Prove that $5\mid x\,$ if $\,x, y \gt 1 $ satisfy $2x^2 - 1=y^{15}$ If $x\gt 1$ and $\,y\gt 1,$ with $ x, y \in \mathbb N$ so  that $(x,y)$ satisfies the equation $$2x^2-1=y^{15},$$ then prove that $5\mid x$.
$\mod {10}$ gave me just what the last digit of y can be.

apply fermat, so $y^{15}\equiv y^3 \bmod 5$. So you have $2x^2\equiv y^3+1$. now cubes $\bmod 5$ are $1,3,2,4,0$. This tells us the statement is false, because when $y^3\equiv 1$ we get $2x^2\equiv 2$ and $x$ can be $1$ or $4 \bmod 5$
