Hello! The given relation can be written as $(ax-1)^2+2ax-a^2=y^2$. Suppose that we can have $\frac{a}{x}<2$.Then $2ax-a^2>0$ thus,since all the numbers are integers, we must necessarily have $(ax-1)^2+2ax-a^2\geq a^2x^2\Leftrightarrow a^2\leq 1$,an obvious contradiction.Thus $\frac{a}{x}\geq 2$. This value can be reached if we choose $(x,y,a)=(n,2n,2n^2-1)$ where $n>100$ is a positive integer.

http://olympiads.mccme.ru/vmo/2015/final/sheets1.pdf 2015 All-Russian Mathematical Olympiad