75th Romanian Mathematical Olympiad

For $a$, $b$, $r$ as in the statement, $r=\frac{(\sqrt{a}-\sqrt{b}{)}^2}{2}$, therefore $r\ge 0$. Moreover, $ab$ must be a square of a natural number $n$, so $r=\frac{a+b-2n}{2}=\frac{m}{2}$, where $m$ is a non-negative integer. Therefore, $r=\frac{m}{2}$, with $m$ a non-negative integer.

Conversely, for $a=b=1$, we obtain that $r=0$ is a solution; for $a=m$ and $b=4m$, with $m$ a positive integer, we obtain $r=\frac{5m}{2}-2m=\frac{m}{2}$. Therefore, the set of solutions is $S=\{\frac{m}{2}\mid m\in \mathbb{N}\}$.