A note on the number of $S$-Diophantine quadruples

Florian Luca; Volker Ziegler

Displaying similar documents to “A note on the number of $S$ -Diophantine quadruples”

Method of infinite ascent applied on $- (2^{p} \cdot A^{6}) + B^{3} = C^{2}$

Susil Kumar Jena (2013)

Communications in Mathematics

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In this paper, the author shows a technique of generating an infinite number of coprime integral solutions for $(A, B, C)$ of the Diophantine equation $- (2^{p} \cdot A^{6}) + B^{3} = C^{2}$ for any positive integral values of $p$ when $p \equiv 1$ (mod 6) or $p \equiv 2$ (mod 6). For doing this, we will be using a published result of this author in The Mathematics Student, a periodical of the Indian Mathematical Society.

On the diophantine equation $x^{2} + 5^{k} 17^{l} = y^{n}$

István Pink, Zsolt Rábai (2011)

Communications in Mathematics

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Consider the equation in the title in unknown integers $(x, y, k, l, n)$ with $x \geq 1$ , $y > 1$ , $n \geq 3$ , $k \geq 0$ , $l \geq 0$ and $gcd (x, y) = 1$ . Under the above conditions we give all solutions of the title equation (see Theorem 1).

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

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

Acta Arithmetica

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Searching for Diophantine quintuples

Mihai Cipu, Tim Trudgian (2016)

Acta Arithmetica

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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 $\frac{q^{n} - 1}{q - 1} = y$

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.