Roots of a quadratic polynomial over a finite field I am having a hard time figuring this problem out. I have not come up with anything substantial in terms of solving it. Any help would be great. 
Show that if $z \in F_{p^2}$ is a root of the polynomial $g = X^2 +aX + b$ where $a, b \in F_p$, then $z^p$ is also a root of $g$. 
Also, verify that $a = - z - z^p$ and that $b = z^{p+1}$ provided that $z \notin F_p$. Is this true if $z \in F_p$? Explain your answer.
 A: Consider the Frobenius morphism $\;\begin{aligned}[t]\mathbf F_{p^2}&\longrightarrow \mathbf F_{p^2}\\x&\longmapsto x^p\end{aligned}$.  Its fixed points are the prime field $\;\mathbf F_p$. Hence, if $z^2+az+b=0$, then 
$$(z^2+az+b)^p=(z^p)^2+az^p+b=0, $$
which proves $z^p$ is a root of $X^2+aX+b$.
On the other hand, the sum of the roots of this polynomial, in any splitting field, is equal to $-a$ and their product is equal to $c$ (Vieta's relations). Thus, if $z^p\neq z$, we have two roots of the equation, and
$$z+z^p=-a, \quad z z^p=z^{p+1}=c.$$
If $z\in \mathbf F_p$, this is no more true since $z$ and $z^p$ are  the same root.
A: For the first part, it is known that since the characteristic of $\Bbb F_{p^2}$ is $p$, then $(x+y)^p = x^p + y^p$. Also, Fermat's little theorem says that for $x \in \Bbb F_p \setminus \{0\}$ one has $x^{p-1} = 1$, which implies that $x^p = x \ \forall \ x \in \Bbb F_p$. Using these, $0 = (g (z))^p = (z^2 + az + b)^p = z^{2p} + a^p z^p + b^p = (z^p)^2 + a z^p + b$, so $z^p$ verifies the equation too.
For the second part, use Viète's relations: $z + z^p = -a$ and $z z^p = b$, provided $z^p$ be the "other" root of the equation besides $z$ (i.e. provided $z^p \ne z$, which is false when $z \in \Bbb F_p$ as shown above and true when $z \notin \Bbb F_p$).
