<s>[INST] 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 $2 \operatorname{mod} 3$?
 [/INST] Theorem: if $q \equiv 2 \pmod 3$ is a prime and $$ x^2 + xy + y^2 \equiv 0 \pmod q \; , \;  $$
then $q$ divides both $x$ and $y$
In particular, under these circumstances, you cannot have $y=1$
Proof: Legendre symbol $$ (-3|q) = (q|3)  $$
using quadratic reciprocity, two cases ($q$ is either $1$ or $3$ $\pmod 4$ )
Also $$ 4( x^2 + xy + y^2) \equiv (2x+y)^2 + 3 y^2 \equiv 0 \pmod q \; , \;  $$
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