Show that the odd prime divisors of $n^2+1$ are of form $4k+1$

Question

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 ?

You almost have it: note that $\phi (p_i) = p_i - 1$ for $p_i$ prime, so $p_i - 1 = 4k$ for some integer $k$, therefore, $p_i$ is of the form $4k+1$. — Darth Geek, Sep 10 '14 at 17:21
@DarthGeek: Please consider making your comment an answer, so that the OP can accept it, and this question will exit the unanswered queue. — Kieren MacMillan, Jun 06 '17 at 19:19
How can we prove the converse statement ? That is , if $p = 4k + 1$ ,then $n^2 \equiv -1 (mod p) $ , where $p$ is prime ? — Vivek, Feb 24 '19 at 04:33
@KierenMacMillan I don't get how $4|\varphi(p)$ comes. Could you shed a light on this way please? I'm asking you because you seem to be the one who most recently visited SE among them :) — VIVID, Sep 15 '20 at 05:26

score 3 · Answer 1 · answered Jun 06 '17 at 22:09

3

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.

answered Jun 06 '17 at 22:09

Darth Geek

1 Answers1