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June 2018 On Terai's conjecture
Xin Zhang
Kodai Math. J. 41(2): 413-420 (June 2018). DOI: 10.2996/kmj/1530496850

Abstract

Let $p$ be an odd prime such that $b^r+1=2p^t$, where $r$, $t$ are positive integers and $b \equiv$ 3,5 (mod 8). We show that the Diophantine equation $x^2+b^m=p^n$ has only the positive integer solution $(x,m,n)=(p^t-1,r,2t)$. We also prove that if $b$ is a prime and $r=t=2$, then the above equation has only one solution for the case $b \equiv$ 3,5,7 (mod 8) and the case $d$ is not an odd integer greater than 1 if $b \equiv$ 1 (mod 8), where $d$ is the order of prime divisor of ideal ($p$) in the ideal class group of $\mathbf{Q}$ ($\sqrt {-q}$).

Citation

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Xin Zhang. "On Terai's conjecture." Kodai Math. J. 41 (2) 413 - 420, June 2018. https://doi.org/10.2996/kmj/1530496850

Information

Published: June 2018
First available in Project Euclid: 2 July 2018

zbMATH: 06936461
MathSciNet: MR3824859
Digital Object Identifier: 10.2996/kmj/1530496850

Rights: Copyright © 2018 Tokyo Institute of Technology, Department of Mathematics

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Vol.41 • No. 2 • June 2018
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