Every odd divisor is $\equiv 1\bmod 3$ 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 $1\pmod 3$. 
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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? 

Hint: Start with odd prime divisors. As you have correctly shown, for any $p\mid n^2+n+1$ we have that
$$n^3\equiv 1\pmod p.$$
Hence $\text{ord}_{p}(n)\mid 3$. Since $n\equiv 2\not\equiv 1\pmod p$ we know that $\text{ord}_{p}(n)=3$. However, we also know that for every $m$ and $x$ coprime to $m$ that $\text{ord}_m(x)\mid \phi(m)$. How can we apply this in this scenario? Once we know the statement holds for every odd prime divisor, what can we say about every odd divisor?
