Frobenius restriction for p=5 and q=3

Given: The distinct odd primes $p=5$ and $q=3$, a fixed primitive fifth root of unity ${\zeta }_5$, the Gauss sum ${\tau }_5$ attached to it, and $p^{\ast }=(-1{)}^{(p-1)\mathrm{/}2}p$ (Quadratic Gauss sum in a prime cyclotomic field).

For distinct odd primes $p,q$, the arithmetic Frobenius of $q$ in $\mathbb{Q}({\zeta }_p)$ acts on the quadratic subfield $\mathbb{Q}(\sqrt{p^{\ast }})$ by ${\tau }_p\mapsto \left(\frac{p^{\ast }}{q}\right){\tau }_p$, and $\left(\frac{p^{\ast }}{q}\right)=\left(\frac{q}{p}\right)$ (Quadratic reciprocity as a Frobenius restriction identity).

For $p=5$ one has $p^{\ast }=(-1{)}^2\cdot 5=5$ and ${\tau }_5^2=5$, so ${\tau }_5=\epsilon \sqrt{5}$ for some $\epsilon \in \{1,-1\}$ (Square of the quadratic Gauss sum). For the standard complex root ${\zeta }_5=e^{2\pi i\mathrm{/}5}$ one has $\epsilon =1$; moreover $\mathbb{Q}({\tau }_5)=\mathbb{Q}(\sqrt{5})$ is the unique quadratic subfield of $\mathbb{Q}({\zeta }_5)$ (Quadratic Gauss sum at p=5, Quadratic subfield generated by the Gauss sum).

Legendre symbols: $\left(\frac{3}{5}\right)=-1$ because the nonzero squares modulo $5$ are $1,4$ and $3$ is not among them, and $\left(\frac{5}{3}\right)=\left(\frac{2}{3}\right)=-1$ because $5\equiv 2(\mathrm{m}\mathrm{o}\mathrm{d}3)$ and the only nonzero square modulo $3$ is $1$ (The Legendre symbol, including its zero value, Quadratic residues and nonresidues modulo an integer).