<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] As the 3rd root of identity $\omega=\mathrm{e}^{2\pi i/3}$ satisfies $x^5+x^4+1=0$, then so does its conjugate $\bar\omega=\mathrm{e}^{-2\pi i/3}$, and hence $(x-\omega)(x-\bar\omega)=x^2+x+1$ divides $x^5+x^4+1$. 
Note that
$$
x^5+x^4+1=(x^2+x+1)(x^3-x+1).
$$
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