2020
DOI: 10.1090/tran/8103
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An old and new approach to Goormaghtigh’s equation

Abstract: We show that if n ≥ 3 n \geq 3 is a fixed integer, then there exists an effectively computable constant c ( n ) c (n) such that if x x , y y , and m m are integers satisfying x m − 1 x − 1 = … Show more

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Cited by 2 publications
(5 citation statements)
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“…As it transpires, this condition allows one to apply a wide variety of effective methods from Diophantine Approximation to the problem, including lower bounds for linear forms in logarithms, Runge's method and the hypergeometric method of Thue and Siegel. Combining results from [4] and [21], we have Theorem 7 (Nesterenko-Shorey, Bennett-Gherga-Kreso). If there is a solution in integers x, y, n and m to equation (1), satisfying (17), then…”
Section: The Case Gcdmentioning
confidence: 77%
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“…As it transpires, this condition allows one to apply a wide variety of effective methods from Diophantine Approximation to the problem, including lower bounds for linear forms in logarithms, Runge's method and the hypergeometric method of Thue and Siegel. Combining results from [4] and [21], we have Theorem 7 (Nesterenko-Shorey, Bennett-Gherga-Kreso). If there is a solution in integers x, y, n and m to equation (1), satisfying (17), then…”
Section: The Case Gcdmentioning
confidence: 77%
“…On the other hand, since n ≥ 18, we may readily show that 6n 3 − 7n 2 + 2n 24(n − 1) 4 x 0 > 1 6(n − 1) 3 for j ∈ {1, 2}, and that 6n 3 − 7n 2 + 2n 24(n − 1) 4 x 0 < 1 6(n − 1) 3 for j = 3 and n ≥ 43, in each case contradicting (30). Finally, if n = 25 and x 0 = 919, we simply check that equation (1) fails to be satisfied.…”
Section: The Case (Mmentioning
confidence: 95%
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“…The above conjecture is a very difficult problem in Diophantine equations. It was solved for some special cases (see [3, 5, 6, 9, 10, 12, 14–18, 22, 23, 26, 2937]). But, in general, the problem is far from solved.…”
Section: Introductionmentioning
confidence: 99%