  4m(2m + 1) ≡ −n(2n + 1) (mod 2c).

From (22) we have

4m(2m + 1) = −n(2n + 1),

which implies n = m = 0.

Suppose that m and n are non-negative integers such that v_{m} = −w_{n}. Then we get in the same way as above that (−1)^{m} = (−1)^{n−1} and therefore 4m(2m + 1) = −n(2n + 1), which implies n = m = 0 and therefore −1 = 1, a contradiction.

Suppose that m and n are positive integers such that v′_{m} = ±w′_{n}. Then, of course, v′_{m} ≡ ±w′_{n} (mod 4c^{2}). By Lemma 7, we have ±\frac{1}{2}(−1)^{n−1} ≡ \frac{1}{2}(−1)^{m} (mod c). Furthermore, Lemma 7 implies

4 (m + 1) (2m + 1)c + \frac{1}{2} ≡ − (n + 1) (2n + 1)c + \frac{1}{2} (mod 4c^{2})

and

(23)  4 (m + 1) (2m + 1) ≡ − (n + 1) (2n + 1) (mod 2c).

From (23) we have

4m(2m + 1) = −n(2n + 1),

which implies n = m = 0 and therefore 4c + \frac{1}{2} = ± (c − \frac{1}{2}) , a contradiction. Suppose now that v_{m} = ±2w′_{n}. Then we get by Lemma 7 that

(−1)^{m}[8m(2m + 1)c + 1] ≡ ±2(−1)^{n} [(n + 1)(2n + 1)c − \frac{1}{2}] (mod 4c^{2}).

From this we conclude (−1)^{m} ≡ ±(−1)^{n−1} (mod c) and therefore (−1)^{m} = ±(−1)^{n−1}. Consequently, we get 4m(2m + 1) = −(n + 1)(2n + 1), a contradiction.

Finally, we consider the case that m and n are positive integers such that 2v′_{m} = ±w_{n}. Then Lemma 7 implies that (−1)^{m} = ±(−1)^{n}. Using Lemma 7 once again we get

2 [4(m + 1)(2m + 1)c + \frac{1}{2}] ≡ −[2n(2n + 1)c − 1] (mod 4c^{2}),

which implies 4(m + 1)(2m + 1) = −n(2n + 1), again a contradiction. Altogether we have proved the statement of the proposition. □

To sum up we have derived the following result:

Corollary 1. Let c ∈ ℂ [T] be a non-constant polynomial, m ∈ ℂ [T] , deg m ≤ deg c. If the system of equations (3) and (4) has primitive solutions (U, V, Z) with deg Z ≤ deg U ≤ deg V, then m = ζ^{2} with ζ ∈ ℂ^{×} and all primitive solutions are given by (U, V, Z) = (±ζ, ±ζ, ±ζ) with mixed signs.

Proof. This follows directly from Propositions 2 and 3. □

In the next section we will use these facts to prove our main theorem.

6. Proof of Theorem 1

Let (x, y) be a solution of the inequality (2), and let U = x^{2} + y^{2}, V = x^{2} + xy − y^{2}, Z = −x^{2} + 4xy + y^{2}. Then (U, V, Z) satisfies the system (3) and (4) for some polynomial m ∈ ℂ[T] such that |m| ≤ |c|.

Assume first that U, V and Z (with deg Z ≤ deg U ≤ deg V) are relatively prime. Then Corollary 1 implies that m = ζ^{2} and in this case we have (U, V, Z) = (±ζ, ±ζ, ±ζ) with mixed signs and some ζ ∈ ℂ^{×}. Therefore, we have

x^{2} + y^{2} = ±ζ,
x^{2} + xy − y^{2} = ±ζ,
−x^{2} + 4xy + y^{2} = ±ζ.

These equations imply that deg x = deg y. Let us assume that deg x = deg y = k ≥ 1 and let the leading coefficient of x be ξ_{x} and the leading coefficient of y be ξ_{y}. Then by comparing the leading coefficients in the above equations, we see that ξ_{x} and ξ_{y} have to satisfy the system

ξ_{x}^{2} + ξ_{y}^{2} = 0,
ξ_{x}^{2} + ξ_{x}ξ_{y} − ξ_{y}^{2} = 0,
−ξ_{x}^{2} + 4ξ_{x}ξ_{y} + ξ_{y}^{2} = 0,

which clearly implies ξ_{x} = ξ_{y} = 0. This gives a contradiction that shows deg x = deg y = 0 or equivalently x, y ∈ ℂ. Clearly, this does not lead to solutions in (ℂ[T] × ℂ[T])\ℂ^{2}.

Assume now that d = gcd(U, V) and deg d > 0. Let U = dU_{1}, V = dV_{1}. Then U_{1} and V_{1} are relatively prime and satisfy

(2c + 1)U_{1}^{2} − 2cV_{1}^{2} = \frac{m}{d^{2}} .

Since deg(m/d^{2}) ≤ deg c, Proposition 2 implies that m/d^{2} = ζ^{2}, i.e. m = ζ^{2}d^{2}. From

4V^{2} + Z^{2} = 5U^{2}

it follows that d|Z, say Z = dZ_{1}. In that way, we have obtained the primitive triple (U_{1}, V_{1}, Z_{1}) satisfying

(2c + 1)U_{1}^{2} − 2cV_{1}^{2} = ζ^{2},
(c − 2)U_{1}^{2} − cZ_{1}^{2} = −2ζ^{2}.

By Proposition 2, it follows that (U_{1}, V_{1}, Z_{1}) = (±ζ, ±ζ, ±ζ) and (U, V, Z) = (±ζd, ±ζd, ±ζd). Therefore, we have

(24)  x^{2} + y^{2} = ±ζd,
(25)  x^{2} + xy − y^{2} = ±ζd,
(26)  −x^{2} + 4xy + y^{2} = ±ζd,

with mixed signs. Now (24) and (25) imply y (2y − x) = 0 or x (2y + x) = 0 (depending on the sign) and therefore (x, y) = (a, 0) or (x, y) = (2a, a) (for the first possible choice of signs in the system above) or (x, y) = (0, a) or (x, y) = (a, −2a) (for the second possible choice of signs in the system above), where a ∈ ℂ[T] with 1 < |a| ≤ |c|^{\frac{1}{4}} , i.e. 0 < deg a ≤ \frac{1}{4} deg c. This completes the proof of our main theorem. □

7. Acknowledgement

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 (No. 20/2004 and 23/2006).

References

[1] A. Baker and H. Davenport, The equations 3x^{2} −2 = y^{2} and 8x^{2} −7 = z^{2}, Quart. J. Math. Oxford Ser. (2) 20 (1969), 129–137.7

[2] A. Dujella, The problem of the extension of a parametric family of Diophantine triples, Publ. Math. Debrecen 51 (1997), 311–322.

[3] A. Dujella and B. Jadrijević, A parametric family of quartic Thue equations, Acta Arith. 101(2) (2001), 159–169.

[4] A. Dujella and B. Jadrijević, A family of quartic Thue inequalities, Acta Arith. 111(1) (2002), 61–76.

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

[6] C. Fuchs and V. Ziegler, On a family of Thue equations over function fields, Monatsh. Math. 147 (2006), 11-23.

[7] C. Fuchs and V. Ziegler, Thomas’ family of Thue equations over function fields, Q. J. Math. (Oxford) 57 (2006), 81-91.

[8] B. P. Gill, An analogue for algebraic functions of the Thue-Siegel theorem, Ann. of Math. (2) 31(2) (1930), 207-218.

[9] C. Heuberger, A. Togbé, and V. Ziegler, Automatic solution of families of Thue equations and an example of degree 8, J. Symbolic Computation 38 (2004), 1145-1163.

[10] B. Jadrijević and V. Ziegler, A system of relative Pellian equations and a family of relative related Thue equations, Int. J. Number Theory, to appear.

[11] G. Lettl, Thue equations over algebraic function fields, Acta Arith. 117 (2004), 107-123.

[12] G. Lettl, Parametrized solutions of Diophantine equations, Math. Slovaca 54 (2004), 465-471.

[13] G. Lettl, A. Pethő, and P. Voutier, Simple families of Thue inequalities, Trans. Amer. Math. Soc. 351 (1999), 1871–1894.

[14] R. C. Mason, On Thue’s equation over function fields, J. London Math. Soc. (2) 24 (1981), 414-426.

[15] R. C. Mason, Diophantine equations over function fields, volume 96 of London Mathematical Society Lecture Note Series, Cambridge University Press, Cambridge, 1984.

[16] B. de Mathan, Approximations diophantiennes dans un corps local, Bull. Soc. Math. France, Suppl. Mem. 21 (1970).

[17] M. Mignotte, A. Pethő, and F. Lemmermeyer, On the family of Thue equations x^{3} − (n − 1)x^{2}y − (n + 2)xy^{2} − y^{3} = k, Acta Arith. 76 (1996), 245–269.

[18] I. Niven, H. S. Zuckerman, and H. L. Montgomery, An Introduction to the Theory of Numbers, John Wiley, New York, 1991.

[19] W. M. Schmidt, Thue’s equation over function fields, J. Austral. Math. Soc. Ser. A 25(4) (1978), 385-422.

[20] W. M. Schmidt, Continued fractions and diophantine approximation, Acta Arith. 95(2) (2000), 139–166.

[21] E. Thomas, Complete solutions to a family of cubic Diophantine equations, J. Number Theory 34 (1990), 235–250.

[22] A. Thue, Über Annäherungswerte algebraischer Zahlen, J. Reine Angew. Math. 135 (1909), 284–305.

[23] H. Stichtenoth, Algebraic function fields and codes, Springer Verlag, Berlin - New York, 1993.

[24] N. Tzanakis, Explicit solution of a class of quartic Thue equations, Acta Arith. 64 (1993), 271–283.

[25] I. Wakabayashi, On a family of quartic Thue inequalities, J. Number Theory 66 (1997), 70–84.

[26] I. Wakabayashi, On a family of quartic Thue inequalities, II, J. Number Theory 80 (2000), 60–88.

[27] I. Wakabayashi, Cubic Thue inequalities with negative discriminant, J. Number Theory 97 (2002), 222-251.

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

Borka Jadrijević University of Split FESB R. Boškovića bb, 21000 Split (Croatia) e-mail: borka@fesb.hr