Cubic forms, powers of primes and the Kraus method

Andrzej Dąbrowski; Tomasz Jędrzejak; Karolina Krawciów

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Solving diophantine equations $x^{4} + y^{4} = q z^{p}$

Luis V. Dieulefait (2005)

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On the diophantine equation $x^{2} = y^{p} + 2^{k} z^{p}$

Samir Siksek (2003)

Journal de théorie des nombres de Bordeaux

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We attack the equation of the title using a Frey curve, Ribet’s level-lowering theorem and a method due to Darmon and Merel. We are able to determine all the solutions in pairwise coprime integers $x, y, z$ if $p \geq 7$ is prime and $k \geq 2$ . From this we deduce some results about special cases of this equation that have been studied in the literature. In particular, we are able to combine our result with previous results of Arif and Abu Muriefah, and those of Cohn to obtain a complete solution for the equation...

Superelliptic equations arising from sums of consecutive powers

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

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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...

On the Diophantine equation $X^{2} - (2^{2 m} + 1) Y^{4} = - 2^{2 m}$

Michael Stoll, P. G. Walsh, Pingzhi Yuan (2009)

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Searching for Diophantine quintuples

Mihai Cipu, Tim Trudgian (2016)

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We consider Diophantine quintuples a, b, c, d, e. These are sets of positive integers, the product of any two elements of which is one less than a perfect square. It is conjectured that there are no Diophantine quintuples; we improve on current estimates to show that there are at most $5.441 \cdot 10^{26}$ Diophantine quintuples.

On the Diophantine equation $x^{2} + 7 = y^{m}$

Samir Siksek, John E. Cremona (2003)

Acta Arithmetica

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On the Diophantine equation $x^{p} + y^{2 p} = z^{2}$

A. Rotkiewicz, A. Schinzel (1987)

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A note on the Diophantine equation $x^{2} + q^{m} = y^{3}$

Hui Lin Zhu (2011)

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The diophantine equation $x^{2} + C = y^{n}$ , II

J. H. E. Cohn (2003)

Acta Arithmetica

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A note on the Diophantine equation $(a^{n} x^{m} \pm 1) / (a^{n} x \pm 1) = y^{n} + 1$

Jiagui Luo (2001)

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On the Diophantine equation $x^{2} + q^{2 m} = 2 y^{p}$

Sz. Tengely (2007)

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A ternary Diophantine inequality over primes

Roger Baker, Andreas Weingartner (2014)

Acta Arithmetica

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Let 1 < c < 10/9. For large real numbers R > 0, and a small constant η > 0, the inequality $| p ₁^{c} + p ₂^{c} + p ₃^{c} - R | < R^{- η}$ holds for many prime triples. This improves work of Kumchev [Acta Arith. 89 (1999)].

On some generalizations of the diophantine equation $s (1^{k} + 2^{k} + \dots + x^{k}) + r = d y^{n}$

Csaba Rakaczki (2012)

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On the equation $a^{2} + b^{2 p} = c^{5}$

Imin Chen (2010)

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On the Diophantine equation $x ² + 2^{α} 13^{β} = y ⁿ$

Florian Luca, Alain Togbé (2009)

Colloquium Mathematicae

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We find all the solutions of the Diophantine equation $x ² + 2^{α} 13^{β} = y ⁿ$ in positive integers x,y,α,β,n ≥ 3 with x and y coprime.

On the Diophantine equation $\frac{q^{n} - 1}{q - 1} = y$

Amir Khosravi, Behrooz Khosravi (2003)

Commentationes Mathematicae Universitatis Carolinae

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There exist many results about the Diophantine equation $(q^{n} - 1) / (q - 1) = y^{m}$ , where $m \geq 2$ and $n \geq 3$ . In this paper, we suppose that $m = 1$ , $n$ is an odd integer and $q$ a power of a prime number. Also let $y$ be an integer such that the number of prime divisors of $y - 1$ is less than or equal to $3$ . Then we solve completely the Diophantine equation $(q^{n} - 1) / (q - 1) = y$ for infinitely many values of $y$ . This result finds frequent applications in the theory of finite groups.

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

Min Tang, Quan-Hui Yang (2013)

Colloquium Mathematicae

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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.