A note on the exponential diophantine equation $(am^2+1)^x+(bm^2-1)^y=(cm)^z$
Volume 149 / 2017
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 $(x, y, z)=(1, 1, 2)$.
