<s>[INST] $ 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? 
 [/INST] The fifth roots of $1$ form a cyclic group of order $5$, so any of them, different from $1$, is primitive. If $w$ is one of the primitive roots, the others are $\omega^2$, $\omega^3=\omega^{-2}$ and $\omega^4=\omega^{-1}=\bar{\omega}$.
It is clear that $(x-\omega)(x-\omega^2)(x-\omega^3)(x-\omega^4)=x^4+x^3+x^2+x+1$. The minimal polynomial of $\omega$ is a factor of this degree $4$ polynomial, so it must have degree $2$ or $4$ (because a degree $3$ polynomial has a real root).
Thus we have to exclude that $\omega$ has degree $2$. If we assume this, then its minimal polynomial must be $(x-\omega)(x-\bar\omega)=x^2-(\omega+\bar\omega)x+1$, which implies $q=\omega+\bar\omega$ is rational.
Therefore also $(x-\omega^2)(x-\omega^3)$ would have rational coefficients and, indeed,
$$
\omega^2+\omega^3=\omega^2+\frac{1}{\omega^2}=
\left(\omega+\frac{1}{\omega}\right)^2-2=q^2-2
$$
Thus
$$
x^4+x^3+x^2+x+1=(x^2-qx+1)(x^2-(q^2-2)x+1)
$$
The coefficient of the term of degree $1$ in the product can be easily computed as
$$
-q-(q^2-2)
$$
so we have $-q^2-q+2=1$, so $q^2+q-1=0$. This is a contradiction, because $\sqrt{5}$ is not rational.
Therefore this polynomial divides $p$, as soon as $p(\omega)=0$.
</s>