Quadratic Gauss sum at p=5

Given: The prime $p=5$, the primitive fifth root of unity $\zeta ={\zeta }_5=e^{2\pi i\mathrm{/}5}$, and the Gauss sum $\tau ={\tau }_5={\sum }_{a<mrow><mi\; mathvariant="normal">m</mi><mi\; mathvariant="normal">o</mi><mi\; mathvariant="normal">d</mi></mrow>5}(a\mathrm{/}5){\zeta }^a$.

The nonzero squares modulo $5$ are $1$ and $4$, so $(1\mathrm{/}5)=(4\mathrm{/}5)=1$ and $(2\mathrm{/}5)=(3\mathrm{/}5)=-1$; hence $\tau =\zeta -{\zeta }^2-{\zeta }^3+{\zeta }^4$ (Quadratic Gauss sum in a prime cyclotomic field, The Legendre symbol, including its zero value).

${\tau }^2=p^{\ast }=(-1{)}^{(5-1)\mathrm{/}2}5=5$ (Square of the quadratic Gauss sum).

For every $b$ not divisible by $5$, the automorphism ${\sigma }_b(\zeta )={\zeta }^b$ satisfies ${\sigma }_b(\tau )=(b\mathrm{/}5)\tau$ (Galois action on the quadratic Gauss sum).

${\mathrm{\Phi }}_5(t)=1+t+t^2+t^3+t^4$ and ${\mathrm{\Phi }}_5(\zeta )=0$ (${\mathrm{\Phi }}_{p^r}(t)={\sum }_{k<p}{t}^{k{p}^{r-1}}$, and ${\mathrm{\Phi }}_{p^r}(t+1)$ is Eisenstein at $p$, Over a field whose characteristic does not divide $n$, the roots of ${\mathrm{\Phi }}_n$ are exactly the primitive roots of unity).

The fifth roots of unity are ${\zeta }^k=\exp (2\pi ik\mathrm{/}5)$, so $\zeta =\exp (2\pi i\mathrm{/}5)$ and ${\zeta }^4={\zeta }^{-1}$; by Euler's formula $\zeta +{\zeta }^{-1}=2\cos (2\pi \mathrm{/}5)$ (The $n$-th roots of a complex number and the $n$ distinct roots of unity for every $n\ge 1$, Euler's formula: $\exp (i\theta )=\cos \theta +i\sin \theta$ for every real $\theta$). Moreover $\cos (2\pi \mathrm{/}5)=\sin (\pi \mathrm{/}2-2\pi \mathrm{/}5)=\sin (\pi \mathrm{/}10)>0$, using the cofunction identity and positivity of sine on $(0,\pi )$ (Cofunction, supplementary, quarter-turn, and reflection identities for the six trigonometric functions, Pi is the first positive zero of sine).

$\mathrm{\mid }\exp (i\theta )\mathrm{\mid }=1$ for real $\theta$ ($\exp (x+iy)=e^x(\cos y+i\sin y)$, $\mathrm{\mid }\exp (x+iy)\mathrm{\mid }=e^x$, and $e^{i\pi }+1=0$).