How does $n+2\mid(n^2+n+1)(n^2+n+2)$ imply $n+2\mid12$? I'm trying to show that if $n+2\mid(n^2+n+1)(n^2+n+2)$, then $n+2 \mid 12$ for all $n\in \mathbb{N}$.

Observe that $\displaystyle n^2+n=n(n+2)-(n+2)+2=(n+2)(n-1)+2$
$\displaystyle\implies (n^2+n+1)(n^2+n+2)=\{\underbrace{(n+2)(n-1)+2}+1\}\{\underbrace{(n+2)(n-1)+2}+2\}\equiv(2+1)(2+2)\pmod{(n+2)(n-1)}\equiv12\pmod{(n+2)} $
