= Solution
Let $\zeta$ be a primitive fifth root of unity. Over $\mathbb F_p$, the <Frobenius automorphism> acts by $\zeta\mapsto\zeta^p$, so the splitting field of $X^5-1$ has degree equal to the <multiplicative order> of $p$ modulo $5$. Its Galois group is generated by Frobenius and is cyclic. The possible orders are therefore $1$ for $p\equiv1$, $4$ for $p\equiv2,3$, and $2$ for $p\equiv4\pmod5$.
Put $\alpha=5^{1/5}$. The polynomial $X^5-5$ satisfies the <Eisenstein criterion> at $5$, so $[\mathbb Q(\alpha):\mathbb Q]=5$. Also $[\mathbb Q(\zeta):\mathbb Q]=\varphi(5)=4$. The intersection of these two fields has degree dividing both $5$ and $4$, hence is $\mathbb Q$. Since the splitting field is $\mathbb Q(\alpha,\zeta)$,
$$
\boxed{[\mathbb Q(\alpha,\zeta):\mathbb Q]=20.}
$$
Label the five roots by $\alpha\zeta^i$, $i\in\mathbb F_5$. The Galois group acts through the <affine group over a finite field>:
$$
i\longmapsto ai+b,
\qquad a\in\mathbb F_5^\times,\quad b\in\mathbb F_5.
$$
Its possible nontrivial cycle types are $5$, $1+2+2$, and $1+4$; in particular, no element has a three-cycle. By the <Dedekind factorization cycle test>, the reduction of $X^5-5$ modulo any prime $p\ne5$ therefore has no irreducible cubic factor.
The multiplier of the Frobenius element is $a=p\pmod5$. If $p\equiv2$ or $3$, then $a$ has order four; every affine map with this multiplier has one fixed point and one four-cycle. Hence $X^5-5$ is a product of a linear factor and an irreducible quartic. If $p\equiv1$, Frobenius is a translation $i\mapsto i+b$: it is the identity when $b=0$, giving complete splitting, and is a five-cycle when $b\ne0$, giving an irreducible polynomial.
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