On the Diophantine equation $x^{p}+2^{2m}=py^{2}$
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Abstract:
Let $p$ be an odd prime. In this paper, using some theorems of Adachi and the author, we prove that if $p \equiv 1(\text {mod }4)$ and $p\nmid B_{(p-1)/2}$, then the equation $x^{p}+1=py^{2}, y\ne 0$, and the equation $x^{p}+2^{2m}=py^{2}, m\in \mathbb {N}, \text { gcd}(x, y )=1, p\mid y$, have no integral solutions respectively. Here $B_{(p-1)/2}$ is $(p-1)/2$th Bernoulli number.References
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Additional Information
- Zhenfu Cao
- Affiliation: Department of Mathematics, Harbin Institute of Technology, Harbin 150001, People’s Republic of China
- Email: zfcao@hope.hit.edu.cn
- Received by editor(s): September 8, 1998
- Published electronically: February 25, 2000
- Communicated by: David E. Rohrlich
- © Copyright 2000 American Mathematical Society
- Journal: Proc. Amer. Math. Soc. 128 (2000), 1927-1931
- MSC (2000): Primary 11D61, 11D41
- DOI: https://doi.org/10.1090/S0002-9939-00-05517-9
- MathSciNet review: 1694856