The equation x+y=1 in finitely generated groups

Beukers, Frits; Schlickewei, Hans Peter

doi:10.4064/aa-78-2-189-199

Cited by 65 publications

(89 citation statements)

References 4 publications

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“…Let K be a discrete valuation field with ring of integers O K , uniformizer π, and residue field k. Let E/K be an elliptic curve given by a Weierstrass equation 4 , and a 6 . Then the condition to have multiplicative reduction is that π a 2 1 + 4a 2 (we set as usual b 2 := a 2 1 + 4a 2 ).…”

Section: 2mentioning

confidence: 99%

“…Let K be a number field. Let A, B ∈ K * , and consider the equation [4], and [22] (1.5)). We will apply this bound in this section for the equation X − Y = 1.…”

Section: 3mentioning

confidence: 99%

“…with res(∆(λ), c 4 (λ)) = 2 4 . It follows that there exists λ ∈ K such that an elliptic curve E/K with a point of order N = 4 can be given by a TOME 61 (2011), FASCICULE 5 Weierstrass equation of the form…”

Section: 3mentioning

confidence: 99%

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Torsion and Tamagawa numbers

Lorenzini¹

2011

Ann. inst. Fourier

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Section: 2mentioning

confidence: 99%

“…Let K be a number field. Let A, B ∈ K * , and consider the equation [4], and [22] (1.5)). We will apply this bound in this section for the equation X − Y = 1.…”

Section: 3mentioning

confidence: 99%

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Torsion and Tamagawa numbers

Lorenzini¹

2011

Ann. inst. Fourier

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“…Here, Schlickewei [4] proved the conjecture to be true. His bound has been improved by Beukers and Schlickewei [1]. They showed that for q = 3 equation (1.1) does not have more than 61 solutions.…”

Section: (12)mentioning

confidence: 99%

Zeros of linear recurrence sequences

Schlickewei¹,

Schmidt²,

Waldschmidt³

1999

manuscripta mathematica

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Abstract. Let f (x) = P 0 (x)α x 0 + · · · + P k (x)α x k be an exponential polynomial over a field of zero characteristic. Assume that for each pair i, j with i = j , α i /α j is not a root of unity. Define = k j =0 (deg P j +1). We introduce a partition of α 0 , . . . , α k into subsets α i0 , . . . , α ik i (1 ≤ i ≤ m), which induces a decomposition of f into f = f 1 +· · ·+f m , so that, for 1 ≤ i ≤ m, α i0 : · · · : α ik i ∈ P k i (Q), while for 1 ≤ i = u ≤ m, the number α i0 /α u0 either is transcendental or else is algebraic with not too small a height. Then we show that for all but at most exp (5 ) 5 solutions x ∈ Z of f (x) = 0, we have f 1 (x) = · · · = f m (x) = 0.

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“…Our approach is similar to that of Birch and Merriman [2], with the necessary modifications. In our proofs we use among other things an upper bound by Beukers and Schlickewei [1,Theorem 1] for the numbers of solutions of the equation x + y = 1 in unknowns x, y from a multiplicative group of finite rank.…”

mentioning

confidence: 99%