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Displaying 201 – 220 of 239

Squares in products with terms in an arithmetic progression

N. Saradha (1998)

Acta Arithmetica

S-unit equations over number fields.

Hans Peter Schlickewei (1990)

Inventiones mathematicae

Superelliptic equations arising from sums of consecutive powers

Michael A. Bennett, Vandita Patel, Samir Siksek (2016)

Acta Arithmetica

Using only elementary arguments, Cassels solved the Diophantine equation (x-1)³ + x³ + (x+1)³ = z² (with x, z ∈ ℤ). The generalization ${(x - 1)}^{k} + x^{k} + {(x + 1)}^{k} = z^{n}$ (with x, z, n ∈ ℤ and n ≥ 2) was considered by Zhongfeng Zhang who solved it for k ∈ 2,3,4 using Frey-Hellegouarch curves and their corresponding Galois representations. In this paper, by employing some sophisticated refinements of this approach, we show that the only solutions for k = 5 have x = z = 0, and that there are no solutions for k = 6. The chief innovation...

Sur la fonction plus grand facteur premier

Michel Langevin (1974/1975)

Séminaire Delange-Pisot-Poitou. Théorie des nombres

The Diophantine equation ${(b n)}^{x} + {(2 n)}^{y} = {((b + 2) n)}^{z}$

Min Tang, Quan-Hui Yang (2013)

Colloquium Mathematicae

Recently, Miyazaki and Togbé proved that for any fixed odd integer b ≥ 5 with b ≠ 89, the Diophantine equation $b^{x} + 2^{y} = {(b + 2)}^{z}$ has only the solution (x,y,z) = (1,1,1). We give an extension of this result.

The Diophantine equation $D x ² + 2^{2 m + 1} = y ⁿ$

J. H. E. Cohn (2003)

Colloquium Mathematicae

It is shown that for a given squarefree positive integer D, the equation of the title has no solutions in integers x > 0, m > 0, n ≥ 3 and y odd, nor unless D ≡ 14 (mod 16) in integers x > 0, m = 0, n ≥ 3, y > 0, provided in each case that n does not divide the class number of the imaginary quadratic field containing √(-2D), except for a small number of (stated) exceptions.

The diophantine equation $x^{2} + 2^{a} \cdot 17^{b} = y^{n}$

Su Gou, Tingting Wang (2012)

Czechoslovak Mathematical Journal

Let $ℤ$ , $ℕ$ be the sets of all integers and positive integers, respectively. Let $p$ be a fixed odd prime. Recently, there have been many papers concerned with solutions $(x, y, n, a, b)$ of the equation $x^{2} + 2^{a} p^{b} = y^{n}$ , $x, y, n \in ℕ$ , $gcd (x, y) = 1$ , $n \geq 3$ , $a, b \in ℤ$ , $a \geq 0$ , $b \geq 0 .$ And all solutions of it have been determined for the cases $p = 3$ , $p = 5$ , $p = 11$ and $p = 13$ . In this paper, we mainly concentrate on the case $p = 3$ , and using certain recent results on exponential diophantine equations including the famous Catalan equation, all solutions $(x, y, n, a, b)$ of the equation $x^{2} + 2^{a} \cdot 17^{b} = y^{n}$ , $x, y, n \in ℕ$ , $gcd (x, y) = 1$ , $n \geq 3$ , $a, b \in ℤ$ , $a \geq 0$ , $b \geq 0$ , are determined....