# Odd prime divisors

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• April 1st 2008, 10:29 AM
tttcomrader
Odd prime divisors
Prove that the odd prime divisors of the integer $n^2 + n + 1$ that are different from 3 are of the form 6k+1.

Proof.

Let p be an odd prime of $n^2 + n + 1$ that is not 3, then $n^2 + n \equiv -1 \ (mod \ p)$

But I'm stuck in here
• April 1st 2008, 11:39 AM
ThePerfectHacker
If $p|n^2+n+1$ then $x^2+x+1\equiv 0 (\bmod p)$ is solvable. This means $(2x+1)^2 \equiv -3(\bmod p)$ is solvable. Thus, $(-3/p) = 1$ this means $(3/p)=(-1/p) = 1$ or $(3/p)=(-1/p) = -1$. Can you find the the primes which make it true?