Show that $2$ is a square in $\mathbb{F}_p$, using an element $\alpha$ obtained from an extension field as a root. This problem asks me to show that $2$ is a square in $\mathbb{F}_p$ iff $p \equiv \pm 1$ mod $8$ through a number of steps, the first step is the following problem:
Let $p$ be an odd prime and let $\alpha$ be the root of $x^4 + 1$ in some extension of $\mathbb{F}_p$, let $y= \alpha + \alpha^{-1}$, show that $y^2 = 2$.
EDIT:
Someone has posted an answer proving the above part of the problem, but I asked them for help solving the remaining parts of the problem. To provide context for the answer they provided, here is the full problem (and my efforts to solve it):
The main problem is to prove that $x^2 = 2$ has a root in $\mathbb{F}_p$ iff $p \equiv \pm 1$ mod $8$. This should be done through a series of steps; part a) is stated above, and part b) and c) are as follows:
b): Show that $y^p = \alpha^p + \alpha^{-p}$
c): Show that $y \in \mathbb{F}_p$ iff $p \equiv \pm 1$ mod $8$.  One may use that a non zero element $a$ in an extension of $\mathbb{F_p}$ lies in $\mathbb{F}_p$ iff $a^{p-1} = 1$
I have proven b), and it turned out to be useful in c) when assuming that $p \equiv \pm 1$ mod $8$ to prove one implication. I also proved that if $a \in \mathbb{F}_p$ then $a^{p-1} = 1$ (by considering the multiplicative group and generating a subgroup whose order divide $p-1$, the usual trick.
What I dont understand is the two other implications, namely $a^{p-1} = 1 \implies a \in \mathbb{F}_p$ and the other part in c), namely $y \in \mathbb{F}_p \implies p \equiv \pm 1$ mod $8$.
 A: Part (a) is actually quite simple; all you need is the algebraic properties of $\alpha$.
We know that $\alpha^4+1=0$, so dividing by $\alpha$ tells us that $\alpha^3+\alpha^{-1}=0$, thus $\alpha^{-1}=-\alpha^3$. We therefore have that
\begin{equation}
y=(\alpha+\alpha^{-1})^2=2+\alpha^2+\alpha^{-2}=2+\alpha^2+\alpha^6=2+\alpha^2(\alpha^4+1)=2
\end{equation}
as desired.
To solve (b), by the freshman's dream, $y^p=\alpha^p+\alpha^{-p}$.
Now, we may approach part (c) by cases:
If $p\equiv \pm 1\pmod 8$, then
\begin{equation}
y^p=\alpha^p+\alpha^{-p}=\alpha^{8n\pm 1}+\alpha^{-8n\mp 1}=(\alpha^8)^n\alpha^{\pm 1}+(\alpha^8)^{-n}\alpha^{\mp 1}=\alpha^{\pm 1}+\alpha^{\mp 1}=y
\end{equation}
(since $\alpha^8=1$) which means that $y$ is a root of $x^p-x$, and thus it must be a member of $\mathbb{F}_p$.
Otherwise, if $p\not\equiv\pm 1\pmod 8$, then we have that $p\equiv \pm 3\pmod 8$, so
\begin{equation}
y^p=\alpha^p+\alpha^{-p}=\alpha^{8n\pm 3}+\alpha^{-8n\mp 3}=(\alpha^8)^n\alpha^{\pm 3}+(\alpha^8)^{-n}\alpha^{\mp 3}=
\alpha^{\pm 3}+\alpha^{\mp 3}=
\alpha^{ 3}+\alpha^{-3}=\alpha^{4}(\alpha^{-1}+\alpha^{-7})=-(\alpha^{-1}+\alpha)=-y
\end{equation}
(since $\alpha^4=-1$ and $\alpha^8=1$) In other words, $y$ is not a root of $x^p-x$, so it must not be a member of $\mathbb{F}_p$.
Therefore, we have shown that $y\in\mathbb{F}_p$ iff $p\equiv\pm 1\pmod 8$.
To conclude, since the polynomial $x^2=2$ can only have two roots in any extension of $\mathbb{F}_p$, and $y,-y$ are precisely those two roots, then $x^2=2$ has a root in $\mathbb{F}_p$ iff $p\equiv \pm 1\pmod 8$.
Addendum:
Why is it true that $y\in\mathbb{F}_p$ iff $y$ is a root of $x^p-x$?
To see this, note that $x^p-x$ is a degree $p$ polynomial in $\mathbb{F}_p$, so it must have at most $p$ roots in any extension of $\mathbb{F}_p$. Also, note that by Fermat's little theorem, every one of the $p$ distinct elements of $\mathbb{F}_p$ is a root of $x^p-x$. This means that the roots of $x^p-x$ in an extension of $\mathbb{F}_p$ are precisely the elements of $\mathbb{F}_p$.
A: This is somewhat similar to C-RAM's solution, but simpler.
We have been given that $\alpha$ is a root of $x^4 + 1$
Thus $\alpha^4 +1 =0$
Clearly, $\alpha \not = 0$
divide by $\alpha ^2$
we get $\alpha^2 + \alpha^{-2} = 0$
I hope you can take it from here.
