If $n$ is a positive integer with $n\equiv 2\pmod 3$ then I want to show that each odd divisor of $n^2+n+1$ is congruent to $1\pmod 3$.
I have done the following:
Let $d$ be an odd divisor of $n^2+n+1$.
Then $$n^2+n+1\equiv 0\bmod d \Rightarrow (n-1)(n^2+n+1)\equiv 0\bmod d \Rightarrow n^3-1\equiv 0 \bmod d \Rightarrow n^3\equiv 1\bmod d$$
Since $n\equiv 2\pmod 3$ we have that $n=2+3k$. Then $$n^3=(2+3k)^3=27k^3+54k^2+36k+8$$
We have that $n^3\equiv 1\bmod d$ therefore we get $$27k^3+54k^2+36k+8\equiv 1\bmod d \Rightarrow 27k^3+54k^2+36k+7\equiv 0\bmod d$$
Is everything correct so far? How could we continue?