$ p \in Q[x] $ has as a root a fifth primitive root of unity, then every fifth primitive root of unity is a root of $p$. I'm extremely stuck. Can't figure it. 
The conjugate is easy: let $w$ be a primitive root of unity, then $w^{-1}$ will also be a root, that's easy. But I'm missing $w^2$ and $w^3$. Why would they be also roots of p? 

Let $R$ be the GCD of $P$ and $X^4+X^3+X^2+X+1$ (considering both polynomials as elements of $\mathbb Q[X]$.
It happens (see here for proof) that $R$ is also the GCD of $P$ and $X^4+X^3+X^2+X+1$ as elements of $\mathbb C[X]$. Since $(X-w)(X-\bar{w})$ divides both, $R$ has degree at least 2.
But $X^4+X^3+X^2+X+1$ is irreducible over $Q$ (Eisenstein can help you with that). Hence $R=X^4+X^3+X^2+X+1$.
Hence $X^4+X^3+X^2+X+1$ divides $P$ and the claim is proved.
