<s>[INST] $\omega^2+\omega+1$divides a polynomial The question is 
Show that $f(n)=n^5+n^4+1$ is not prime for $n>4$.
The solution is given as
Let $\omega$ be the third root of unity. Then $\omega^2+\omega+1=0$. Since $\omega^5+\omega^4+1=\omega^2+\omega+1$, we see that $\omega^2+\omega+1$ is a *factor of the polynomial. So *$n^2+n+1|n^5+n^4+1$.
Which polynomial are we referring to in the bold typeface above? And how is $n^2+n+1|n^5+n^4+1$ true due to $\omega^5+\omega^4+1=\omega^2+\omega+1$?
 [/INST] This is rather poor phrasing (in the highlighted solution). What it is saying is that that $\omega,$ which is a root of the irreducible polynomial $x^2+x+1$ is also a root of $x^5+x^4+1,$ and so the gcd of the two polynomials is not $1,$ and since the first polynomial is irreducible, it must divide the second. Now, that said, you can verify this without any fanciness by long division.
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