When is $n^{2015}+n+1$ prime?

\begin{eqnarray*} n^{2015} +n+1 &=& n^{2015} -n^2+n^2+n+1 \\ &=& n^2(n^{2013} -1) + n^2+n+1 \\ &=& n^2(n^3-1)\underbrace{(n^{2010}+n^{2007}+...+n^3+1)}_a +n^2+n+1 \\ &=& (n^2+n+1)(an^2(n-1)+1) \\ &=& \end{eqnarray*} Since $n^2+n+1\geq 3$ this could be prime only if $an^2(n-1)+1 =1$ and this is only when $n=1$.

$2015\equiv 2\pmod{3}$ implies that $\Phi_3(n)=n^2+n+1$ is a divisor of $n^{2015}+n+1$, so...