48th Austrian Mathematical Olympiad Regional Competition (Qualifying Round)

If $n$ is odd, then both $a$ and $b$ are odd and therefore $n=a^2+b^2$ is even, contradiction. Therefore, $n$ is even and $a=2$. This also shows that $b$ is even. Furthermore, $b\mid (n-b^2)=a^2=4$. Thus $b\in \{2,4\}$, which results in $n=8$ and $n=20$, respectively. $\square$