Show that the odd prime divisors of $n^2+n+1$ are of form $6k+1$ (exclude 3) Show that the odd prime divisors of $n^2+n+1$ are of form $6k+1$ (exclude 3)

I have started like below:
$n^2+n+1\equiv 0 \pmod {p_i}$
$(n+1)^2\equiv n \pmod {p_i}$  
Any hints/help on how to proceed from here ?

Let $p$ be an odd prime divisor of $n^2+n+1$
Then   $n^2+n+1  \equiv 0 \pmod{p}
\implies n^3-1  \equiv 0 \pmod{p}
\implies n^3  \equiv 1 \pmod{p}$
So $ord_pn=3$ (obvious!) and so $3|\phi(p)=p-1$
Note that $p-1$ is even. So $2|p-1$.
Since $gcd(2, 3)=1$, so $6|p-1\implies p-1=6k, k\in\Bbb{Z}\implies p=6k+1$
