= Solution
The quadratic field $K=\mathbb Q(\sqrt{-7})$ is the unique quadratic subfield of $\mathbb Q(\zeta_7)$. Its nontrivial character is the quadratic character
$$
\chi_7(a)=\left(\frac a7\right).
$$
Under the <Artin reciprocity map>, Frobenius at an unramified prime $p$ restricts trivially to $K$ exactly when $\chi_7(p)=1$. The quadratic residues modulo $7$ are $1,2,4$, while $5$ is a nonresidue. Thus
$$
\chi_7(5)=-1,
$$
so the Artin symbol at $5$ is the nonidentity element of $\operatorname{Gal}(K/\mathbb Q)$. Therefore $5$ has residue degree two and is inert in $K$.
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