
Odd prime divisors
Prove that the odd prime divisors of the integer $\displaystyle n^2 + n + 1 $ that are different from 3 are of the form 6k+1.
Proof.
Let p be an odd prime of $\displaystyle n^2 + n + 1 $ that is not 3, then $\displaystyle n^2 + n \equiv 1 \ (mod \ p)$
But I'm stuck in here

If $\displaystyle pn^2+n+1$ then $\displaystyle x^2+x+1\equiv 0 (\bmod p)$ is solvable. This means $\displaystyle (2x+1)^2 \equiv 3(\bmod p)$ is solvable. Thus, $\displaystyle (3/p) = 1$ this means $\displaystyle (3/p)=(1/p) = 1$ or $\displaystyle (3/p)=(1/p) = 1$. Can you find the the primes which make it true?