Without loss of generality, let's assume $x \le y$ which means $\sqrt{x-\frac{1}{5}} \le \sqrt{y-\frac{1}{5}}$. Therefore, $$\sqrt{x-\frac{1}{5}}\le\frac{\sqrt{5}}{2}\implies x = 1 \implies y=2$$ Therefore all solutions are $\boxed{(1, 2), (2, 1)}$.