European Mathematical Cup

If $a$ is a square of an integer, any its power is also square of an integer. If $a$ is not a perfect square, number of its positive divisors is even. We can prove this by pairing divisors of $a$ as $d$ and $\frac{a}{d}$. A divisor $d$ won't be paired with itself because that would imply $a=d^2$. This proves that $d(a)$ is even and hence $a^{d(a)}$ is a perfect square for every positive integer $a$. The extension in the problem is hence a sum of two squares. Every number of the form $4t+3$ can't be written as a sum of two squares because 0 and 1 are the only quadratic residues modulo 4, so it is impossible for a sum of two squares to give remainder 3 modulo 4.