TITLE: Find all $n$ for which $n^8 + n + 1$ is prime
QUESTION [6 upvotes]: Find all $n$ for which $n^8 + n + 1$ is prime. I can do this by writing it as a linear product, but it took me a lot of time. Is there any other way to solve this? The answer is $n = 1$.

REPLY [9 votes]: HINT:
If $w$ is a complex cube root of unity and $f(x)=x^8+x+1$
$f(w)=(w^3)^2\cdot w^2+w+1=0$
So $(x^2+x+1)|(x^8+x+1)$

REPLY [4 votes]: Since $n^2+n+1$ divides $n^8+n+1$ and $1<n^2+n+1<n^8+n+1$ for $n>1$, then $n=1$ is the unique solution (which indeed gives a prime).