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A note on the exponential diophantine equation $(am^2+1)^x+(bm^2-1)^y=(cm)^z$

Volume 149 / 2017

Xiaowei Pan Colloquium Mathematicum 149 (2017), 265-273 MSC: Primary 11D61. DOI: 10.4064/cm6878-10-2016 Published online: 21 June 2017

Abstract

Let $a, b, c, m$ be positive integers such that $a+b=c^{2}$, $2\nmid c$, $m \gt 1$ and $m\equiv \pm 1\ ({\rm mod}\,c)$. We prove that if $a\equiv 4$ or $5\ ({\rm mod}\, 8)$, $((a+1)/c)=-1$ and $m \gt 6c^2\log c$, where $((a+1)/c)$ is the Jacobi symbol, then the equation $(am^2+1)^x+(bm^2-1)^y=(cm)^z$ only has the positive integer solution $(x, y, z)=(1, 1, 2)$.

Authors

  • Xiaowei PanDepartment of Health Management
    Xi’an Medical University
    Xi’an, 710021 Shaanxi, China
    e-mail

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