, 2015

Suppose for the sake of contradiction that $\operatorname{gcd}\left(c^{2} + d^{2}, a^{2} + b^{2}\right) = 1$. We first notice that
$$(a c + b d)^{2} + (a d - b c)^{2} = \left(a^{2} + b^{2}\right)\left(c^{2} + d^{2}\right).$$
Since $a c + b d$ is divisible by $a^{2} + b^{2}$, it follows that $(a d - b c)^{2}$ is also divisible by $a^{2} + b^{2}$.
Let $p$ be a prime divisor of $a^{2} + b^{2}$ and $n = v_{p}\left(a^{2} + b^{2}\right)$. We have
$$v_{p}\left((a c + b d)^{2} + (a d - b c)^{2}\right) = n$$
since $c^{2} + d^{2}$ is not divisible by $p$. Note that,
$$v_{p}\left((a c + b d)^{2}\right) = 2 v_{p}(a c + b d) \geq 2 v_{p}\left(a^{2} + b^{2}\right) = 2 n > n.$$
This implies that $v_{p}\left((a d - b c)^{2}\right) = n$, which means that $n$ is even. This proves that $a^{2} + b^{2} = u^{2}$ for some positive integer $u$, and therefore both $a c + b d$ and $a d - b c$ are divisible by $u$.
But
$$a\left(c^{2} + d^{2}\right) = c(a c + b d) + d(a d - b c)$$
and
$$b\left(c^{2} + d^{2}\right) = d(a c + b d) - c(a d - b c)$$
are also divisible by $u$. Because $\operatorname{gcd}\left(u, c^{2} + d^{2}\right) = 1$, we deduce that $u$ divides both $a$ and $b$. But $u^{2} = a^{2} + b^{2}$. We have a contradiction.