Originally Posted by

**kingwinner** **Prove that for ANY integer n, $\displaystyle n^2+n+1$ has no prime divisors of the form 6m-1.**

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__Attempt:__

Let p be a prime divisor of $\displaystyle n^2+n+1$.

Then $\displaystyle n^2+n+1$ ≡ 0 (mod p)

=> $\displaystyle n^3 -1 = (n-1)(n^2+n+1)$ ≡ 0 (mod p)

=> $\displaystyle n^3$ ≡ 1 (mod p)

Let $\displaystyle e_p(n)$ = order of n mod p

=> $\displaystyle e_p(n)$|3

=> $\displaystyle e_p(n)$= 1 or 3

Now I'm stuck here. How to finish the proof from here?

Any help is greatly appreciated!

[also under discussion in math links forum]