The Diophantine equation $y^2 = Dx^4 + 1$, II

J. Cohn

Displaying similar documents to “The Diophantine equation $y^{2} = D x^{4} + 1$ , II”

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

J. H. E. Cohn (2003)

Colloquium Mathematicae

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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.

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 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.

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).

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.

Displaying similar documents to “The Diophantine equation y 2 = D x 4 + 1 , II”

The Diophantine equation D x ² + 2 2 m + 1 = y ⁿ

A ternary Diophantine inequality over primes

On the Diophantine equation q n - 1 q - 1 = y

On the diophantine equation x 2 + 5 k 17 l = y n

Method of infinite ascent applied on - ( 2 p · A 6 ) + B 3 = C 2

Displaying similar documents to “The Diophantine equation $y^{2} = D x^{4} + 1$ , II”

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

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

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

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