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ON THE EXPONENTIAL DIOPHANTINE EQUATION $\mathop{({m}^{2} + 1)}\nolimits ^{x} + \mathop{(c{m}^{2} - 1)}\nolimits ^{y} = \mathop{(am)}\nolimits ^{z} $

Part of: Diophantine equations

Published online by Cambridge University Press:  26 March 2014

TAKAFUMI MIYAZAKI  and
NOBUHIRO TERAI
TAKAFUMI MIYAZAKI
Affiliation:
Department of Mathematics, College of Science and Technology, Nihon University, Suruga-Dai, Kanda, Chiyoda, Tokyo 101-8308, Japan email [email protected]
NOBUHIRO TERAI*
Affiliation:
Division of Information System Design, Ashikaga Institute of Technology, 268-1 Omae, Ashikaga, Tochigi 326–8558, Japan
*
✉[email protected]
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Abstract

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Let $m$, $a$, $c$ be positive integers with $a\equiv 3, 5~({\rm mod} \hspace{0.334em} 8)$. We show that when $1+ c= {a}^{2} $, the exponential Diophantine equation $\mathop{({m}^{2} + 1)}\nolimits ^{x} + \mathop{(c{m}^{2} - 1)}\nolimits ^{y} = \mathop{(am)}\nolimits ^{z} $ has only the positive integer solution $(x, y, z)= (1, 1, 2)$ under the condition $m\equiv \pm 1~({\rm mod} \hspace{0.334em} a)$, except for the case $(m, a, c)= (1, 3, 8)$, where there are only two solutions: $(x, y, z)= (1, 1, 2), ~(5, 2, 4). $ In particular, when $a= 3$, the equation $\mathop{({m}^{2} + 1)}\nolimits ^{x} + \mathop{(8{m}^{2} - 1)}\nolimits ^{y} = \mathop{(3m)}\nolimits ^{z} $ has only the positive integer solution $(x, y, z)= (1, 1, 2)$, except if $m= 1$. The proof is based on elementary methods and Baker’s method.

Keywords

exponential Diophantine equationinteger solutionlower bound for linear forms in two logarithms

MSC classification

Secondary: 11D61: Exponential equations
Type
Research Article
Information
Bulletin of the Australian Mathematical Society , Volume 90 , Issue 1 , August 2014 , pp. 9 - 19
Copyright
Copyright ©2013 Australian Mathematical Publishing Association Inc. 

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