Reducibility of $x^3+nx+1$ over $ \Bbb Z$
If it is reducible over $\Bbb{Z}$ then it has a root in $\Bbb{Z}$, say $k\in\Bbb{Z}$. Then $k^3+nk+1=0$ so $$-1=k^3+nk=k(k^2+n),$$ which shows that $k$ divides $-1$, so $k=\pm1$. Solving the two equations $$1^3+n\cdot1+1=0\qquad\text{ and }\qquad (-1)^3+n\cdot(-1)+1=0,$$ yields $n=-2$ and $n=0$ as the only values for which the polynomial is reducible over $\Bbb{Z}$.