A note on the Diophantine equation $(x^k-1)(y^k-1)^2=z^k-1$
A note on the Diophantine equation $(x^k-1)(y^k-1)^2=z^k-1$
We prove that, for $k \geq 10$, the Diophantine equation $(x^k-1)(y^k-1)^2=z^k-1$ in positive integers $x,y,z,k$ with $z > 1$, has no solutions satisfying $1 < x \leq y$ or $1 < y <x \leq((y^k-1)^{k-2}+1)^{\frac{1}{k}}$.