TITLE: Show that the odd prime divisors of $n^2+1$ are of form $4k+1$
QUESTION [9 upvotes]: Show that the odd prime divisors of $n^2+1$ are of form $4k+1$

I have below so far
$n^2 + 1 \equiv  0 \mod p_i$
$n^4  \equiv  1 \mod p_i$
$4 \mid \phi(p_i)$
I am not sure where to go from here. Any hint/help ?

REPLY [3 votes]: As requested I'm including the answer given in the comment as an answer.
Since $4\mid \phi(p_i) = p_i -1$ then $p_i-1 = 4k$ for some integer $k$. And thus $p_i$ is of the form $4k+1$ as desired.