$(r^2-s^2)^2-(5\cdot\min\{r,s\})=2015$. Find all positive integer solution of this equation.

Question

I know the $\min\{x,y\}$ means the minimum value of $x$ and $y$. and it can be expressed as, $\min\{x,y\}= \frac12\left( x+y-\sqrt{(x-y)^2}\right)$

Answer 1

By symmetry, assume $r>s$. There is the solution $(7,2)$ when $r^2-s^2=45$ Otherwise, $$r^2-s^2\geq50\\ 5s\geq485\\ s\geq97\\ r\geq98\\ (r^2-s^2)^2-5s\geq(2r-1)^2-5r=4r^2-9r+1>3r^2>2015 $$