Chabauty methods using elliptic curves

Nils Bruin

June 22, 1999

Abstract In this article, we consider algebraic curves over ℚ that cover an elliptic curve over some extension of ℚ. We show how we can use the arithmetic on that elliptic curve to obtain information on the rational points on the cover. We apply this method to curves arising from the diophantine equations x^{2} ± y^{4} = z^{5} and determine all solutions with coprime, integral x, y, z. To do this, we determine the rational points on several curves of genus 5.

1 Introduction

Since 1983 (see [Fal83]), it is known that an algebraic curve of genus g ≥ 2 over a number field has only finitely many rational points. The proof of this theorem does not help in actually determining them, however. A much older, partial, proof by Chabauty (see [Cha41]) does suggest a way of bounding the number of rational points.

The proof applies to curves C of genus g over a number field K such that the rank of the Mordell-Weil group of the Jacobian variety rk(Jac_{C}(K)) is smaller than g. Let p be a finite prime of K. The proof is based on p-adic analytic geometry. It uses that the dimension of the p-adic closure of a finitely generated group of rank r on a p-adic abelian variety has dimension ≤ r. Therefore, the p-adic closure of the Mordell-Weil group is a proper submanifold of Jac_{C}(K_{p}).

If we embed C in Jac_{C} over K (assuming C(K) ≠ ∅), then C(K) will land in Jac_{C}(K). As a consequence, the image of C(K) will be contained in the intersection of the image of C(K_{p}) and the p-adic closure of Jac_{C}(K). The latter is an intersection of a p-adic curve and a manifold of codimension ≥ 1. Chabauty proves that the intersection will be 0-dimensional and, since Jac_{C}(K_{p}) is compact, it is finite.

This intersection can often be determined explicitly and the number of points gives a bound on the size of C(K). Recently, Chabauty-methods have proved quite succesful in giving sharp bounds on C(K). See for instance [FPS97], [Fly97], [CF96], [Wet97] and [Bru97]. These papers concentrate on the application of this argument to the simplest non-trivial case g = 2. Even then, the computations necessary on Jac_{C} are quite involved.

The variety Jac_{C} need not be absolutely simple. A part of it might be isogenous to a product of elliptic curves (over an extension of the base field). In [FW99], it is described how this can be used for curves of genus 2 that cover an elliptic curve over the base field with degree 2. In this article, we describe a general way to take advantage of elliptic factors in Jac_{C}.

Let K be a number field. In our applications, we have K = ℚ. Suppose we have a cover Φ : C → ℙ_{1} over K of genus g > 1 and suppose that there is an elliptic cover φ : E → ℙ_{1} over a finite extension L of K such that Φ factors through φ, as depicted in the following diagram.

C  E ℙ_{1}_{_{Φ} u} _{‚ ‚ ‚ ›}_{π} A A A D_{φ}

If we want to determine C(K), then we can use this factorisation of covers in the following way. Suppose that P ∈ C(K). Then, clearly, π(P) ∈ E(L) and φ(π(P)) ∈ ℙ_{1}(K). Thus, Φ(C(K)) is contained in φ(E(L)) ∩ ℙ_{1}(K). We treat the case where deg(φ) = 2 in detail.

As an example, we prove the following two theorems, that play a role in the ongoing analysis of the generalised Fermat equation x^{r} + y^{s} = z^{t}, x, y, z ∈ ℤ, gcd(x, y, z) = 1 (see [Bru97] and [Bru99]).

Theorem 1.1 If x, y, z ∈ ℤ satisfy x^{2} + y^{4} = z^{5} and gcd(x, y, z) = 1 then xyz = 0.

Theorem 1.2 The only integer, pairwise prime integer solutions to x^{2} − y^{4} = z^{5} are

(x, y, z) ∈ {(±1, 0, 1), (0, ±1, −1), (±122, ±11, 3), (±7, ±3, −2)}.

In order to reduce these theorems to the problem of finding rational points on algebraic curves, we examine primitive solutions to equations of the form F (s, t) = Dy^{m}. This kind of equations was already treated in [DG95], but we reformulate some of their results more explicitly. These lead to the problem of finding rational points on curves of genus 5.

This article contains material from [Bru99].

2 Preliminaries

Let L be a number field with ring of integers O = O_{L}. Let 𝔭 be a finite prime of L. Then we write ν_{𝔭} for the 𝔭-adic logarithmic valuation and char(𝔭) for the rational prime lying below 𝔭. In this article, we consider elliptic curves E over L. In this section we fix some notation. See [Sil86] for a thorough treatment. We work with elliptic curves in twisted Weierstrass form

E : γY^{2} = X^{3} + a_{2}X^{2} + a_{4}X + a_{6},

with γ, a_{2}, a_{4}, a_{6} ∈ O. The projective closure of this curve has a unique point ∞. The curve E is an abelian variety with a group law that is characterised by the fact that three points P_{1}, P_{2}, P_{3} sum to 0 (= ∞) if and only if they are collinear.

Take 𝔭 ∤ 2 to be a finite prime of L. The curve E has good reduction at 𝔭 if the curve E mod 𝔭 over O/𝔭 defined by reducing E coefficient wise mod 𝔭, is again a curve of genus 1. We have a reduction map E(L_{𝔭}) → (E mod 𝔭)(O/𝔭), which induces the short exact sequence of group homomorphisms

0 ⟶ E^{(1)}(L_{𝔭}) ⟶ E(L_{𝔭}) ⟶^{mod 𝔭} (E mod 𝔭)(O/𝔭) ⟶ 0,

where E^{(1)}(L_{𝔭}) is called the kernel of reduction. Consider the function Z := X/Y on E. Then E^{(1)}(L_{𝔭}) = {G ∈ E(L_{𝔭}) : Z(G) ∈ 𝔭O_{𝔭}}. Suppose that 𝔭 is not too highly ramified (say, unramified). As described in [Sil86, Chapter IV], we have a group homomorphism E^{(1)}(L_{𝔭}) → 𝔭O_{𝔭}. This can be expanded as a power series in Z and we have Log_{𝔭}(Z) = Z mod (Z^{2}). Furthermore, by [Sil86, Proposition IV.5.5] we have that Log_{𝔭}(G) mod 𝔭^{2} = Z(G) mod 𝔭^{2} for G ∈ E^{(1)}(L_{𝔭}).

We will use

Lemma 2.1 Let E be an elliptic curve over a number field L and let p ∈ ℤ be a prime > 2, unramified in O_{L}/ℤ, such that E has good reduction at 𝔭_{i} | p and #(E mod 𝔭_{i})(O/𝔭_{i}) is prime to p for i = 1, …, m. Let B_{1}, …, B_{r} ∈ E(L) with B_{j} = 0 mod 𝔭_{i} such that ⟨B_{1}, …, B_{r}⟩ in E(L) is of finite index divisible by p, then there are n_{1}, …, n_{r} ∈ ℤ_{p} with (n_{1}, …, n_{r}) ≠ (0, …, 0) mod p such that n_{1}Z(B_{1}) + ⋯ + n_{r}Z(B_{r}) = 0 mod 𝔭_{i}^{2} for i = 1, …, m.

Proof: The conditions in the lemma imply that there exists a G ∈ E(L) and n_{1}, …, n_{r} ∈ ℤ, not all divisible by p, such that n_{1}B_{1} + ⋯ + n_{r}B_{r} = pG. Let i ∈ {1, …, m}. Note that pG ∈ E(L) ∩ E^{(1)}(L_{𝔭_{i}}). Since the reduction group has order prime to p, we have that G = 0 mod 𝔭_{i}. By the good reduction properties, we have n_{1}Log_{𝔭_{i}}(B_{1}) + ⋯ + n_{r}Log_{𝔭_{i}}(B_{r}) = pLog_{𝔭_{i}}(G). The statement follows by observing that Z = Log_{𝔭_{i}} mod 𝔭_{i}^{2} and that Z(E^{(1)}(L_{𝔭_{i}})) = 0 mod 𝔭_{i}. □

Furthermore, we apply

Lemma 2.2 Let K be a number field, let F, G ∈ O_{K}[X, Y] be coprime homogeneous polynomials, m ∈ ℤ_{>0} and D ∈ O_{K}. Suppose that S is a set of primes such that res(F (X, 1), G(X, 1)), D ∈ O_{S}^{∗}. If x, y, z ∈ K with (x, y, z) S-primitive such that

F (x, y)G(x, y) = Dz^{m},

then there are z_{1}, z_{2} ∈ K, with (z_{1}, z_{2}) S-primitive and δ_{1}, δ_{2} ∈ K(S, m) with δ_{1}δ_{2}/D ∈ (K^{∗})^{m} such that

\begin{aligned}F (x, y) & = & δ_{1}z_{1}^{m}, \\ G(x, y) & = & δ_{2}z_{2}^{m}, \\ z & = & \sqrt[m]{\frac{δ_{1}δ_{2}}{D}} z_{1}z_{2}. \end{aligned}

Proof: Let 𝔭 be a prime of K outside S. Note that since F and G have integral coefficients and D ∈ O_{𝔭}^{∗}, we have that mν_{𝔭}(z) = ν_{𝔭}(F (x, y)G(x, y)/D) ≥ min(ν_{𝔭}(x), ν_{𝔭}(y)). Therefore (x, y) is S-primitive as well. So, x mod 𝔭 ≠ 0 or y mod 𝔭 ≠ 0. Assume the latter. Since res(F (X, 1), G(X, 1)) ∈ O_{𝔭}^{∗} we have that F (X, 1) mod 𝔭 and G(X, 1) mod 𝔭 cannot both be zero. It follows that ν_{𝔭}(F (x/y, 1)) = 0 or ν_{𝔭}(G(x/y, 1)) = 0. By homogeneity, we have that ν_{𝔭}(F (x, y)) = ν_{𝔭}(F (x/y, 1)) + deg(F)ν_{𝔭}(y) = ν_{𝔭}(F (x/y, 1)) and the same for G. Since

mν_{𝔭}(z) = ν_{𝔭}(Dz^{m}) = ν_{𝔭}(F (x, y)G(x, y)) = ν_{𝔭}(F (x, y)) + ν_{𝔭}(G(x, y)),

we see that ν_{𝔭}(F (x, y)), ν_{𝔭}(G(x, y)) ∈ mℤ for all 𝔭 ∤ S. The case x mod 𝔭 ≠ 0 follows by symmetry and the fact that res(F (X, 1), G(X, 1)) = ±res(F (1, X), G(1, Y)). □

3 Elliptic covers of degree 2

Let E be an elliptic curve over a number field L. In this section we determine what degree 2 covers φ : E → ℙ_{1} look like. Suppose that E is a twisted Weierstrass model over the ring of integers O_{L} of a number field L.

E : γY^{2}D = X^{3} + a_{2}X^{2}D + a_{4}XD^{2} + a_{6}D^{3}

Suppose that φ is a degree 2 cover E → ℙ_{1} over L. Then we can choose a model (φ_{1}(X, Y, D) : φ_{2}(X, Y, D)), with φ_{1}, φ_{2} ∈ O_{L}[X, Y, D] homogeneous polynomials of equal degree. By choosing affine coordinates on ℙ_{1}, we write φ = φ_{1}/φ_{2}. Since deg(φ) = 2, there are at most two points G_{1}, G_{2} ∈ E(L̄) such that φ(G_{1}) = φ(G_{2}) = 0. These two points determine the intersection of φ_{1}(X, Y, D) = 0 with E in ℙ_{2}. If G_{1} = G_{2}, then φ_{1}(X, Y, D) = 0 should be tangent to E in G_{1}. Along the same lines, there are two points G_{3}, G_{4} with φ(G_{3}) = φ(G_{4}) = ∞. Up to scalar multiplication, φ is determined by the lines through G_{1} and G_{2} and through G_{3} and G_{4}. We can assume φ_{1} = c_{11}X + c_{12}Y + c_{13}D and φ_{2} = c_{21}X + c_{22}Y + c_{23}D, with c_{ij} ∈ O_{L}. Note that φ_{1}(X, Y, D) = 0 has 3 points of intersection with E and so has φ_{2}(X, Y, D) = 0. For φ to have degree 2, we must have that the unique point G_{φ} with φ_{1}(G_{φ}) = φ_{2}(G_{φ}) = 0 lies on E. If we define

\begin{matrix}G_{φ,1} = & c_{12}c_{23} − c_{13}c_{22} \\ G_{φ,2} = & c_{13}c_{21} − c_{11}c_{23} \\ G_{φ,3} = & c_{11}c_{22} − c_{12}c_{21} \end{matrix}

then we have G_{φ} = (G_{φ,1} : G_{φ,2} : G_{φ,3}). The map τ = τ_{φ} : E → E that interchanges the elements of the fibers of φ is an involution, i.e. τ ∈ Aut(E) (where E is the algebraic curve corresponding to E) and τ ∘ τ = id. From [Sil86, Corollary III.10.2] we know that Aut(E)[2] = {[1], [−1]} and from [Sil86, Example III.4.7] that an automorphism of E as a curve is the composition of an automorphism of E as an elliptic curve with a translation. Translations over 2-torsion points are involutions, but they give unramified covers. Thus, there is a G_{τ} ∈ E(L̄) such that τ(G) = G_{τ} − G. Note that G_{2} = τ(G_{1}) = G_{τ} − G_{1}. Therefore G_{1}, G_{2} and −G_{τ} are collinear. Note however that G_{φ} is collinear with G_{1} and G_{2} as well. It follows that G_{τ} = −G_{φ} and thus that τ is defined over L. We will either assume that G_{τ} ≠ G_{φ} or that G_{τ} = G_{φ} = ∞. If G_{φ} = G_{τ}, then we can take G_{φ} = ∞ by choosing the distinguished point on the algebraic curve corresponding to E.

We now derive some expressions that allow us to calculate 𝔭-adic approximations to φ. Let 𝔭 be a prime of O_{L}. We call 𝔭 a good prime with respect to φ : E → ℙ_{1} if

• E has good reduction at 𝔭

• φ_{1} mod 𝔭 and φ_{2} mod 𝔭 have degree 1 and are linearly independent

• if G_{φ} ≠ −G_{φ}, then G_{φ} mod 𝔭 ≠ −G_{φ} mod 𝔭

• ν_{𝔭}(char(𝔭)) < char(𝔭) − 1.

Suppose that 𝔭 is such a prime. Then Exp_{𝔭} : 𝔭O_{𝔭} → E^{(1)}(L_{𝔭}) is a group isomorphism with the property that Z(Exp_{𝔭}(z)) = z mod 𝔭^{2}. Let G ∈ E(L_{𝔭}) with G mod 𝔭 ≠ G_{τ} mod 𝔭. Then, by choosing coordinates on ℙ_{1} (i.e., by interchanging φ_{1} and φ_{2} if necessary), we can assume that φ(G) mod 𝔭 ≠ ∞. Then φ(G + Exp_{𝔭}(z)) is a power series with coefficients in L and convergent on 𝔭O_{𝔭} with values in O_{𝔭}. We derive some approximations to these power series. Suppose that z ∈ 𝔭O_{𝔭}. If G = ∞ and G_{τ} ≠ G_{φ} then

φ(Exp_{𝔭}(z)) = φ(∞) + \frac{G_{φ,3}}{c_{22}^{2}} z mod 𝔭^{2}.

Put F′(x) = 3x^{2} + 2a_{2}x + a_{4}. If G = (x, y) and G mod 𝔭 ≠ ∞, then

φ((x, y) + Exp_{𝔭}(z)) = φ(x, y) + \frac{F′(x)(xG_{φ,3} − G_{φ,1}) − 2γy(yG_{φ,3} − G_{φ,2})}{γ(c_{21}x + c_{22}y + c_{23})^{2}}z mod 𝔭^{2}.

Now suppose that G_{τ} = G_{φ} = ∞. Then φ(X : Y : D) = (c_{11}X + c_{13}D)/(c_{21}X + c_{23}D) and τ = [−1]. Consequently, φ(Exp_{𝔭}(Z)) and φ((x, 0) + Exp_{𝔭}(Z)) will be power series in Z^{2}. Using higher order terms, we derive

φ(Exp_{𝔭}(z)) = \frac{c_{11}}{c_{21}} + \frac{G_{φ,2}}{γc_{21}} z^{2} mod 𝔭^{3},
φ((x, 0) + Exp_{𝔭}(z)) = φ(x, 0) − \frac{F′(x)G_{φ,2}}{γ(c_{21}x + c_{23})^{2}} z^{2} mod 𝔭^{3}.

4 Rationality Restrictions on Elliptic Covers

Let ℚ ⊂ K ⊂ L be number fields and let φ : E → ℙ_{1} be an elliptic cover defined over L. In this section we propose a method for determining the L-rational points G on E such that φ(G) is K-rational. Note that, although φ is just defined over L, the answer to this question requires ℙ_{1} to be viewed as a curve over K and not over L. The method we explain here might give a sharp bound on the number of such G if rk(E(L)) < [L : K]. We remind the reader that not much is lost by the assumption K = ℚ.

By the Mordell-Weil theorem ([Sil86, VIII]), E(L) is a finitely generated abelian group. Suppose that E(L) = ⟨G_{1}, …, G_{r}, G_{r+1}, …, G_{r+t}⟩, where ⟨G_{1}, …, G_{r}⟩ ≃ ℤ^{r} and ⟨G_{r+1}, …, G_{r+t}⟩ is finite.

We choose a prime p of O_{K} such that all 𝔭 | p of O_{L} are unramified in O_{L}/O_{K}, ν_{𝔭}(char(𝔭)) < char(𝔭) − 1, E has good reduction at 𝔭 and φ mod 𝔭 : (E mod 𝔭) → ℙ_{1} is again a cover.

Choose B_{1}, …, B_{r} ⊂ E(L) such that

⟨B_{1}, …, B_{r}⟩ = ⋂_{𝔭|p}(E^{(1)}(L_{𝔭}) ∩ E(L)) .

Since E(L)/⟨B_{1}, …, B_{r}⟩ is finite, we need only finitely many G_{0} ∈ E(L) to cover E(L) with translates G_{0} + ⟨B_{1}, …, B_{r}⟩.

We fix G_{0} and try to determine how many points G of the form G = G_{0} + n_{1}B_{1} + ⋯ + n_{r}B_{r} exist such that φ(G) ∈ ℙ_{1}(K). Note that φ(G) is K-rational if and only if 1/φ(G) is. If 𝔭, 𝔮 | p such that φ(G_{0}) mod 𝔭 = ∞ and φ(G_{0}) mod 𝔮 ≠ ∞, then there is no G = G_{0} + n_{1}B_{1} + ⋯ n_{r}B_{r} with φ(G) ∈ ℙ_{1}(K), as this would imply φ(G_{0}) mod 𝔭 = φ(G_{0}) mod 𝔮. Therefore, by changing from φ to 1/φ if necessary, which corresponds to a K-rational coordinate transformation on ℙ_{1}, we can assume that φ(G_{0}) mod 𝔭 ≠ ∞ for any 𝔭 | p. Since B_{1}, …, B_{r} ∈ E^{(1)}(L_{𝔭}) for all 𝔭 | p, we have

n_{1}B_{1} + ⋯ + n_{r}B_{r} = Exp_{𝔭}(n_{1}Log_{𝔭}(B_{1}) + ⋯ + n_{r}Log_{𝔭}(B_{r})).

Consequently, we can write

θ_{𝔭}^{G_{0}}(n_{1}, …, n_{r}) = φ (G_{0} + Exp_{𝔭} (∑ n_{i}Log_{𝔭}(B_{i}))) ∈ L[[n_{1}, …, n_{r}]],

which is convergent for (n_{1}, …, n_{r}) ∈ (O_{𝔭})^{r} and has values in O_{𝔭}. If φ(G_{0} + ∑ n_{i}B_{i}) ∈ ℙ_{1}(K), then, identifying ℙ_{1}(L) \ {∞} with L, we have θ_{𝔭}^{G_{0}}(n_{1}, …, n_{r}) ∈ O_{p} ⊂ O_{𝔭}. If 𝔮 | p as well then θ_{𝔭}^{G_{0}}(n_{1}, …, n_{r}) = θ_{𝔮}^{G_{0}}(n_{1}, …, n_{r}). These requirements can be expressed in power series over K in the following way. Let I = [L_{𝔭} : K_{p}] and let 1, α, …, α^{I−1} be an O_{p}-basis of O_{𝔭}. Then there are unique θ_{𝔭,i}^{G_{0}} ∈ K_{p}[[n_{1}, …, n_{r}]] such that

θ_{𝔭}^{G_{0}} = θ_{𝔭,0}^{G_{0}} + αθ_{𝔭,1}^{G_{0}} + ⋯ + α^{I−1}θ_{𝔭,I−1}^{G_{0}}.

The statement φ(G_{0} + ∑ n_{i}B_{i}) ∈ ℙ_{1}(K) translates into θ_{𝔭,i}^{G_{0}} and θ_{𝔭,0}^{G_{0}} − θ_{𝔮,0}^{G_{0}} having a simultaneous zero in (n_{1}, …, n_{r}) for all 𝔭, 𝔮 | p and i ≥ 1. Taking all these conditions together, this corresponds to some θ^{G_{0}} ∈ K_{p}[[n_{1}, …, n_{r}]]^{[L:K]−1} vanishing in (n_{1}, …, n_{r}). If p splits completely (i.e. L_{𝔭} = K_{p} for all 𝔭 | p) then it is particularly easy to this power series. Suppose that 𝔭_{1}, …, 𝔭_{m} | p. Then

θ^{G_{0}}(n_{1}, …, n_{r}) = ( θ_{𝔭_{2}}^{G_{0}}(n_{1}, …, n_{r}) − θ_{𝔭_{1}}^{G_{0}}(n_{1}, …, n_{r}) ; ⋮ ; θ_{𝔭_{m}}^{G_{0}}(n_{1}, …, n_{r}) − θ_{𝔭_{1}}^{G_{0}}(n_{1}, …, n_{r})).

It is often possible to give a bound on the number of zeros that such a power series has if r < m. The following lemma is an example of the kind of arguments that might apply.

Lemma 4.1 Let O_{𝔭} be a complete local ring with maximal ideal 𝔭 and

f = (f_{1}, …, f_{m}) ∈ (O_{𝔭}[[X_{1}, …, X_{r}]])^{m},

convergent on O_{𝔭}^{r}. Write X = (X_{1}, …, X_{r}). If one of the following conditions holds,

• f(X_{1}, …, X_{r}) = b + A X mod 𝔭, where A is an m × r matrix over O_{𝔭} such that A mod 𝔭 has rank r

• f_{i}(0, …, 0) = 0, \frac{∂f_{i}}{∂X_{j}} (0, …, 0) = 0 and f_{i}(X_{1}, …, X_{r}) = X^{t} A_{i} X mod 𝔭 for all i, j, where the A_{1}, …, A_{m} are symmetric r × r matrices such that the projective variety in ℙ_{r−1} described by {X^{t}(A_{i} mod 𝔭)X = 0}_{i=1,... ,m} has no points over O_{𝔭}/𝔭,

then f has at most one zero in O_{𝔭}^{r}.

Proof: Let u be a uniformiser of O_{𝔭}. Consider the first case. Note that g(X) = f(X) − b − A X ∈ (𝔭O_{𝔭}[[X_{1}, …, X_{r}]])^{m}. If x ∈ (O_{𝔭})^{r} and f(x) = 0 then we have that A x = −b mod 𝔭. By assumption, there is at most one such x mod 𝔭. It remains to show that f cannot have two zeros reducing to the same vector mod 𝔭. Suppose that there is an e ≥ 1 and x, y ∈ (O_{𝔭})^{r}, y ≠ 0 mod 𝔭, such that f(x) = f(x + u^{e}y) = 0. Subtraction yields 0 = −u^{e}A y + g(x) − g(x + u^{e}y). By the assumption on the rank of A and y ≠ 0 mod 𝔭, it follows that −u^{e}A y ≠ 0 mod 𝔭^{e+1}, but since all coefficients of g are divisible by 𝔭, we have that g(x) = g(x + u^{e}y) mod 𝔭^{e+1}. It follows that such y, e cannot exist.

For the second case, suppose that there is a y ∈ O_{𝔭}^{r} with y mod 𝔭 ≠ 0 and an e ≥ 0 such that f(u^{e}y) = 0. Then 0 = f_{i}(y) = u^{2e}y^{t}A_{i}y mod 𝔭^{2e+1}. It follows that y reduces to a point on {X^{t}(A_{i} mod 𝔭)X = 0}_{i=1,... ,m}. □

We apply these ideas to the case where deg(φ) = 2. We adopt the notation from Section 3 and we assume that the 𝔭 | p are good with respect to φ : E → ℙ_{1}. If we stay away from G_{φ} mod 𝔭 then the formulas given there lead to

θ_{𝔭}^{∞} = φ(∞) + \frac{G_{φ,3}}{c_{22}^{2}} ∑_{i=1}^{r} n_{i}Z(B_{i}) mod 𝔭^{2},

θ_{𝔭}^{(x,y)} = φ(x, y) + \frac{F′(x)(xG_{φ,3} − G_{φ,1}) − 2γy(yG_{φ,3} − G_{φ,2})}{γ(c_{21}x + c_{22}y + c_{23})^{2}} ∑_{i=1}^{r} n_{i}Z(B_{i}) mod 𝔭^{2},

which enables us to compute θ^{∞} mod p^{2} and θ^{(x,y)} mod p^{2}. For the case G_{φ} = G_{τ} = ∞ we have

θ_{𝔭}^{∞} = \frac{c_{11}}{c_{21}} + \frac{G_{φ,2}}{γc_{21}} ∑_{i,j=1}^{r} n_{i}n_{j}Z(B_{i})Z(B_{j}) mod 𝔭^{3},

θ_{𝔭}^{(x,0)} = φ(x, 0) − \frac{F′(x)G_{φ,2}}{γ(c_{21}x + c_{23})^{2}} ∑_{i,j=1}^{r} n_{i}n_{j}Z(B_{i})Z(B_{j}) mod 𝔭^{3}.

5 PARAMETRISATION OF F (X, Y) = DZ^{M} 8

which enable us to compute θ^{∞} mod p^{3} and θ^{(x,0)} mod p^{3} in these cases. Note that the fact that φ is even in this case, guarantees that only monomials of even degree occur in θ^{∞} and θ^{(x,0)}. Furthermore, since ν_{𝔭}(Z(B_{i})) ≥ 1, we only need Z(B_{i}) mod 𝔭^{2} to compute any of these approximations. We summarise this information in

Z(B)/u_{p} = (\begin{matrix}Z(B_{1})/u_{p} mod 𝔭_{1} & ⋯ & Z(B_{r})/u_{p} mod 𝔭_{1} \\ ⋮ & ... & ⋮ \\ Z(B_{1})/u_{p} mod 𝔭_{m} & ⋯ & Z(B_{r})/u_{p} mod 𝔭_{m} \end{matrix} )

where u_{p} is some fixed uniformiser for p in K. (Since the 𝔭_{i} are unramified over p, u_{p} is also a uniformiser for 𝔭_{i} in L).

For simplicity, we assumed that we have generators of E(L). Note that E^{(1)}(L_{𝔭}) is isomorphic to 𝔭O_{𝔭} and as such has an O_{𝔭}-module structure. In particular, it is an O_{p}-module. In fact, instead of generators of E(L), we only need E(L) mod 𝔭 and a set {B_{1}, …, B_{r}} ∈ E(L) that generates an O_{p}-module in E^{(1)}(L_{𝔭}) containing ⋂(E^{(1)}(L_{𝔭})∩E(L)). This means that we only have to prove that

char(p) ∤ [⋂_{𝔭|p}(E^{(1)}(L_{𝔭}) ∩ E(L)) : ⟨B_{1}, …, B_{r}⟩] ,

which is much easier to establish. For instance, This follows from Lemma 2.1 if all L_{𝔭} = K_{p} and Z(B)/u_{p} mod p has rank r. If E(L) mod 𝔭 = ⟨G_{1}, …, G_{r+t}⟩, then representatives of E(L) mod 𝔭 are easily generated.

We assumed that we used the information at all primes of L above p. The argument might already work if we just use the information at 𝔭_{1}, …, 𝔭_{m} | p with ∑_{i=1}^{m}[L_{𝔭} : K_{p}] > r. Then, it is sufficient to take B_{1}, …, B_{r} to generate a subgroup of ⋂_{i=1}^{m}(E^{(1)}(L_{𝔭_{i}}) ∩ E(L)) of index prime to char(p). Thus, we can bound the number of G ∈ E(L) with K-rational image under φ in a p-adic neighbourhood by bounding the number of zeros of θ^{G_{0}}. This can be done using Lemma 4.1. We have the following.

Lemma 4.2 Let K, L, p, 𝔭_{1}, …, 𝔭_{m} and φ : E → ℙ_{1} be defined as above. Let G ∈ E(L).

• If θ^{G} mod p ≠ 0, then φ(G) mod p is not hit by φ(E(L)) ∩ ℙ_{1}(K).

• If φ(G) ∈ ℙ_{1}(L) and θ^{G}/p satisfies the first condition in Lemma 4.1, then G and τ_{φ}(G) are the only G′ ∈ E(L) such that φ(G′) ∈ ℙ_{1}(K) and φ(G) mod p = φ(G′) mod p.

• If G = G_{φ} = G_{τ} = (0 : 1 : 0) and θ^{G}/p^{2} satisfies the second condition in Lemma 4.1, then G is the only G′ ∈ E(L) such that φ(G′) ∈ ℙ_{1}(K) and φ(G) mod p = φ(G′) mod p.

5 Parametrisation of F (x, y) = Dz^{m}

In this section we show how Lemma 2.2 leads to an effective and practical way to find the parametrising curves for the S-primitive solutions of a weighted homogeneous equation of the form F (x, y) = Dz^{n} (where F is a square free homogeneous polynomial of degree n ≥ 2) over a number field K. First we investigate the underlying geometry of the parametrising curves.

Let K be a number field and let F (X, Y) ∈ O_{K}[X, Y] and D ∈ O_{K}. Let S be a set of primes such that disc(F (X, 1)), disc(F (1, Y)), D ∈ O_{S}^{∗}. The presence of D allows us to assume that F is monic in X. Let L be a splitting field of F (X, 1) over K. We have α_{1}, …, α_{n} ∈ L such that F (X, Y) = ∏_{i=1}^{n}(X − α_{i}Y). Note that σ ∈ Gal(L/K) acts as a permutation on the α_{i} and use this to fix Gal(L/K) ↪ S_{n}. We write ^{σ}α_{i} = α_{σ(i)}. Suppose that x, y, z is an S-primitive solution in K. Lemma 2.2 gives that we have δ_{1}, …, δ_{n} ∈ L(S, m) with (δ_{1} ⋯ δ_{n})/D ∈ (K^{∗})^{m} and S-primitive (z_{1}, …, z_{n}) ∈ L^{n} such that

\begin{aligned}x − α_{i}y & = & δ_{i}z_{i}^{m} \\ z & = & \sqrt[m]{\frac{δ_{1}···δ_{n}}{D}} z_{1} ⋯ z_{n}. \end{aligned}

Since x, y ∈ K, we can assume, without loss of generality, that ^{σ}δ_{i} = δ_{σ(i)} and ^{σ}z_{i} = z_{σ(i)} for σ ∈ Gal(L/K). If F is irreducible over K then Gal(L/K) acts transitively on the α_{i}, so then δ_{1} determines all δ_{i}. See Lemma 5.2 for details.

If we eliminate x and y from these equations, then we see that (z_{1}, …, z_{n}) must be a zero of the ideal

I_{δ} := {(α_{i} − α_{j})(δ_{k}Z_{k}^{m} − δ_{l}Z_{l}^{m}) − (α_{k} − α_{l})(δ_{i}Z_{i}^{m} − δ_{j}Z_{j}^{m})}_{i,j,k,l}

such that its image under Φ_{δ} : (Z_{1}, …, Z_{n}) ↦ \frac{α_{j}δ_{i}Z_{i}^{m}−α_{i}δ_{j}Z_{j}^{m}}{δ_{i}Z_{i}^{m}−δ_{j}Z_{j}^{m}} is K-rational (corresponding to x/y), where the definition of Φ_{δ} is independent of the actual choice of i, j because of the relations generating I_{δ}. Also note that the zero-locus of I_{δ} does not intersect any Z_{i} = Z_{j} = 0, since the α_{i} are distinct.

We claim that the model C_{δ} described by I_{δ} is a smooth projective model of a curve over L in the ℙ_{n−1}. For n = 2 we have nothing to prove, since I_{δ} = 0, so C_{δ} = ℙ_{1}, which is smooth. Otherwise, we have that, away from Z_{i} = Z_{k} = 0,

d(\frac{Z_{i}}{Z_{k}} ) = \frac{(α_{k} − α_{i})δ_{j}}{(α_{k} − α_{j})δ_{i}} \frac{mZ_{j}^{m−1}}{mZ_{i}^{m−1}} d(\frac{Z_{j}}{Z_{k}} ) ,

so Z_{j}/Z_{k} can be used as a uniformiser there.

Let ζ be a primitive m-th root of unity. The variety C_{δ} has several automorphisms. Consider τ_{i} : ℙ_{n−1} ↦ ℙ_{n−1} defined by Z_{i} ↦ ζZ_{i}. Note that τ_{n} = (τ_{1} ∘ ⋯ ∘ τ_{n−1})^{−1}. It is straightforward to check that Φ_{δ} : C_{δ} → ℙ_{1} is finite of degree m^{n−1} and galois with galois group ⟨τ_{1}, …, τ_{n}⟩.

To conclude that C_{δ} is actually geometrically irreducible, we consider the following argument. Since C_{δ} is smooth, it is a disjoint union of components. Each τ_{i} has a fixed point, so the component containing that point is mapped to itself by τ_{i}. Since C_{δ} is a galois cover of the (connected) projective line, we have that the abelian galois group ⟨τ_{1}, …, τ_{n}⟩ acts transitively on the set of components of C_{δ}. Consequently, the τ_{i} act as the trivial permutation on the components, so there is only one.

Since ^{σ}I_{δ} = I_{δ} and ^{σ}Φ_{δ} = φ_{δ} for σ ∈ Gal(L/K), we see that both are defined over K, so Φ_{δ} : C_{δ} → ℙ_{1} is in fact a model of a cover over K. Furthermore, C_{δ} has good reduction at primes outside S ∪ {p | m}.

We can now calculate the genus of C_{δ} using Riemann-Hurwitz (see [Sil86, Theorem II.5.9]. Note that #Φ_{δ}^{−1}({a}) = m^{n−1} if a ∉ {α_{1}, …, α_{n}} and #Φ_{δ}^{−1}({α_{i}}) = m^{n−2}. As a consequence, ∑_{P ∈C(K̄)}(e_{P} (Φ_{δ}) − 1) = nm^{n−2}(m − 1). Since genus(ℙ_{1}) = 0 we get

genus(C_{δ}) = 1 + m^{n−2} (\frac{1}{2}n(m − 1) − m) .

Now suppose that x, y, z is an S-primitive K-rational solution to F (x, y) = Dz^{m} and that a = x/y (if y = 0, then a is the point ∞ ∈ ℙ_{1}(K)). Suppose P ∈ Φ_{δ}^{−1}({a}). If σ ∈ Gal(K̄/K) then ^{σ}Φ_{δ}(P) = ^{σ}a = a. Since ^{σ}Φ_{δ} = Φ_{δ}, it follows that there is a τ_{σ} ∈ Gal(C/ℙ_{1}) such that ^{σ}P = τ_{σ}(P). It is easy to check that ξ_{P} : σ ↦ τ_{σ} is a cocycle. By [Sil86, Theorem X.2.2] there is a curve C_{P} over K and a degree 1 cover ψ : C_{P} → C_{δ} (not necessarily over K) such that ξ_{P} = (σ ↦ ^{σ}ψψ^{−1}). Since

σ(ψ^{−1}(P)) = ^{σ}ψ^{−1}(τ_{σ}(P)) = ^{σ}ψ^{−1}^{σ}ψψ^{−1}(P) = ψ^{−1}(P),

we see that ψ^{−1}(P) ∈ C_{P} (K). Furthermore, since Φ_{δ} is τ-invariant, Φ_{P} := Φ_{δ} ∘ ψ^{−1} : C_{P} → ℙ_{1} is a cover over K and a ∈ Φ_{P} (C_{P} (K)). We see that the C_{P} form a parametrising set of curves for the S-primitive solutions. (Note that the C_{δ} themselves are twists of each other). To see that we only need a finite number of C_{P}, we need that C_{P} has again good reduction outside S and that the number of twists with this property is finite. This follows from [Sil86, Lemma X.4.3]. Alternatively, finiteness follows from Lemma 2.2 together with Lemma 5.2. Summarising:

Theorem 5.1 Let K, F (x, y) = Dz^{m} and S be as above. Then there is a finite number of galois-covers Φ_{P} : C_{P} → ℙ_{1} over K with Gal(C_{P} /ℙ_{1}) ≅ (ℤ/mℤ)^{n−1}, where C_{P} is of genus 1 + \frac{1}{2}m^{n−2}(n(m − 1) − m) and has good reduction outside S ∪ \{m\} such that

⋃_{C_{P}} Φ_{P} (C_{P} (K)) = {(x : y) : F (x, y) = Dz^{m} with x, y, z ∈ K and (x, y, z) S-primitive}.

The C_{P} are all birationally equivalent over K̄ and the Φ_{P} are ramified exactly above the points (x : y) for which F (x, y) = 0.

While the model C_{δ} is well suited to analyse the underlying geometry of the problem, it is not very useful for explicitly determining a set of curves. This is partly because the model itself is a priori given over L and that we conclude that C_{δ} is defined over K by Galois invariance. This means that even the question P ∈ C_{δ}(K) is not easily answered. Furthermore, while by the correspondence proved in [Sil86, Theorem X.2.2], determining the appropriate twists is effective, it is not a very practical procedure. We can do better.

First, note that if F = F_{1}F_{2} with F_{1}, F_{2} ∈ O_{K}[X, Y] then we can apply Lemma 2.2 to obtain a finite number of systems of equations over O_{K} of the form

\{ \begin{matrix}F_{1}(x, y) & = & δz_{1}^{m} \\ F_{2}(x, y) & = & Dδ^{m−1}z_{2}^{m}. \end{matrix}

6 SUBCOVERS FOR F (X, Y) = DZ^{2} 11

Therefore, it is enough to deal with the case that F is irreducible over K. Let α be a root of F (X, 1) and let L = K(α). Then, applying Lemma 2.2 over L we see that for an S-primitive solution x, y, z there exists a δ ∈ L(S, m) and a_{0}, …, a_{n−1} ∈ K such that

\begin{aligned}x − αy & = & δ (a_{0} + a_{1}α + ⋯ + a_{n−1}α^{n−1})^{m} \\ z & = & \sqrt[m]{\frac{N_{L/K}(δ)}{D}}N_{L/K} (∑_{i=0}^{n−1} a_{i}α^{i}) . \end{aligned}

We have unique forms b_{δ,i} ∈ K[X_{0}, …, X_{n−1}], homogeneous of degree m, such that

∑_{i=0}^{n−1} b_{δ,i}(a_{0}, …, a_{n−1})α^{i} = δ (a_{0} + a_{1}α + ⋯ + a_{n−1}α^{n−1})^{m} .

Consequently, (x : y) = (b_{δ,0}(a_{0}, …, a_{n−1}) : −b_{δ,1}(a_{0}, …, a_{n−1})) and b_{δ,i}(a_{0}, …, a_{n−1}) should vanish for i = 2, …, n − 1. This gives us

Lemma 5.2 Let K, F, D and S be as above. Suppose that F is irreducible over K, that α is a root of F (X, 1) and that L = K(α). Suppose that x, y, z ∈ K are S-primitive and satisfy F (x, y) = Dz^{m}. Then there are a_{0}, …, a_{n−1} ∈ K and δ ∈ L(S, m) with N_{L/K}(δ)/D ∈ (K^{∗})^{m} such that for the b_{δ,i} as defined above, we have

\begin{aligned}(x : y) & = & (b_{δ,0}(a_{0}, …, a_{n−1}) : −b_{δ,1}(a_{0}, …, a_{n−1})) \\ b_{δ,i}(a_{0}, …, a_{n−1}) & = & 0 for i = 2, …, n − 1. \end{aligned}

This shows that for irreducible F, models of the C_{P} mentioned in Theorem 5.1 are given by ideals of the form

I_{P} = (b_{δ,2}(X_{0}, …, X_{n−1}), …, b_{δ,n−1}(X_{0}, …, X_{n−1}))

for appropriate values of δ ∈ L(S, m) and that Φ_{P} takes the form (b_{δ,0}(X_{0}, …, X_{n−1}) : −b_{δ,1}(X_{0}, …, X_{n−1})). These models have the advantage of being completely explicit, over K, and efficiently computable.

6 Subcovers for F (x, y) = Dz^{2}

Consider the situation of Section 5 with m = 2. In principle, Theorem 5.1 guarantees that primitive solutions of the equation F (x, y) = Dz^{2} over a number field K are parametrised by a finite number of curves C_{P} over K. If the genus of those curves is > 1, in which case deg(F) ≥ 5, then it may be possible to determine C_{P} (K) for each of these using an effective Chabauty-method. Since the genus of C_{P} may be quite high, this would involve computations on high dimensional abelian varieties. We follow another approach.

7 THE EQUATIONS X^{2} ± Y^{4} = Z^{5} 12

We use the notation of Section 5 and we will put n = 2 and write Φ = Φ_{P}. Recall that Φ is galois with galois group ⟨τ_{1}, …, τ_{n}⟩. Furthermore, note that H := ⟨τ_{1} ∘ τ_{2}, τ_{2} ∘ τ_{3}, τ_{5}, …, τ_{n}⟩ is a normal subgroup of Gal(C_{δ}/ℙ_{1}) of index 2. Consequently, Φ : C_{P} → ℙ_{1} splits in C_{P} → E_{P} →^{φ} ℙ_{1}, induced by the map H\. : C_{δ} → E_{δ}. In general, dividing out a variety by a group of automorphisms gives a variety again. See [Sil86, Exercise 3.13] or [Mum70, §7]. This is quite a deep result. In this special case, observe that H\. is induced by the map (Y_{1} : …: Y_{n}) ↦ (Y_{1}^{2}Y_{4} : Y_{2}^{2}Y_{4} : Y_{1}Y_{2}Y_{3} : Y_{4}^{3}), which can be seen from the fact that it is invariant under H and induces a map of degree 2^{n−1} on C_{δ}. The curve E_{δ} is the image of C_{δ} under this map. That the image gives a smooth model is not important for our purposes and is left to the reader.

This construction is nicely summarised in the following commutative diagram.

C_{δ}  C_{P} E_{δ}  E_{P} ℙ_{1}_{ȷ ȷ fl_{H\.}}  _{444444 6} h h h h h h k_{Φ} u ^{ψ}  _{[ []}_{\begin{aligned}& ψ^{−1}Hψ\. \\ π \end{aligned}}_{‚ ‚ ‚ ‚ ‚‚ ›}  A A A A A A D_{φ}

From degree and ramification behaviour, it follows that φ : E_{P} → ℙ_{1} is a double cover, which is ramified exactly above α_{1}, …, α_{4}. Therefore it is of genus 1 and has a model of the form

E_{P} : γY^{2} = (X − α_{1})(X − α_{2})(X − α_{3})(X − α_{4}) = R(X),

where φ = X. This model is not smooth at ∞. We denote the branches at infinity by ∞^{+} and ∞^{−}, classified by Y/X^{2}. It does have good reduction at primes outside S ∪ {2} in the sense that the locally regular charts (X, Y) and (1/X, Y/X^{2}) in reduction cover the curve with locally regular charts again.

Let K(R) denote the field of definition of R(X). If Φ(P) = X(π(P)) ∈ ℙ_{1}(K), then γY (π(P))^{2} ∈ K(R), the field of definition of R. Therefore, it suffices to consider only γ that are representatives of K(R)(S, 2). Note that the choices of δ are absorbed in γ.

Now P ∈ C_{P} (K) leads to a point G = π(P) ∈ E_{P} (K(R)) with φ(G) = Φ(P) ∈ ℙ_{1}(K). If E_{P} has a K(R)-rational point, then we can make it into an elliptic curve. This places us in the situation of Section 4.

Lemmas 7.3, 7.4 and 7.5 apply this procedure to some equations where deg(F) = 5. In those cases, genus(C_{δ}) = 5. For a more general treatment of curves of genus 5 admitting maps to curves of genus 1, see [ACGH85, exercise section VI.F].

7 The Equations x^{2} ± y^{4} = z^{5}

In this section we determine the primitive solutions to x^{2} + y^{4} = z^{5} and x^{2} − y^{4} = z^{5}. We use the same methods as in the previous sections. Lemmas 7.1 and 7.2 show that we have to determine {2, 5}-primitive solutions fo equations of the form y^{2} = F (s, t), where F is a homogeneous form of degree 5.

Lemma 7.1 Let x, y, z ∈ ℤ be coprime integers satisfying x^{2} + y^{2} = z^{5}. Then there exist s, t ∈ ℤ_{{2,5}} with (s, t) mod p ≠ (0, 0) for any p ∤ 10 such that

\{ \begin{matrix}x & = & t(t^{4} − 10t^{2}s^{2} + 5t^{4}) \\ y & = & s(s^{4} − 10s^{2}t^{2} + 5t^{4}) \\ z & = & s^{2} + t^{2} \end{matrix}

Proof: Let i^{2} = −1. Then x^{2} + y^{2} = (x + iy)(x − iy). Since x and y are coprime, we have (x + iy, x − iy) | 2. Consequently, x + iy = δ(s + it)^{5}, where δ is some fifth power free 2-unit in ℤ[i] such that Norm(δ) is a fifth power. Since there 2 = −i(1 + i)^{2}, it follows that δ is a unit. Since every unit in ℤ[i] is a fifth power, we can assume that δ = 1. □

Lemma 7.2 Let x, y, z ∈ ℤ be coprime integers satisfying x^{2} − y^{2} = z^{5}. Then there exist s, t ∈ ℤ_{{2,5}} with (s, t) ≠ (0, 0) mod p for any prime p ∤ 10 such that

\{ \begin{matrix}±x & = & \frac{1}{2}(s^{5} + t^{5}) \\ y & = & \frac{1}{2}(s^{5} − t^{5}) \\ ±z & = & st \end{matrix} or \{ \begin{matrix}±x & = & s^{5} + 8t^{5} \\ y & = & s^{5} − 8t^{5} \\ ±z & = & 2st \end{matrix}

Proof: By factorisation, it follows that there are s, t ∈ ℤ_{{2,5}} and a fifth power free δ ∈ ℤ such that, neglecting the sign of y, x+y = δs^{5} and x−y = δ^{4}t^{5}. Since x and y are coprime, we can take δ | 2. Note that (δ, s, t) ↦ (−δ, −s, t) corresponds to (x, y, z) ↦ (x, y, −z). So, if we neglect the sign of z, we can assume that δ is positive. Taking δ = 1, 2 gives the relations mentioned above. The map (s, t) ↦ (−s, −t) induces (x, y, z) ↦ (−x, −y, z), so the mentioned relations also take into account the sign of y. □

According to Theorem 5.1, solutions to equations as mentioned in the lemmas above, are parametrised by the rational points on genus 5 curves. By Section 6, these curves cover elliptic curves. The next three lemmas determine these elliptic curves including the induced cover of the projective line.

Lemma 7.3 The {2, 5}-primitive solutions to y^{2} = s(s^{4} − 10s^{2}t^{2} + 5t^{4}) have s/t = φ(P), where φ : E_{P} → ℙ_{1} is a double cover of genus 1 over a field L and P ∈ E_{P} (L). It suffices to take E_{1}, …, E_{4} as described in Table 1.

Proof: Let β^{4} − 5β^{2} + 5 = 0. The field ℚ(β) is galois and

X^{4} − 10X^{2} + 5X = (X + 2β − β^{3})(X − 2β + β^{3})(X − 4β + β^{3})(X + 4β − β^{3}).

Put α = β^{3} − 2β. Using Lemma 5.2, it follows that with S = {2, 5}, we have

s = Norm(δ)a_{4}^{2}
s − αt = δ(a_{0} + a_{1}β + a_{2}β^{2} + a_{3}β^{3})^{2}

where δ ∈ L(S, 2). It follows that (s/t)^{4} − 10(s/t)^{2} + 5 = Norm(δ)(y/(a_{4}t^{2}))^{2} as well. As is easily checked, the curve X^{4} − 10X^{2} + 5 = D Y^{2} has no ℚ-rational points for

j  E_{j}  φ_{j}(X, Y) L
1  Y^{2}=X^{4} − 10X^{2} + 5  X  ℚ
2  5Y^{2}=X^{4} − 10X^{2} + 5  X  ℚ
3 (8β − 2β^{3} − 6)Y^{2}=β^{3}X^{3} + (4β^{2} − 5)X^{2} + (β^{3} − 4β)X − 1 1/X ℚ(β)
4 (2β^{3} − 8β − 6)Y^{2}=β^{3}X^{3} + (4β^{2} − 5)X^{2} + (β^{3} − 4β)X − 1 1/X ℚ(β)
5  5Y^{2}=X^{4} + X^{3} + X^{2} + X + 1  X  ℚ
6 2(ζ − ζ^{2} − 1)Y^{2}=X^{4} − ζX^{3} + ζ^{2}X^{2} − ζ^{3}X + ζ^{4}  X  ℚ(ζ)
7 2(1 − ζ + ζ^{2})Y^{2}=X^{4} − ζX^{3} + ζ^{2}X^{2} − ζ^{3}X + ζ^{4}  X  ℚ(ζ)
8  Y^{2}=X^{4} + α^{3}X^{3} + 2αX^{2} + 2α^{4}X + 4α^{2}  X  ℚ(α)
9 (α^{3} + α^{2} − 1)Y^{2}=X^{4} + α^{3}X^{3} + 2αX^{2} + 2α^{4}X + 4α^{2}  X  ℚ(α)
L ℤ-basis of O_{L} disc(O_{L}/ℤ)  O_{L}^{∗}  reg(O_{L}^{∗}) h(O_{L})
ℚ(β) 1, β, β^{2}, β^{3}  2000  ⟨−1, η_{1}, η_{2}, η_{3}⟩  1.8528  1
ℚ(ζ)  1, ζ, ζ^{2}, ζ^{3}  125  ⟨ζ, 1 − ζ⟩  0.9624  1
ℚ(α) 1, α, α^{2}, α^{3}, α^{4}  50000  ⟨−1, 1 − α, α^{4} − α^{3} + α^{2} − 1⟩ 4.8349  1

\begin{aligned}β^{4} − 5β^{2} + 5 & = & 0 & η_{1} & = & 2 − β^{2} \\ ζ^{4} − ζ^{3} + ζ^{2} − ζ + 1 & = & 0 & η_{2} & = & 2 + β − β^{2} \\ α^{5} − 2 & = & 0 & η_{3} & = & −3 + 3β + β^{2} − β^{3} \end{aligned}

L p  defining relation
ℚ(β) 41 β = 3 mod 𝔭_{p,1}
β = 18 mod 𝔭_{p,2}
β = 23 mod 𝔭_{p,3}
β = 38 mod 𝔭_{p,4}
2 2O_{L} = (1 + β − β^{2})O_{L}^{2} = 𝔭_{2}^{2}
5 5O_{L} = βO_{L}^{5} = 𝔭_{5}^{4}
ℚ(ζ) 31  ζ = 15 mod 𝔭_{p,1}
ζ = 23 mod 𝔭_{p,2}
ζ = 27 mod 𝔭_{p,3}
ζ = 29 mod 𝔭_{p,4}

L p  defining relation
ℚ(ζ) 2 2O_{L} = 𝔭_{2}
5 5O_{L} = (1 + ζ + ζ^{2} + ζ^{3})O_{L}^{4} = 𝔭_{5}^{4}
ℚ(α) 151 α = 22 mod 𝔭_{p,1}
α = 25 mod 𝔭_{p,2}
α = 49 mod 𝔭_{p,3}
α = 90 mod 𝔭_{p,4}
α = 116 mod 𝔭_{p,5}
2 2O_{L} = αO_{L}^{5} = 𝔭_{2}^{5}
5 5O_{L} = (α + 1)O_{L}^{5} = 𝔭_{5}^{5}

Table 1: Parametrising curves and their fields of definition

D = 2, 10, −2, −5, −10. For D = 1, 5, the curves are mentioned in the lemma. For D = −1, however, we get an elliptic curve of rank 1, which is not usable if we want a finite bound on the number of solutions. Therefore, we examine the case where −Norm(δ) is a square in more detail. As it turns out, all units in O have positive norm. Generators of the prime ideal 𝔭_{2} | 2, for instance 1+β−β^{2}, have norm −4. Since 5O = 𝔭_{5}^{4}, we can take δ to be 1+β−β^{2} times a square-free unit. This gives 16 possibilities. For each δ, we write out the 5 equations for s, t with respect to the basis 1, …, β^{3} and we eliminate s, t from them. This leaves us with 3 quadratic forms in a_{0}, …, a_{4}. By homogeneity, if they have a common rational zero, then there also is a solution (a_{0}, …, a_{4}) ∈ ℤ^{5} with gcd(a_{0}, …, a_{4}) = 1.

Thus, there should exist a solution (a_{0}, …, a_{4}) ∈ ℤ_{p}^{5} with (a_{0}, …, a_{4}) ≠ (0, …, 0) mod p for any prime p as well. If we test this for both p = 2 and p = 5, we see that only for δ = 3β^{3} + 6β^{2} − 4β − 9, −3β^{3} + 6β^{2} + 4β − 9 do we have solutions locally. If we combine this with our original relations on s, t and choose appropriate representatives of L(S, 2), we get that for some y_{1} ∈ ℚ(β) and x = s/t, we have

x(x^{4} − 10x^{2} + 5)/(x + β^{3} − 4β) = 6 ± 2β(β^{2} − 4)y_{1}^{2}.

This leads to the remaining two curves. □

Lemma 7.4 The {2, 5}-primitive solutions to 2y^{2} = s^{5} − t^{5} have s/t = φ(P), where φ : E_{P} → ℙ_{1} is a double cover of genus 1 over a field L and P ∈ E_{P} (L). It suffices to take E_{5}, E_{6}, E_{7} as in Table 1.

Proof: Let ζ^{4} − ζ^{3} + ζ^{2} − ζ + 1 = 0. Then

X^{5} − 1 = (X − 1)(X + ζ)(X − ζ^{2})(X + ζ^{3})(X − ζ^{4}).

Put α = −ζ. It follows that

s − t = 2Norm(δ)a_{4}^{2}
s + ζt = δ(a_{0} + a_{1}ζ + a_{2}ζ^{2} + a_{3}ζ^{3})^{2}

where δ ∈ L(S, 2). It follows that (s/t)^{4} + (s/t)^{3} + (s/t)^{2} + (s/t) + 1 = (y/a_{4}t^{2}))^{2}/Norm(δ) as well. As is easily checked, the curve X^{4} +X^{3} +X^{2} +X +1 = DY^{2} has ℚ-rational points for D = 1, 5 only. For D = 5, the curve is mentioned in the lemma. For D = 1 we find a curve of positive rank, so we examine the case where Norm(δ) is a square in more detail. We can take δ to be a multiplicative combination of {2, ζ^{3} + ζ − 1, ζ}. Filtering at 2 and 5 gives that, without loss of generality, δ = ζ^{3} − 1, 1 − ζ^{3}. It follows that for some y_{1} ∈ ℚ(ζ) and x = s/t, we have

(x^{5} − 1)/(x + ζ) = 2Norm(δ)/δy_{1}^{2}.

This leads to the remaining two curves. □

Lemma 7.5 The {2, 5}-primitive solutions to y^{2} = s^{5} − 8t^{5} have s/t = φ(P), where φ : E_{P} → ℙ_{1} is a double cover of genus 1 over a field L and P ∈ E_{P} (L). It suffices to take E_{8}, E_{9} as in Table 1.

Proof: Let α^{5} − 2 = 0. Then

X^{5} − 8 = (X − α^{3})(X^{4} + α^{3}X^{3} + 2αX^{2} + 2α^{4}X + 4α^{2}).

It follows that

s − α^{3}t = δ(a_{0} + a_{1}α + a_{2}α^{2} + a_{3}α^{3} + a_{4}α^{4})^{2}
y^{2} = Norm(δ)Norm(a_{0} + a_{1}α + a_{2}α^{2} + a_{3}α^{3} + a_{4}α^{4})^{2}

\begin{aligned}E_{j} : γ_{j}Y^{2} & = & X^{3} − 5X^{2} + 5X \\ γ_{3} & = & 8β − 2β^{3} − 6 \\ φ_{3}(X, Y) & = & \frac{(6β − 2β^{3})X + 3β^{3} − 10β}{5} \\ γ_{6} & = & 2(ζ^{2} − ζ) \\ φ_{6}(X, Y) & = & \frac{X − 2ζ^{2} + ζ − 2}{X + ζ^{3} + ζ^{2} − ζ − 1} \\ γ_{7} & = & −2(ζ^{2} − ζ) \\ φ_{7}(X, Y) & = & \frac{X − 2ζ^{2} + ζ − 2}{X + ζ^{3} + ζ^{2} − ζ − 1} \\ γ_{8} & = & 1 \\ φ_{8}(X, Y) & = & \frac{−α^{3}X + 2α^{3}Y}{4X − 5} \\ γ_{9} & = & α^{3} + α^{2} − 1 \\ φ_{9}(X, Y) & = & \frac{2(6α^{4} − 4α^{3} + 5α^{2} + 11α − 13)X + 12Y − 10(3α^{4} − α^{3} + 4α − 6)}{(−4α^{3} + 2α^{2} + 2α + 2)X − 12Y − 5(3α^{4} − 3α^{3} + α^{2} − 4)} \end{aligned}

Table 2: Description of covers with respect to models E_{j}

where δ ∈ L(S, 2). From the second equation, it follows that δ is of square norm. Since 2 and 5 completely ramify, this leaves δ ∈ {1, α − 1, 1 + α + α^{3}, α^{4} − α^{3} + α^{2} − 1}. Filtering at 2 only leaves δ = 1, α^{4} − α^{3} + α^{2} − 1. For x = s/t and some y_{1} ∈ ℚ(α) we have

(x^{5} − 8)/(x − α^{3}) = Norm(δ)/δy_{1}^{2}.

This leads to the curves in the lemma. □

Lemma 7.6 The covers E_{3}, E_{6}, …, E_{9} in Table 1 are birational to the covers E_{j} : γY^{2} = X^{3} − 5X^{2} + 5X described in Table 2.

Proof: It is straightforward to check this. □

Having established elliptic covers that parametrise {2, 5}-primitive solutions in some way, we now determine their Mordell-Weil groups (up to finite index). We do this using a 2-isogeny-descent (see [Sil86, Proposition X.4.9] or [Bru99, Section 4.3]). We only give results. For more detailed information, see [Bru99, Section 4.8] and the electronic references given there.

Lemma 7.7 Let j = 3, 6, 7, 8, 9. Consider E_{j}, G_{i}, p and M be as given in Table 2 and let L be the field of definition of E_{j}. then the group spanned by the G_{i} span a subgroup of finite index prime to 2 and p in E_{j}(L), containing E_{j}^{tor}(L). Let 𝔭 be a prime of L above p. Then the reduction of the G_{i} span E_{j}(L) mod 𝔭. The rows of M form linear combinations of the G_{i} that span the intersections of the kernels of reduction mod 𝔭 | p.

Proof: We get that the index is finite and prime to 2 by doing a 2-isogeny-descent (see [Sil86, Proposition X.4.9] or [Bru99, Section 4.3]). For more detailed information, see

\begin{matrix}j & G_{i} & p & M^{t} & Z(B)/p \\ 3 & \begin{matrix}(10 + 5β − 2β^{2} − β^{3}, 20 + 15β − 5β^{2} − 4β^{3}) \\ (β^{3}, 0) \\ (0, 0) \end{matrix} & 41 & ( \begin{matrix}110 \\ 0 \\ 0 \end{matrix} ) & ( \begin{matrix}21 \\ 35 \\ 2 \\ 18 \end{matrix} ) \\ 6 & \begin{matrix}(1 − 3ζ^{2} + 3ζ^{3}, 1 + 3ζ − 2ζ^{2} + ζ^{3}) \\ (1 − 1ζ^{2} + ζ^{3}, ζ^{3}) \\ (2 − ζ^{2} + ζ^{3}, 0) \\ (0, 0) \end{matrix} & 31 & ( \begin{matrix}4 & 0 \\ 8 & 16 \\ 0 & 0 \\ 0 & 0 \end{matrix} ) & ( \begin{matrix}3 & 3 \\ 12 & 29 \\ 12 & 15 \\ 0 & 17 \end{matrix} ) \\ 7 & \begin{matrix}(2 + ζ^{2} − ζ^{3}, 1 − ζ − ζ^{3}) \\ (−1 + 3ζ^{2} − 3ζ^{3}, 1 − 7ζ + 8ζ^{2} − 4ζ^{3}) \\ (2 − ζ^{2} + ζ^{3}, 0) \\ (0, 0) \end{matrix} & 31 & ( \begin{matrix}8 & 0 \\ 4 & 8 \\ 0 & 0 \\ 0 & 0 \end{matrix} ) & ( \begin{matrix}21 & 8 \\ 2 & 2 \\ 0 & 15 \\ 22 & 27 \end{matrix} ) \\ 8 & \begin{matrix}(1 − 2α + α^{3}, 3 − 2α^{2} − 2α^{3} + α^{4}) \\ (4, 2) \\ (0, 0) \end{matrix} & 151 & ( \begin{matrix}80 & 0 \\ 0 & 20 \\ 0 & 1 \end{matrix} ) & ( \begin{matrix}51 & −1 \\ 43 & −1 \\ 146 & −1 \\ 127 & −1 \\ 92 & −1 \end{matrix} ) \\ 9 & \begin{matrix}(1 + α + α^{2} − α^{3} + α^{4}, 1 − 5α + 6α^{2} − 4α^{3} + 2α^{4}) \\ (3 + 5α + 4α^{2} + 2α^{3}, 9 + 13α + 7α^{2} + 5α^{3} + 2α^{4}) \\ (0, 0) \end{matrix} & 151 & ( \begin{matrix}120 & 0 \\ 120 & 720 \\ 0 & 0 \end{matrix} ) & ( \begin{matrix}49 & 64 \\ 78 & 22 \\ 47 & 145 \\ 57 & 135 \\ 94 & 60 \end{matrix} ) \end{matrix}

Table 3: Curves E_{j} and data on the Mordell-Weil groups

[Bru99, Section 4.8] and the electronic references given there. The torsion can be bounded by using that, under mild conditions, reduction modulo a prime is injective on torsion. Therefore, the torsion is a subgroup of the group of rational points over the residue field of most primes.

The matrix M gives us a subgroup of ⋂_{𝔭|p}(E_{j}(L) ∩ E_{j}^{(1)}(L_{𝔭}). Applying Lemma 2.1 to Z(B)/p (the matrix of Z-coordinates of generators of that group) gives that the index is prime to p as well. That the G_{i} span E_{j}(L) mod 𝔭 can be checked by computing the index of the reduction of the G_{i} in the finite group (E_{j} mod 𝔭)(O/𝔭). □

Having established knowledge of the appropriate Mordell-Weil groups, we can proceed and apply the method described in Section 4.

Proposition 7.8 E_{1}(ℚ) = {∞^{+}, ∞^{−}}.

Proof: Obviously, the points mentioned are rational. It is straightforward to show that this curve is birational to the elliptic curve 400D1 in [Cre92]. This elliptic curve has only 2 rational points, which shows that the list is complete. □

Proposition 7.9 E_{2}(ℚ) = {(0, 1), (0, −1)}.

Proof: Obviously, the points mentioned are rational. It is straightforward to show that this curve is birational to the elliptic curve 400F1 in [Cre92]. This elliptic curve has only 2 rational points, which shows that the list is complete. □

Proposition 7.10 E_{5}(ℚ) = {∞^{+}, ∞^{−}}.

Proof: Obviously, the points mentioned are rational. It is straightforward to show that this curve is birational to the elliptic curve 200D1 in [Cre92]. This elliptic curve has only 2 rational points, which shows that the list is complete. □

Proposition 7.11 The following statements are true.

• The points P ∈ E_{3}(ℚ(β)) with X(P) ∈ ℙ_{1}(ℚ) have X(P) = 0.

• The points P ∈ E_{4}(ℚ(β)) with X(P) ∈ ℙ_{1}(ℚ) have X(P) = 0.

• The points P ∈ E_{6}(ℚ(ζ)) with X(P) ∈ ℙ_{1}(ℚ) have X(P) = ±1.

• The points P ∈ E_{7}(ℚ(ζ)) with X(P) ∈ ℙ_{1}(ℚ) have X(P) ∈ \{1, \frac{1}{3}, 3\}.

• The points P ∈ E_{8}(ℚ(α)) with X(P) ∈ ℙ_{1}(ℚ) have X(P) ∈ {∞, 0, −2}.

• The points P ∈ E_{9}(ℚ(α)) with X(P) ∈ ℙ_{1}(ℚ) have X(P) ∈ {∞}.

Proof: We apply the method proposed in Section 4 to each of the curves individually. We work with the twisted Weierstrass models E_{i} from Table 2 instead of the models E_{i}. The map X : E_{i} → ℙ_{1} corresponds to φ_{i} : E_{i} → ℙ_{1}. Note that the map (X, Y, β) → (−X, Y, −β) gives a map between the covers φ_{4} : E_{4} → ℙ_{1} and φ_{3} : E_{3} → ℙ_{1}. Consequently, the statement on E_{4} follows from the one on E_{3}. The method is almost automatic, so we only give the argument for E_{7}, which illustrates the different facets of the method nicely. The arguments for the other curves are similar. Many of the computational steps cannot be reproduced here on paper. Instead, a program is available that does the computations for you. It is written for KASH 2.0 (see [DFK^{+}97]) and is available from the preprint server of the Mathematical Institute of Leiden University or through the author’s homepage.

We have K = ℚ, L = ℚ(ζ), E = E_{7} and φ = φ_{7}. According to Lemma 7.7, we have that ⟨G_{1}, …, G_{4}⟩ generate a subgroup of E(L) with appropriate properties. For the sake of simplicity of the argument, we pretend that they actually generate the whole of E(L).

We choose p = 31 and 𝔭_{1}, …, 𝔭_{4} = 𝔭_{31,1}, …, 𝔭_{31,4}. It is straightforward to compute that ⟨G_{1}, …, G_{4}⟩ ∩ E^{(1)}(L_{𝔭_{i}}) = ⟨2G_{1} + G_{2} + G_{4}, 4G_{2} + G_{4}⟩, ⟨2G_{1} + 3G_{2} + G_{4}, 4G_{2} + G_{1} + G_{2}⟩, ⟨2G_{1} + G_{2} + G_{3}, 4G_{2} + G_{3} + G_{4}⟩, ⟨2G_{1} + 3G_{2}, 4G_{2} + G_{4}⟩ for i = 1, 2, 3, 4 respectively. Consequently, B_{1} = 8G_{1} + 4G_{2}, B_{2} = 8G_{2} generate the intersection of the kernels of reduction.

In order to determine φ(E(L)) ∩ ℙ_{1}(ℚ), we first look at the reduction mod 31. If P ∈ φ(E(L)) ∩ ℙ_{1}(ℚ), then certainly P mod 31 ∈ φ(E(L)) mod 𝔭_{i}. The latter is something

j  G  φ(G) θ^{G}(n_{1}, …, n_{r})
3  ∞  ∞ 41^{2}( \begin{matrix}23n_{1}^{2} \\ n_{1}^{2} \\ 36n_{1}^{2} \end{matrix} ) mod 41^{3}

6  ∞  1 31^{2}( \begin{matrix}n_{1}^{2} + 8n_{1}n_{2} + 19n_{2}^{2} \\ 27n_{1}^{2} + 17n_{1}n_{2}^{2} + 28n_{2}^{2} \\ 8n_{1}^{2} + 16n_{1}n_{2} + 16n_{2}^{2} \end{matrix} ) mod 31^{3}

G_{1} + G_{3} + G_{4} −1 31( \begin{matrix}9n_{1} + 22n_{2} \\ 11n_{1} + 22n_{2} \\ 29n_{1} \end{matrix} ) mod 31^{2}

7  ∞  1 31^{2}( \begin{matrix}10n_{1}n_{2} + n_{2}^{2} \\ 11n_{1}^{2} + n_{1}n_{2} + 23n_{2}^{2} \\ 2n_{1}^{2} + 24n_{1}n_{2} + 24n_{2}^{2} \end{matrix} ) mod 31^{3}

G_{1} + G_{2} + G_{4}  \frac{1}{3} 31( \begin{matrix}26n_{1} + 22n_{2} \\ 16n_{1} + 18n_{2} \\ 5n_{1} + 14n_{2} \end{matrix} ) mod 31^{2}

G_{1} − G_{2} + G_{4} 3 31( \begin{matrix}30n_{1} + 5n_{2} \\ 7n_{1} + 10n_{2} \\ 12n_{1} + 29n_{2} \end{matrix} ) mod 31^{2}

8  ∞  ∞ 151( \begin{matrix}77n_{1} + 10n_{2} \\ 4n_{1} + 46n_{2} \\ 17n_{1} + 85n_{2} \\ 134n_{1} + 14n_{2} \end{matrix} ) mod 151^{2}

G_{2}  0 151( \begin{matrix}103n_{1} + 139n_{2} \\ 2n_{1} + 35n_{2} \\ 119n_{1} + 90n_{2} \\ 109n_{1} + 13n_{2} \end{matrix} ) mod 151^{2}

−2G_{1}  −2 151( \begin{matrix}78n_{1} + 29n_{2} \\ 79n_{1} + 3n_{2} \\ 44n_{1} + 45n_{2} \\ 28n_{1} + 2n_{2} \end{matrix} ) mod 151^{2}

9  ∞  −1 151( \begin{matrix}147n_{1} + 57n_{2} \\ 31n_{1} + 84n_{2} \\ 16n_{1} + 42n_{2} \\ 24n_{1} + 119n_{2} \end{matrix} ) mod 151^{2}

Table 4: Fibers of rational points we can compute explicitly. We find

φ(E(L)) mod 𝔭_{1} = {0, 1, 2, 3, 4, 8, 9, 13, 15, 16, 17, 21, 24, 25, 26, 27, 29, ∞}
φ(E(L)) mod 𝔭_{2} = {1, 2, 3, 4, 5, 6, 7, 9, 10, 11, 14, 15, 16, 18, 19, 21, 25, 27}
φ(E(L)) mod 𝔭_{3} = {1, 2, 3, 5, 7, 8, 9, 16, 17, 18, 19, 20, 21, 23, 25, 26, 28, 29}
φ(E(L)) mod 𝔭_{4} = {0, 1, 2, 3, 4, 5, 6, 7, 8, 11, 12, 15, 16, 21, 22, 23, 29, ∞}.

It follows that φ(E(L)) ∩ ℙ_{1}(ℚ) mod 31 ⊂ {1, 2, 3, 16, 21}. Upon closer inspection, we see that if G ∈ E(L) and φ(G) mod 𝔭_{1} = 2, then G ∈ {G_{3}} + E^{(1)}(L_{𝔭_{1}}). Similarly, for φ(G) mod 𝔭_{4} = 2, we get G ∈ {2G_{2}, 2G_{2}+G_{4}}+E^{(1)}(L_{𝔭_{4}}). But then we have that 2G_{2}−G_{3} or 2G_{2}+G_{4}−G_{3} is in E^{(1)}(L_{𝔭_{1}})∩E^{(1)}(L_{𝔭_{4}}) = ⟨2G_{1}+G_{2}, G_{2}, G_{4}⟩. This is clearly not the case. A similar argument rules out 16. Thus we see that φ(E(L) ∩ ℙ_{1}(ℚ)) mod 31 ⊂ {1, 3, 1/3}. We know that equality holds, since G = 0, G_{1} − G_{2} + G_{4}, G_{1} + G_{2} + G_{4} realise these values.

It remains to show that these are the only points (apart from the points G_{τ} − G) with rational image under φ. We can do so by applying Lemma 4.2. For this, we need Z(B_{1}), Z(B_{2}) mod 𝔭_{i}^{2}. We can compute these either by performing the group addition with exact precision and then reduce mod 𝔭_{i}^{2} or (more efficiently), do the group operations with finite 𝔭_{i}-adic precision. Either way, we find the values mentioned in Table 3. By applying Lemma 2.1 to these, we get that ⟨G_{1}, …, G_{4}⟩ has index prime to 31 in E(L), as stated in Lemma 7.7.

Using these values we can compute approximations to θ^{G_{0}}(n_{1}, n_{2}), as shown in Table 4. In is easily checked that in each case, one of the criteria in Lemma 4.2 applies. We conclude that P ∈ φ(E(L)) ∩ ℙ_{1}(ℚ) with P mod 31 ∈ {1, 3, 21} are P = 1, 3, 1/3. □

Proof of Theorem 1.1: Consider x^{2} + y^{4} = z^{5}. Lemma 7.1 together with Lemma 7.3 show that the curves E_{1}, …, E_{4} from Table 1 parametrise the primitive solutions and in what way the parameter values s/t can be recovered from the points P ∈ E_{i}(L) with φ_{i}(P) ∈ ℙ_{1}(ℚ). Propositions 7.8 through 7.11 give those points. We see all points must have s/t = 0, ∞, so we have that either s = 0 or t = 0 which leads to y = 0 or x = 0. □

Proof of Theorem 1.2: Consider x^{2} − y^{4} = z^{5}. Lemmas 7.2, 7.4 and 7.5 show that E_{5}, …, E_{9} determine all possible solutions. Propositions 7.10 through 7.11 give the possible candidates and the values of s/t belonging to them. The values s/t = ∞, 1, −1 in Lemma 7.4 lead to solutions with z = 0, y = 0 or x = 0. The values s/t = 3, \frac{1}{3} lead to x = ±122, y = ±11, z = 3.

The points on E_{8}(L) lead to s/t = 1, ∞, −2. These correspond to (x, y, z) = (0, ±1, −1) and (±16, ±4, 0). While (−2, 2(α − α^{2} − α^{3})) is a genuine point on E_{8}(L), we have that λ(2 + α^{3}) is not a square in L for any λ ∈ ℚ^{∗}. We therefore see that no rational s, t with s/t = −2 exist that satisfy Lemma 7.5. The point on E_{9}(L) leads to (x, y, z) = (±7, ±3, −2). □

Acknowledgements

The results presented in this article are part of [Bru99], which is a thesis resulting from 4 years of research at Leiden University under supervision of F. Beukers and R. Tijdeman, who helped with valuable discussions. Furthermore, E. Schaefer and H. W. Lenstra Jr. offered many useful comments. The many computations made heavy use of the freely available computer algebra package KASH/KANT ([DFK^{+}97]). The maintaining group was of great help with their quick responses to bug reports and requests for new features.

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Address for correspondence: Nils Bruin, Math. Institute, Leiden University, P.O. Box 9512, 2300 RA Leiden, The Netherlands; e-mail: nbruin@wi.leidenuniv.nl