Prove that $\left(\frac2p\right) = 1$ if $p \equiv 1,7 \pmod 8$ and $\left(\frac2p\right) = -1$ if $p \equiv 3,5 \pmod 8$ using ring theory

Question

Let $p$ be an odd prime number and let $\alpha = [X] \in R=\mathbb F_p[X]/\langle X^4+1\rangle$, and $y = \alpha + \alpha^{-1}$

I've proven:

1) $\alpha$ is a primitive eight root of unity in $R$.

2) $y^2 = 2$ and $y^p = \alpha^p + \alpha^{-p}$.

$\alpha^{-p} = {(\alpha^{-1})}^{p}$ right ?

3) $y^p = y$ if $p \equiv 1,7 \pmod 8$ and $y^p = -y$ if $p\equiv 3,5 \pmod 8$.

However I need help proving that $\left(\frac2p\right) = 1$ if $p \equiv 1,7 \pmod 8$ and $\left(\frac2p\right) = -1$ if $p \equiv 3,5 \pmod 8$. This should be accomplished using what I've already proven according to the text.

Please dont forget comment on my "$\alpha^{-p} = {(\alpha^{-1})}^{p}$ right ?".

Thanks.

Answer 1

Yes, right, $\alpha^{-p} = \left(\alpha^{-1}\right)^p$.

Regarding the Legendre symbol, note that

$$y^p = \left(y^2\right)^{(p-1)/2}\cdot y = 2^{(p-1)/2}\cdot y = \left(\frac{2}{p}\right)\cdot y.$$