Can a divisor of $n^2 +n+1$ be $2\operatorname{mod} 3$? Let $n$ be a natural number.  It seems that a divisor of $n^2+n+1$ cannot be $2 \operatorname{mod} 3$. 
Couldn't $n^2+n+1$ have an even number of divisors which are $2 \operatorname{mod} 3$?

Let $p$ be a prime $\equiv 2\bmod 3$.  Suppose $p|(n^2+n+1)$.  Then also $p|(n^3-1)$ since $n^3-1=(n-1)(n^2+n+1)$.
So $n^{p-1}\equiv 1\bmod p$ from Fermat's Little Theorem and also $n^3\equiv 1\bmod p$ from above.  Since $p$ is taken to be $\equiv 2\bmod 3$, the latter implies $n^{p+1}\equiv 1\bmod p$.  Then
$n^{3+(p-1)-(p+1)}=n\equiv 1\bmod p$
which is contradictory because $1^2+1+1\equiv 3\not\equiv 0\bmod p$.
