$x=y^p-y$, $y \in \mathbb{F}_{p^2}$. Prove: $x^2 \in \mathbb{F}_p$

Question

Im practicing for an exam, can anyone help me with this?

Let $p$ be prime and let $x=y^p-y$, for $y \in \mathbb{F}_{p^2}$. Prove: $x^2 \in \mathbb{F}_p$

Answer 1

Since no one else seems inclined to answer and @user8268's comment is the key to the solution, here is my own comment converted into an answer. I have marked it as Community wiki.

Since $x = y^p-y \in \mathbb F_{p^2}$ has one conjugate $x^p = (y^p-y)^p = y^{p^2}-y^p = y - y^p = -x$, the minimal polynomial of $x$ in $\mathbb F_p[z]$ is $(z-x)(z-(-x)) = (z-x)(z+x) = z^2-x^2$ which shows that $-x^2$, and hence $x^2$, belongs to $\mathbb F_p$.