Find all natural numbers $n$, such that polynomial $n^7+n^6+n^5+1$ would have exactly 3 divisors. 
Find all natural numbers $n$, such that polynomial $$n^7+n^6+n^5+1$$ would have exactly 3 divisors.

What I found out is that this polynomial must be a square of prime number, otherwise it will have more than 3 divisors. Also, n must be an even number, otherwise this polynomial  would be divisible by 4 (if it's an odd number, then  it would be possible and would fit conditions only and only if n=1). My guts are telling me that somehow I have to prove that it's impossible with even numbers too, though I have no idea how.

You can factorize this polynomial like this: $$(n+1)(n^2+1)(n^4-n+1)$$ so this one has almost always at least 3 divisors.
Actualy, for $n>1$ we have $n+1>2$, $n^2+1>2$ and $n^4-n+1>2$ so it could be only $n=1$ which works.

How I got this this factorization:
\begin{eqnarray} (n^7+n^5)+(n^6+1) &=& n^5(n^2+1) + (n^2+1)(n^4-n^2+1)\\
 &=& (n^2+1)(n^5+n^4-n^2+1)\\
 &=& (n^2+1)\Big(n^4(n+1)-(n-1)(n+1)\Big)\\
&=&(n+1)(n^2+1)(n^4-n+1)
\end{eqnarray}
