Find all $n$s

[Phlembac Adib Hasan]

Find all positive integer $n$ such that $n^{14}+(n^2+1)(n+1)$ is a prime.

[Phlembac Adib Hasan]

Nobody interested?Well, I'm giving the solution.

Try to prove $n^4+n^3+n^2+n+1|n^{14}+(n^2+1)(n+1)$.

[Nadim Ul Abrar]

How did you figure that out ?

$(n-1)[n^{14}+(n^2+1)(n+1)]=n^{15}-n^{14}+(n-1)(n^3+n^2+n+1)=n^{15}-n^{14}+n^4-1$

now
Since $ x^5\equiv 1mod x^5-1$, $n^{15}-n^{14}+n^4-1 \equiv 0 mod x^5-1$ :p

[Phlembac Adib Hasan]

How did you figure that out ?

$(n-1)[n^{14}+(n^2+1)(n+1)]=n^{15}-n^{14}+(n-1)(n^3+n^2+n+1)=n^{15}-n^{14}+n^4-1$

now
Since $ x^5\equiv 1mod x^5-1$, $n^{15}-n^{14}+n^4-1 \equiv 0 mod x^5-1$ :p

This problem is self-made.Let $f(n)=n^{14}+(n^2+1)(n+1)$.Note that unity has five fifth roots namely $1,\omega,\omega^2,\omega^3,\omega^4$.First expand that thing.Then notice the powers form a complete residue system of $5$.Then this idea comes to put values of the fifth roots of unity to that function.