THOMAS’ FAMILY OF THUE EQUATIONS OVER IMAGINARY QUADRATIC FIELDS

CLEMENS HEUBERGER, ATTILA PETHŐ, AND ROBERT F. TICHY

Dedicated to Professor H. G. Zimmer on his 65^{th} birthday

Abstract. We consider the family of relative Thue equations

x^{3} − (t − 1)x^{2}y − (t + 2)xy^{2} − y^{3} = μ, where the parameter t, the root of unity μ and the solutions x and y are integers in the same imaginary quadratic number field.

We prove that there are only trivial solutions (with |x|, |y| ≤ 1), if |t| is large enough or if the discriminant of the quadratic number field is large enough or if Re t = −1/2 (there are a few more solutions in this case which are explicitly listed). In the case Re t = −1/2, an algebraic method is used, in the general case, Baker’s method yields the result.

1. Introduction

Let F ∈ ℤ[x, y] be an irreducible form of degree at least 3 and m be a nonzero integer. Then the Diophantine equation

F (x, y) = m

is called a Thue equation in honor of A. Thue [21] who proved that it has only finitely many solutions x, y ∈ ℤ. Upper bounds for the solutions have been given using A. Baker’s [1] theory on linear forms in logarithms of algebraic numbers, cf. Bugeaud and Győry [4]. Algorithms to solve single Thue equations have been developed, we refer to Bilu and Hanrot [3].

In the last decade, several families of parametrized Thue equations have been investigated; a survey containing further references is given in [10]. Furthermore, we refer to the more recent results [23], [20], [22], [6], [12], [5], [8], and [11].

Relative Thue equations, i. e. Thue equations with coefficients in an algebraic number field, where solutions come from the ring of integers of the same number field, have also been successfully solved, we refer to Gaál and Pohst [9] for further references.

However, to our knowledge, no infinite family of relative Thue equations has been solved up to now. The aim of this paper is to solve such a family. As the ground field, we take an imaginary quadratic number field. We consider the form x^{3} − (t − 1)x^{2}y − (t + 2)xy^{2} − y^{3}. If t ∈ ℤ, it generates Shanks’ [17] simplest cubic number fields. The corresponding family of absolute Thue equations was the first family of positive discriminant which has been solved, cf. Thomas [18] and Mignotte [14]. We consider the same form with an imaginary quadratic integral parameter t. This setting has been chosen in order to avoid the unit rank to increase.

Our main result is the following: Theorem 1. Let t be an integer in an imaginary quadratic number field, ℤ_{ℚ(t)} be the ring of integers of ℚ(t) and μ be a root of unity in ℤ_{ℚ(t)}.

If |t| ≥ 3 023 195 238, then the only solutions x, y in ℤ_{ℚ(t)} to the family of relative Thue equations

(1)  x^{3} − (t − 1)x^{2}y − (t + 2)xy^{2} − y^{3} = μ

are listed in Table 1.

The first and the third author were supported by grant S8307-MAT of the Austrian Science Fund.

The second author was partially supported by Hungarian National Foundation for Scientific Research Grants No 29330 and 38225.

x y μ_{0 1 −1}_{−1 0 −1}_{1 −1 −1}_{0 −1 1}_{−1 1 1}_{1 0 1}_{0 −i −i}_{−i i −i} x  y μ

i  0 −i
0  i  i
−i  0  i
i  −i  i
0  −ω_{3} −1
0 −1 + ω_{3} −1
−ω_{3}  ω_{3} −1
1 − ω_{3}  0 −1

x  y μ_{−1 + ω_{3} 1 − ω_{3} −1}_{ω_{3}  0 −1}_{0 1 − ω_{3} 1}_{0  ω_{3} 1}_{−ω_{3}  0 1}_{1 − ω_{3} −1 + ω_{3} 1}_{−1 + ω_{3}  0 1}_{ω_{3}  −ω_{3} 1}

Table 1. Solutions (if contained in ℚ(t)) to (1) for all t, where ω_{3} = (1 + \sqrt{−3})/2.

For t ∈ ℤ^{≥0}, Theorem 1 has been proved by Thomas [18]. Writing F_{t}(X, Y) = X^{3}−(t−1)X^{2}Y − (t + 2)XY^{2} − Y^{3} and noting that F_{−t−1}(Y, X) = −F_{t}(X, Y), his result is easily generalized to negative rational integers t. We note that Thomas [18] and Mignotte [14] solved the rational integer case completely; they found some extra solutions for |t| ≤ 4. Therefore, we will assume t ∉ ℤ in the remainder of this paper.

The condition |t| ≥ 3 023 195 238 is certainly fulfilled if the discriminant of the base field is sufficiently large (in absolute value). It is well known that if D_{K} < −4 then the only roots of unity in K are ±1. Hence, if |D_{K}| > 4 we may take μ = ±1 and get immediately the following corollary. Corollary 2. Let k be an imaginary quadratic number field with discriminant ≤ −3.66 · 10^{19} and t ∈ k \ ℤ be an algebraic integer. Then the only solutions x, y ∈ ℤ_{k} to

x^{3} − (t − 1)x^{2}y − (t + 2)xy^{2} − y^{3} = ±1

are ±{(0, 1), (−1, 0), (1, −1)}.

The proof of Theorem 1 is based on Baker’s theory on linear forms in logarithms of algebraic numbers. This yields the rather large constant 3.03 · 10^{9}. Since the parameter t is complex, a complete enumeration of the remaining cases seems to be hopeless. However, if Re t = −1/2, we can use an algebraic argument and solve the equation completely for these values of the parameter. Theorem 3. Let t ≠ (−1 ± 3\sqrt{−3})/2 be an integer in an imaginary quadratic number field with Re t = −1/2 and μ be a root of unity in ℚ(t). Then the only solutions x, y ∈ ℤ_{ℚ(t)} to (1) are listed in Table 1 or in Table 2.

\begin{matrix}t & (x, y) ∈ \\ − ω_{3} & ±\{(3 − 6ω_{3}, −1 + 3ω_{3}), (−2 + 3ω_{3}, 3 − 6ω_{3}), (−1 + 3ω_{3}, −2 + 3ω_{3})\} \\ −1 + ω_{3} & ±\{(−3 + 6ω_{3}, 2 − 3ω_{3}), (1 − 3ω_{3}, −3 + 6ω_{3}), (2 − 3ω_{3}, 1 − 3ω_{3})\} \\ −ω_{7} & ±\{(−1 + 2ω_{7}, −ω_{7}), (−ω_{7}, 1 − ω_{7}), (1 − ω_{7}, −1 + 2ω_{7})\} \\ −1 + ω_{7} & ±\{(1 − 2ω_{7}, −1 + ω_{7}), (−1 + ω_{7}, ω_{7}), (ω_{7}, 1 − 2ω_{7})\} \\ −ω_{19} & ±\{(−1 + 2ω_{19}, −1 − ω_{19}), (−1 − ω_{19}, 2 − ω_{19}), (2 − ω_{19}, −1 + 2ω_{19})\} \\ −1 + ω_{19} & ±\{(1 − 2ω_{19}, −2 + ω_{19}), (−2 + ω_{19}, 1 + ω_{19}), (1 + ω_{19}, 1 − 2ω_{19})\} \end{matrix}

Table 2. All solutions for Re t = −1/2 to (1), where ω_{D} = (1 + \sqrt{−D})/2.

The case t = (−1 ± 3\sqrt{−3})/2 has to be excluded because the form F_{t} is the cube of a linear polynomial in that case.

The remainder of this paper is organized as follows: In Section 2, we collect elementary properties and asymptotic expressions for the algebraic numbers generating the splitting field of F_{t}. Small solutions are dealt with in Section 3. The Galois structure of the family will be determined in Section 4. It will turn out that the case Re t = −1/2 has a special Galois group. The unit structure of the relevant order will be discussed in Section 5. Approximation properties of solutions are the topic of Section 6. “Stable growth” will be proved in Section 7, which will exclude all medium sized solutions. The large solutions are excluded in Section 8 using a corollary of Mignotte [15] of the lower bound for linear forms in two logarithms by Laurent, Mignotte, and Nesterenko [13]. Finally, we will prove Theorem 3 in Section 9. Acknowledgement. We are grateful to Peter Kirschenhofer for raising this interesting problem, which he posed during an oral discussion.

2. Elementary Properties

Let t ∈ ℂ and

(2)  F_{t}(X, Y) := X^{3} − (t − 1)X^{2}Y − (t + 2)XY^{2} − Y^{3},
(3)  f_{t}(X) := F_{t}(X, 1).

It is straightforward to verify the identities

(4) F_{t}(Y, −(X + Y)) = F_{t}(−(X + Y), X) = F_{t}(X, Y)  and  f_{t} (−1 − \frac{1}{X} ) = \frac{f_{t}(X)}{X^{3}}.

Let α^{(1)} = α = α(t) be a root of f_{t}(X) = 0. Then (4) implies that

(5)  α^{(2)} := −1 − \frac{1}{α}  and  α^{(3)} := −1 − \frac{1}{α^{(2)}} = −\frac{1}{α + 1}

are roots of f_{t}. The roots α, α^{(2)}, and α^{(3)} are pairwise distinct for t ≠ (−1 ± 3\sqrt{−3})/2. Indeed, assuming α = α^{(2)} implies α^{2} + α + 1 = f_{t}(α) = 0, which is possible only if t = (−1 ± 3\sqrt{−3})/2. If t = (−1 ± 3\sqrt{−3})/2, we have f_{t}(X) = (X − (t − 1)/2)^{3}.

Assume that α ∈ ℝ. Then we get 0 = f_{t}(α) − \overline{f_{t}}(α) = −(t − \overline{t})(α^{2} + α), which implies t ∈ ℝ. We will use the following variant of the usual O-notation: For two functions g(t) and h(|t|)

and a positive number t_{0} we will write g(t) = L_{t_{0}}(h(|t|)) if |g(t)| ≤ h(|t|) for all t with absolute value at least t_{0}. We will use this notation in the middle of an expression in the same way as it is usually done with the O-notation. Lemma 4. Let t ∈ ℂ. Then there is a root α of f_{t} such that

(6)  α = t + \frac{2}{t} − \frac{1}{t^{2}} − \frac{3}{t^{3}} + L_{6} (\frac{5}{|t|^{7/2}} ) .

Proof. Let h(z) = f(z + t + 2t^{−1} − t^{−2} − 3t^{−3}) and h_{1}(z) := h(z) − h(0). For |t| ≥ 6 and |z| = 5 |t|^{−7/2}, we obtain

|h(z) − h_{1}(z)| = |h(0)| < \frac{7.38}{|t|^{2}} < \frac{5}{|t|^{3/2}} − \frac{10.26}{|t|^{5/2}} < |h_{1}(z)| .

By Rouché’s theorem, h and h_{1} have the same number of zeros in the disc {z : |z| < 5 |t|^{−7/2}}. Since h_{1}(0) = 0, this implies the theorem. □

Similarly, we will use

(7)  α = t + \frac{2}{t} − \frac{1}{t^{2}} − \frac{3}{t^{3}} + \frac{5}{t^{4}} + \frac{7}{t^{5}} − \frac{26}{t^{6}} − \frac{10}{t^{7}} + L_{20} (\frac{33}{|t|^{15/2}} ) .

Denoting the root calculated in Lemma 4 by α^{(1)} = α and using (5), we get the following estimates for the other roots:

(8a)  α^{(2)} = −1 − \frac{1}{α^{(1)}} = −1 − \frac{1}{t} + L_{6} (\frac{102}{5 |t|^{5/2}} ) ,

(8b)  α^{(3)} = −\frac{1}{α^{(1)} + 1} = −\frac{1}{t} + \frac{1}{t^{2}} + L_{6} (\frac{183}{10 |t|^{5/2}} ) .

3. Small Solutions

Lemma 5. Let |t| > 4. All solutions (x, y) ∈ ℤ_{ℚ(t)} to (1) with |y| < \sqrt{5}, |x + y| < \sqrt{5}, or |x| < \sqrt{5} are listed in Table 1.

Proof. We will first consider solutions with |y| < \sqrt{5}.

Let some imaginary quadratic root of unity μ and some imaginary quadratic integer y with |y| < \sqrt{5} be given such that [ℚ(μ, y) : ℚ] ≤ 2. It is clear that there is only a finite number of such pairs (y, μ). We are looking for quadratic integers x and t such that [ℚ(t, x, y, μ) : ℚ] ≤ 2 and F_{t}(x, y) = μ. The latter condition is equivalent to

(9)  x(x^{2} − (t − 1)xy − (t + 2)y^{2}) = μ + y^{3}.

If μ + y^{3} ≠ 0, we conclude that N(x) | N(μ + y^{3}), where N( ·) = N_{ℚ(t)/ℚ}( ·) denotes the norm function. This yields a finite number of possible values for x. We only consider those x such that [ℚ(μ, x, y) : ℚ] ≤ 2 and such that (μ + y^{3})/x is an algebraic integer.

Then (9) implies t(−xy − y^{2}) = (μ + y^{3})/x − x^{2} − xy + 2y^{2}. If (−xy − y^{2}) = (μ + y^{3})/x − x^{2} − xy + 2y^{2} = 0, we found a solution for all values of t and list it in Table 1. Otherwise, if xy + y^{2} ≠ 0, we calculate an explicit value for t. We check whether this t is an algebraic integer. It turns out that this process only yields values of t with |t| ≤ 4.

If μ + y^{3} = 0, we may either have x = 0 and arbitrary t (in the appropriate field) and the solution is listed in Table 1, or (9) results in

(x + y)(x − yt) = 2y^{2}.

This implies N(x + y) | N(2y^{2}) = 4. Therefore, we only have a finite number of choices for x + y. For each choice, we get a value of t. It turns out that all those t have absolute value at most 4.

Let now (x, y) be a solution of (1) with |x| < \sqrt{5}. By (4), (−(x + y), x) is a solution contained in Table 1. The case |x + y| < \sqrt{5} is dealt with similarly. It can be checked that for each solution (x, y) listed in Table 1, (y, −(x + y)) and (−(x + y), x) are also contained in the table. □

4. Computation of the Galois Group

For the remainder of this paper, we fix a positive squarefree integer D and assume that t ∈ ℤ_{k}\ℤ, where ℤ_{k} is the ring of integers in the algebraic number field k := ℚ(\sqrt{−D}). Lemma 6. For t ≠ (−1 ± 3\sqrt{−3})/2, f_{t} is irreducible over k.

Proof. Assume first |t| ≥ 6. From (8) and α^{(3)} ∉ ℝ we see that 0 < |α^{(3)}| ≤ 1/2 which implies that α^{(3)} ∉ ℤ_{k}. However, α^{(3)} is an algebraic integer. This yields α^{(3)} ∉ k. From (5) we conclude that α^{(j)} ∉ k for j = 1, …, 3.

For |t| < 6, the assertion has been checked using Pari [2]. □

Let K := ℚ(α). From (5) and t−1 = α^{(1)}+α^{(2)}+α^{(3)} we see that K = ℚ(α^{(1)}, α^{(2)}, α^{(3)}, \sqrt{−D}). For t ≠ (−1 ± 3\sqrt{−3})/2, we have [K : ℚ] = 6 and the minimal polynomial of α over ℚ is f_{t}\overline{f_{t}}. Theorem 7. Let t ∈ ℤ_{k} \ ℤ, L := ℚ(α, \overline{α}) and let G := Gal(L/ℚ). Let τ denote the complex conjugation.

(1) If t ∈ T_{1} := \{(−1 ± 3\sqrt{−3})/2\}, then L = ℚ(\sqrt{−3}) and G = ⟨τ⟩ ≃ C_{2}.

(2) If t ∈ T_{2} := \{±\sqrt{−7}, −1 ± \sqrt{−7}\}, then L = K = ℚ(α) and G = ⟨σ⟩ ≃ C_{6} with σ = (α^{(1)}\overline{α^{(3)}}α^{(2)}\overline{α^{(1)}}α^{(3)}\overline{α^{(2)}}).

(3) If

Re t = −1/2 and t ∉ T_{1}

or if

t ∈ T_{3} := \{(−5 ± \sqrt{−11})/2, (3 ± \sqrt{−11})/2, (−13 ± \sqrt{−23})/2, (−3 ± \sqrt{−23})/2,

(1 ± \sqrt{−23})/2, (11 ± \sqrt{−23})/2, (−7 ± \sqrt{−31})/2, (5 ± \sqrt{−31})/2\}, then L = K = ℚ(α) and G = ⟨σ, τ⟩ ≃ S_{3} with σ = (α^{(1)}α^{(2)}α^{(3)})(\overline{α^{(1)}}\overline{α^{(3)}}\overline{α^{(2)}}). There are four fields between ℚ and K, namely the quadratic field k = ℚ(\sqrt{−D}) and three cubic fields. One of the cubic fields is real.

(4) If Re t ≠ −1/2 and t ∉ T_{2} ∪ T_{3}, then [L : K] = 3 and G = ⟨σ, σ′, τ⟩ with |G| = 18, σ = (α^{(1)}α^{(2)}α^{(3)}) and σ′ = (\overline{α^{(1)}}\overline{α^{(2)}}\overline{α^{(3)}}). Furthermore, the only field between ℚ and K is k.

Proof. We assume |t| ≥ 21, leaving the remaining cases to Pari [2]. We note that the complex conjugation τ is an element of G.

Assume first that G is cyclic, generated by some σ. Without loss of generality, we may assume that σ^{2} is the automorphism which induces (5). It is clear that τ = σ^{3}, which implies σ = (α^{(1)}\overline{α^{(3)}}α^{(2)}\overline{α^{(1)}}α^{(3)}\overline{α^{(2)}}). Let

u_{1} := (4 − ∏_{j=1}^{3}(α^{(j)} + \overline{α^{(j)}}))^{2} − 2 ∏_{j=1}^{3}((α^{(j)})^{2} + (\overline{α^{(j)}})^{2}).

Since u_{1} is invariant under σ and u_{1} ∈ ℤ_{L}, we have u_{1} ∈ ℤ. Using (7), we get |u_{1}| < 1 for |t| ≥ 21. This implies u_{1} = 0. Eliminating α and \overline{α} from the set of equations u_{1} = 0, f(α) = 0 and \overline{f(α)} = 0, we get D = 0 or Im t = 0, which has been excluded.

Next, we assume that L = K but G is not cyclic. Let σ ∈ G be the automorphism which maps α^{(1)} to α^{(2)}. Since σ(\overline{α^{(1)}}) ≠ \overline{α^{(1)}} because |G| = 6, the only possibility for σ is σ = (α^{(1)}α^{(2)}α^{(3)})(\overline{α^{(1)}}\overline{α^{(3)}}\overline{α^{(2)}}). We note that τσ = σ^{2}τ and get G = ⟨σ, τ⟩. We consider θ_{j} := σ^{j}(α + \overline{α}), j = 1, 2, 3, and

u_{2} := 1 − θ_{1}θ_{2}θ_{3} − θ_{1}θ_{2} − θ_{2}θ_{3} − θ_{3}θ_{1} − t − \overline{t}.

As above, we obtain u_{2} = 0 for |t| ≥ 20. This yields Re t = −1/2.

Conversely, if Re t = −1/2, we get \overline{f_{t}}(−1 − α) = 0 and we obtain the claimed Galois group. Finally, \overline{α} ∉ K yields [L : K] = 18 and the asserted Galois group. The assertions on the fields between ℚ and K follow from Galois Theory. □

For the case Re t = −1/2, we will need the following relation. Lemma 8. For Re t = −1/2, |t| ≥ 6, and α = α^{(1)} as calculated in Lemma 4, we have

\overline{α^{(1)}} = −1 − α^{(1)}.

Proof. Using (6), we get

|\overline{α^{(1)}} − (−1 − α^{(1)})| = |\frac{3 + 10t + 10t^{2}}{(t + t^{2})^{3}} + L_{6} (\frac{10}{|t|^{7/2}} )| ≤ 0.86.

In the proof of Theorem 7, we noticed that −1 − α^{(1)} is a root of \overline{f_{t}}. From (8) we see that |\overline{α^{(i)}} − \overline{α^{(1)}}| > 1 for i = 2, 3. This yields the assertion. □

5. Unit group

Let (x, y) ∈ ℤ_{k} be a solution to (1). From

(10)  ∏_{l=1}^{3}(x − α^{(l)}y) = F_{t}(x, y) = μ

we conclude that β^{(l)} := x − α^{(l)}y, l = 1, 2, 3, are units in 𝔒 := ℤ_{k}[α].

Therefore, we investigate the unit structure of 𝔒. It can easily be checked that

1 = α(α^{2} − (t − 1)α − (t + 2)) = (α + 1)(2 + tα − α^{2}),

which implies that α and (α + 1) are units in 𝔒.

Lemma 9. Let |t| ≥ 20 and

ζ = \begin{cases}(1 + \sqrt{−3})/2 & if D = 3, \\ i & if D = 1, \\ −1 & otherwise. \end{cases}

Then the index I of ⟨ζ, α, α + 1⟩ in the unit group 𝔒^{×} can be bounded by

(11)  I ≤ 5.03 log^{2} |t| .

Proof. Assume that ρ ∈ 𝔒 is a root of unity. Since ℚ(ρ) is Galois and since by Theorem 7, the only Galois subfield between ℚ and K is k, we see that ρ ∈ k. It is clear that the torsion group of ℤ_{k}^{×} (and therefore of 𝔒^{×}) is generated by ζ.

Using (7), we calculate

(12)  0.97 log^{2} |t| ≤ − det (\begin{matrix}log|α^{(1)}| & log|α^{(1)} + 1| \\ log|α^{(2)}| & log|α^{(2)} + 1| \end{matrix}) ≤ 1.03 log^{2} |t| .

This implies that α and α + 1 are independent units of 𝔒.

By Friedman [7, Theorem B], the regulator of K can be bounded by Reg ℤ_{K} ≥ 0.2052. From Pohst and Zassenhaus [16, p. 361], we conclude that

I = [𝔒^{×} : ⟨ζ, α, α + 1⟩] ≤ [ℤ_{K}^{×} : ⟨ζ, α, α + 1⟩] = \frac{Reg ⟨ζ, α, α + 1⟩}{Reg ℤ_{K}} ≤ \frac{1.03 log^{2} |t|}{0.2052} ≤ 5.03 log^{2} |t| .

□

6. Approximation Properties

We call a solution (x, y) to (1) a solution of type j, if

|β^{(j)}| = min_{l=1,2,3}|β^{(l)}|.

From

|y||α^{(l)} − α^{(j)}| ≤ |x − α^{(l)}y| + |x − α^{(j)}y| ≤ 2|β^{(l)}|,

and (10) we conclude that

(13)  |β^{(j)}| = \frac{1}{∏_{l≠j}|β^{(l)}|} ≤ \frac{4}{∏_{l≠j} |α^{(l)} − α^{(j)}|} · \frac{1}{|y|^{2}} = \frac{4}{|f′_{t}(α^{(j)})|} · \frac{1}{|y|^{2}} .

Lemma 10. Let |t| ≥ 20 and (x, y) be a solution to (1) of type j. Then (−(x+y), x) is a solution to (1) of type j′ := (j + 1) mod 3.

Proof. We may assume |y| ≥ \sqrt{5} by Lemma 5. By (4), (−(x + y), x) is also a solution to (1). From (5) and (13) we see that

|\frac{−(x + y)}{x} − α^{(j′)}| = |\frac{α^{(j)} − x/y}{α^{(j)} · x/y} | ≤ \frac{4}{|y|^{3} · |f′_{t}(α^{(j)})|} · \frac{1}{|α^{(j)}|} · \frac{1}{|α^{(j)}| − 4/(|y|^{3} · |f′_{t}(α^{(j)})|)}.

From (7), the assumption |y| ≥ \sqrt{5}, and (13) we conclude that

min_{l} |\frac{−(x + y)}{x} − α^{(l)}| = |\frac{−(x + y)}{x} − α^{(j′)}| .

□

Lemma 10 shows that if there is a solution to (1) which is not listed in Table 1, then there is also a solution (x, y) of type 1 which is not contained in Table 1.

7. Stable Growth

Let (x, y) be a solution to (1) of type 1 which is not listed in Table 1 and |t| ≥ 20. By Lemma 5, we may assume |y| ≥ \sqrt{5}. For l = 2, 3, we have

(14)  |β^{(l)}| = |y| · |α^{(1)} − α^{(l)}| · |1 + \frac{x/y − α^{(1)}}{α^{(1)} − α^{(l)}} | .

Taking logarithms and using (13) and (7), we get

(15)  log|β^{(l)}| = log |y| + log|α^{(1)} − α^{(l)}| + L_{20} (\frac{c_{l}}{|t|^{3} |y|^{3}} )

for c_{2} = 4.6 and c_{3} = 4.37. Taking this for l = 3 and l = 2 and eliminating log |y|, we get

(16)  |Λ| := |log |\frac{β^{(2)}}{β^{(3)}} | − log |\frac{α^{(1)} − α^{(2)}}{α^{(1)} − α^{(3)}} || ≤ \frac{8.97}{|t|^{3} |y|^{3}} .

From (14) we see that Λ = 0 would imply

\frac{x/y − α^{(1)}}{α^{(1)} − α^{(2)}} = \frac{x/y − α^{(1)}}{α^{(1)} − α^{(3)}} .

This would lead to x/y = α^{(1)} or α^{(3)} = α^{(2)}, which are both contradictions. Therefore we have Λ ≠ 0.

Lemma 9 implies that (β^{(l)})^{I} = ζ^{u_{0}}(α^{(l)})^{u_{1}}(α^{(l)} +1)^{u_{2}} for some integers u_{0}, u_{1}, and u_{2}. Taking logarithms yields

(17)  log|β^{(l)}| = \frac{u_{1}}{I} log|α^{(l)}| + \frac{u_{2}}{I} log|α^{(l)} + 1|,  l = 1, 2, 3.

From (16), we obtain

|Λ| = |\frac{u_{1}}{I} log |\frac{α^{(2)}}{α^{(3)}} | + \frac{u_{2}}{I} log |\frac{α^{(2)} + 1}{α^{(3)} + 1}| − log |\frac{α^{(1)} − α^{(2)}}{α^{(1)} − α^{(3)}} || ≤ \frac{8.97}{|t|^{3} |y|^{3}} .

Taking advantage of (5), this linear form can be rewritten as a linear form in two logarithms:

(18)  |I · Λ| = |v_{1} log |α| + v_{2} log |1 + \frac{1}{α}|| ≤ \frac{8.97 · I}{|t|^{3} |y|^{3}} ,

where

v_{1} = u_{1} − u_{2} and v_{2} = 2u_{1} + u_{2} − I

are integers.

If Re t = −1/2, we have

|1 + \frac{1}{α}|^{2} = 1 + \frac{1}{α} + \frac{1}{\overline{α}} + \frac{1}{α\overline{α}} = 1 + \frac{α + \overline{α} + 1}{α\overline{α}} = 1

by Lemma 8. This implies log |1 + α^{−1}| = 0. Therefore, (18), (11), and (7) yield |v_{1}| < 1 for |t| ≥ 20. This is a contradiction to Λ ≠ 0. Hence, there are no solutions which are not listed in Table 1 for Re t = −1/2 and |t| ≥ 20. This proves Theorem 3 for |t| ≥ 20.

It is now crucial to observe that log |α| ∼ log |t|, whereas log |1 + α^{−1}| = O(|t|^{−1}). This means that v_{2} has to be much bigger than v_{1}.

Assume that v_{1} = 0. Using (7), we get

log |1 + α^{−1}| = \frac{1}{2} (\frac{1}{t} + \frac{1}{\overline{t}}) − \frac{1}{4} (\frac{1}{t^{2}} + \frac{1}{\overline{t}^{2}} ) + L_{20} (\frac{93}{1000 |t|^{2}} ) .

Simple estimates show that this implies

(19)  |log |1 + α^{−1}|| ≥ \frac{0.4}{|t|^{2}}

for Re t ≠ −1/2. Combining this with (18), (11), and the assumption v_{1} = 0 yields

(20)  |v_{2}| ≤ \frac{113 log^{2} |t|}{|y|^{3} |t|}.

For |t| ≥ 351, this bound is smaller than 1. This yields a contradiction to Λ ≠ 0.

Therefore, we get v_{1} ≠ 0. This implies |v_{1}| ≥ 1 and from (18), (19), and (7), we obtain

|v_{2}| ≥ \frac{1}{log |1 + α^{−1}|} (|v_{1} log|α|| − |Λ|) ≥ \begin{matrix}log & |α| \\ log & |1 + α^{−1}| \end{matrix} − \frac{8.97 · I}{0.4 |y|^{3} |t|} ≥ 0.89 |t| log |t| .

We solve the system (17) of linear equations in u_{1}/I and u_{2}/I by Cramer’s rule and using (15). We obtain

−\frac{v_{2}}{I} = (\frac{3}{log |t|} + L_{20} (\frac{2.93}{|t| log |t|})) log |y| + L_{20}(4.3).

Combining this with (20) and (11) yields

(21)  log |y| ≥ log y_{0} := 5.6 · 10^{−2} · |t| − 1.37 · log |t| .

We have proved the property which is called “stable growth” by E. Thomas [19]: Proposition 11. Let |t| ≥ 351 and x, y ∈ ℤ_{ℚ(t)} be a solution to (1) which is not listed in Theorem 1. Then

max\{|x| , |y| , |x + y|\} ≥ \frac{1.057^{|t|}}{|t|^{1.37}} .

8. Lower Bound for |Λ|

We use the following refinement, due to Mignotte [15], of a theorem of Laurent, Mignotte, and Nesterenko [13] on linear forms in two logarithms. For any non-zero algebraic number γ of degree d over ℚ, whose minimal polynomial over ℤ is c ∏_{j=1}^{d}(X − γ^{(j)}), we denote by

h(γ) = \frac{1}{d} (log |c| + ∑_{j=1}^{d} log max(1, |γ^{(j)}|))

its absolute logarithmic height. Lemma 12 (Mignotte [15, Theorem 2]). Consider the linear form

Λ = b_{2} log α_{2} − b_{1} log α_{1},

where b_{1} and b_{2} are positive integers, α_{1} and α_{2} are non-zero algebraic numbers of absolute value at least 1, and log α_{1} and log α_{2} are any values of their logarithms.

Suppose that α_{1} and α_{2} are multiplicatively independent. Put

D = [ℚ(α_{1}, α_{2}) : ℚ]/[ℝ(α_{1}, α_{2}) : ℝ].

Let a_{1}, a_{2}, h, k be real positive numbers, and ρ a real number > 1. Put λ = log ρ, χ = h/λ and suppose that χ ≥ χ_{0} for some number χ_{0} ≥ 0 and that

h ≥ D (log (\frac{b_{1}}{a_{2}} + \frac{b_{2}}{a_{1}} ) + log λ + f(⌈K_{0}⌉)) + 0.023,
a_{i} ≥ max{1, ρ |log α_{i}| − log |α_{i}| + 2Dh(α_{i})}, (i = 1, 2)
a_{1} · a_{2} ≥ λ^{2},

where

f(x) = log \frac{(1 + \sqrt{x − 1})\sqrt{x}}{x − 1} + \frac{log x}{6x(x − 1)} + \frac{3}{2} + log \frac{3}{4} + \frac{log \frac{x}{x−1}}{x − 1}

and

K_{0} = \frac{1}{λ} (\frac{\sqrt{2 + 2χ_{0}}}{3} + \sqrt{\frac{2(1 + χ_{0})}{9} + \frac{2λ}{3} (\frac{1}{a_{1}} + \frac{1}{a_{2}} ) + \frac{4λ\sqrt{2 + χ_{0}}}{3\sqrt{a_{1}a_{2}}}} )^{2} a_{1}a_{2}.

Put

v = 4χ + 4 + 1/χ, m = max{2^{5/2}(1 + χ)^{3/2}, (1 + 2χ)^{5/2}/χ}.

Then we have the lower bound

log |Λ| ≥ − \frac{1}{λ} (\frac{v}{6} + \frac{1}{2}\sqrt{\frac{v^{2}}{9} + \frac{4λv}{3} (\frac{1}{a_{1}} + \frac{1}{a_{2}} ) + \frac{8λm}{3\sqrt{a_{1}a_{2}}}} )^{2} a_{1}a_{2}
− max\{λ(1.5 + 2χ) + log(((2 + 2χ)^{3/2} + (2 + 2χ)^{2}\sqrt{k^{∗}})A + (2 + 2χ)), D log 2\},

where

A = max\{a_{1}, a_{2}\} and k^{∗} = \frac{1}{λ^{2}} (\frac{1 + 2χ}{3χ})^{2} + \frac{1}{λ} (\frac{2}{3χ} + \frac{2}{3} · \frac{(1 + 2χ)^{1/2}}{χ} ) .

Let σ := sign(|α^{(2)}|−1). We will apply Lemma 12 to the linear form in logarithms σ2IΛ, which can be written as

σ2IΛ = σv_{1} log(α^{(1)}\overline{α^{(1)}}) − (−v_{2}) log(α^{(2)}\overline{α^{(2)}})^{σ}

by (18) and (5). We note that log(α^{(1)}\overline{α^{(1)}}) and log(α^{(2)}\overline{α^{(2)}}) are linearly independent over ℚ by (12).

From now on, we will assume |t| ≥ 10^{9} in order to obtain sharper bounds.

We have

D = 9
h(α^{(1)}\overline{α^{(1)}}) = h(α^{(2)}\overline{α^{(2)}})

≤ \frac{2}{3}(| log|α^{(1)}|| + | log|α^{(2)}||) ≤ \frac{2}{3} log |t| + \frac{1}{3t} + \frac{1}{3\overline{t}} + L_{10^{9}} (4.87 · 10^{−2} · \begin{matrix}log & |t| \\ |t| & 2 \end{matrix}) .

We choose ρ = 30.0658 and obtain

a_{1} = 12 log |t| + 6 (\frac{1}{t} + \frac{1}{\overline{t}}) + \frac{62.76}{|t|} + 0.877 · \begin{matrix}log & |t| \\ |t| & 2 \end{matrix},
a_{2} = 74.14 · log |t| + 6 (\frac{1}{t} + \frac{1}{\overline{t}}) + 7.31 · \begin{matrix}log & |t| \\ |t| & 2 \end{matrix},
0 < σv_{1} ≤ \frac{14.7}{|t|} log |y| + 19.5 · \frac{log |t|}{|t|},
−v_{2} ≤ (\frac{2500}{171} log |t| + 1.46 · \frac{log |t|}{|t|}) log |y| + (19.50 log^{2} |t| + 2.42 · \frac{log^{2} |t|}{|t|}) ,
h = 9 log log |y| + (7.519 + 177.2 · \frac{log |t|}{|t|} + \frac{3.72}{|t|}) ,
K_{0} ≥ 4 862 274.

Since the only available asymptotic information on |y| is (21), we use this bound everywhere but in the asymptotic main term. After some calculations, we finally get

(22)  log |2IΛ| ≥ − log^{2} log |y| · (3073 log^{2} |t| + 7505 log |t| + 941.1\sqrt{log |t|} + 44510).

We consider the function

h(y, t) := 3 log |y| − log^{2} log |y| · (3073 log^{2} |t| + 7505 log |t| + 941.1\sqrt{log |t|} + 44510)

+ 3 log |t| − 2 log log |t| − 4.384.

By (22), (18), and (11), we have

(23)  h(y, t) ≤ 0

for |t| ≥ 10^{9}.

Since

\frac{∂}{∂ log |y|}h(y, t) = 3 − 2(3073 log^{2} |t| + 7505 log |t| + 941.1\sqrt{log |t|} + 44510)\begin{matrix}log & log & |y| \\ log & |y| \end{matrix} > 0 for |y| ≥ y_{0} (y_{0} has been defined in (21)) and |t| ≥ 2.15 · 10^{8}, we have h(y, t) ≥ h(y_{0}, t). But h(y_{0}, t) > 0 for |t| ≥ 3 023 195 238, which is a contradiction to (23). This proves Theorem 1.

9. Proof of Theorem 3

Theorem 3 has already been proved for |t| ≥ 20 in Section 7. For the remaining small values of |t|, the asymptotic expansions do not help. Especially, we have to redefine the order of the roots of f_{t}, since α^{(1)} was defined via an asymptotic expansion.

We easily calculate

f̂_{t_{2}}(x_{2}) := f_{−1/2+it_{2}}(−1/2 + ix_{2}) = −x_{2}^{3} + t_{2}x_{2}^{2} − \frac{9}{4}x_{2} + \frac{t_{2}}{4} .

Since discr f̂_{t_{2}} < 0 for t_{2} ≠ ±3\sqrt{3}/2, we conclude that there is exactly one root of f_{−1/2+it_{2}} with real part equal to −1/2. This root will be denoted by α := α^{(1)}. Since α^{(1)} + \overline{α^{(1)}} = −1, we have

|α^{(1)} + 1| = |α^{(1)}|.

The relations (5) give us the other roots of f_{t}.

The type of a solution is defined as in Section 6. However, we cannot use Lemma 10.

For each 1 ≤ |t| ≤ 20 with Re t = −1/2 and for each pair (j, l) with 1 ≤ l ≠ j ≤ 3, we calculate α explicitly and set

ĉ_{tjl} := |\frac{4}{f′_{t}(α^{(j)})} · \frac{1}{α^{(j)} − α^{(l)}} · \frac{1}{11^{3/2}} | , c_{tjl} := max\{log(1 + ĉ_{tjl}), − log(1 − ĉ_{tjl})\}.

By (13) and (14), we get

log|β^{(l)}| = log |y| + log|α^{(j)} − α^{(l)}| + ε_{tjl}

for some |ε_{tjl}| ≤ c_{tjl} and for all solutions (x, y) to (1) of type j with |y| ≥ \sqrt{11}.

As in Section 7, we obtain

|I · Λ| = |v_{1} log|α^{(j)}| + v_{2} log |1 + \frac{1}{α^{(j)}} || ≤ c_{tj} := \frac{log^{2} |α|}{0.2052} · ∑_{\begin{matrix}l=1 \\ l≠j \end{matrix}}^{3} c_{tjl}.

We note that log|1 + 1/α^{(1)}| = log|α^{(2)}| = 0 and log|α^{(1)}| = − log|1 + 1/α^{(2)}| = − log|α^{(3)}| = log|1 + 1/α^{(3)}| = log|α|. This yields

|w_{j} log |α|| ≤ c_{tj},

where w_{1} = v_{1}, w_{2} = −v_{2}, and w_{3} = v_{2} − v_{1} are integers.

However, it is checked numerically that c_{tj}/log |α| < 1 for all relevant t and j. Therefore, we get w_{j} = 0, which implies Λ = 0. But this is impossible by the argument in Section 7.

Finally, we have to find all solutions with |y| < \sqrt{11}. The procedure described in Lemma 5 exactly yields the solutions listed in Table 2.

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THOMAS’ FAMILY OF THUE EQUATIONS OVER IMAGINAR