I am trying to solve this problem:
Let $p \neq 5$ an odd prime and let $\omega$ a $5$th root of $1$ in some extension of $\mathbb{F}_p$. Consider $\alpha = \omega+\omega^{-1}$ and prove:
a) $(2\alpha+1)^2=5$
b)$5$ is a square in $\mathbb{F}_p \Longleftrightarrow \alpha^p=\alpha \Longleftrightarrow p \equiv \pm 1 \ mod \ 5.$
For (a) since $\omega$ is a $5$th square we know that $1+\omega+\omega^2+\omega^3+\omega^4=0$ and then $$(2\alpha+1)^2 = 4\alpha^2+4\alpha+1 = 4(\omega^2+\omega^3+2)+4(\omega+\omega^4)+1 = 9-1=5.$$
However, for part (b) following the hints from @reuns I could almost finish it, but there is still a detail I have to work on.
Since $(2\alpha+1)^5=5$ we have $\alpha = \frac{-1\pm \sqrt{5}}{2}$ then if $\sqrt{5}\in \mathbb{F}_p$ then $\alpha \in \mathbb{F}_p$ and then $\alpha$ is a root of $x^p-x$, that is, $\alpha^p=\alpha$. Conversely if $\alpha$ is a root of $x^p-x$ and we know that this is the product of all factor $(x-\beta)$ with $\beta \in \mathbb{F}_p$ then $\alpha \in \mathbb{F}_p$ and since $2\alpha+1=\pm\sqrt{5}$, then $5$ is a square in $\mathbb{F}_p$.
For the other part if $p \equiv 1 \ mod \ 5$ than $\mathbb{F}_p(\omega)=\mathbb{F}_p$ and then since $\alpha \in \mathbb{F}_p(\omega)$ we have $\alpha \in \mathbb{F}_p$ and then $\alpha^p=\alpha$. If $p\equiv -1 \ mod \ 5$ then $p^2 \equiv 1 \ mod \ 5$ and then $\mathbb{F}_p(\omega)=\mathbb{F}_{p^2}$. Now since $\alpha$ is a root of the quadratic polynomial $p(x)=x^2+x-1$ over $\mathbb{F}_p$ then $m_{\alpha,\mathbb{F}_p}(x)|p(x)$ and then $deg(m_{\alpha,\mathbb{F}_p})=1$ or $2$. If this is one we are done since it will implie that $\alpha \in \mathbb{F}_p$, but if this is $2$ then $\mathbb{F}_p(\alpha)=\mathbb{F}_{\omega}=\mathbb{F}_{p^2}.$
Any hint about how to conclude?
Note: I have a qualifying exam about this things in somedays, so I would like to have some hints about how to give one or two more steps and then I would like to try to solve by myself as much as I can!
Can somebody give me a hint to move?
Thank you very much!