Second supplement from Frobenius on Q(zeta_8)

Given: An odd prime $q$, a primitive eighth root of unity $\zeta ={\zeta }_8$, and the element $t:=\zeta +{\zeta }^{-1}\in \mathbb{Q}(\zeta )$.

The index $f=8$ is reduced, and $q\nmid 8$; hence the arithmetic Frobenius of $q$ in $\mathbb{Q}({\zeta }_8)\mathrm{/}\mathbb{Q}$ is the power map ${\sigma }_q(\zeta )={\zeta }^q$ (Arithmetic Frobenius is the power map in an unramified cyclotomic field, The cyclotomic extension $K({\mu }_n)$ as a splitting field of $t^n-1$).

${\zeta }^2={\zeta }_4$ has order $4$, so ${\zeta }^2+{\zeta }^{-2}=i+(-i)=0$ and $t^2={\zeta }^2+2+{\zeta }^{-2}=2$; in particular $t$ is a square root of $2$ and $\mathbb{Q}(t)=\mathbb{Q}(\sqrt{2})$ is a degree-two subfield of $\mathbb{Q}({\zeta }_8)$ (Ring of integers of every cyclotomic field, The cyclotomic extension $K({\mu }_n)$ as a splitting field of $t^n-1$).

For every odd integer $m$, writing the residue of $m$ modulo $8$ gives ${\zeta }^m+{\zeta }^{-m}=t$ when $m\equiv \pm 1(\mathrm{m}\mathrm{o}\mathrm{d}8)$ and ${\zeta }^m+{\zeta }^{-m}=-t$ when $m\equiv \pm 3(\mathrm{m}\mathrm{o}\mathrm{d}8)$, because ${\zeta }^4=-1$ and ${\zeta }^3=-{\zeta }^{-1}$. For residues $1,7$ the integer $(m^2-1)\mathrm{/}8$ is even, and for residues $3,5$ it is odd. Thus ${\sigma }_q(t)=(-1{)}^{(q^2-1)\mathrm{/}8}t$. [F1, algebra]

Euler's criterion: for every integer $a$ and odd prime $q$, $(a\mathrm{/}q)\equiv a^{(q-1)\mathrm{/}2}(\mathrm{m}\mathrm{o}\mathrm{d}q)$, and $(a\mathrm{/}q)\in \{-1,0,1\}$ (Euler's criterion: $(a\mathrm{/}p)\equiv a^{(p-1)\mathrm{/}2}(\mathrm{m}\mathrm{o}\mathrm{d}p)$, The Legendre symbol, including its zero value).