Prove that the odd prime divisors of the integer $n^2 + 1$ are of the form $4k + 1$.
If odd prime $p$ is a factor of $n^2+1$ then
$$ n^2 \equiv -1 \pmod p $$
From Proposition (7.11) extended by the result from Exercise (7.1).6 we know that $(\frac{-1}{p})=1 \iff p \equiv 1 \pmod 4$. That is, $p=4k+1$ for some integer $k$.