I have a question about an exercise from Borel's Linear Algebraic Groups book. Let $p \neq 2$ a prime number with the property that $2$ modulo $p$ is not a square. We going to study the finite group $\operatorname{PSL}_2(\mathbb{F}_p) \cong \operatorname{SL}_2(\mathbb{F}_p)/ \pm I_2$.
Let $\overline{A} \in \operatorname{PSL}_2(\mathbb{F}_p)$ be any element of order $p$, ie $(\overline{A})^p= I_2$ in $\operatorname{PSL}_2(\mathbb{F}_p)$.
Claim: $\overline{A}$ has a representant $A$ in $\operatorname{SL}_2(\mathbb{F}_p)$ which has an eigenvalue $\lambda_A=1$.
I not know how to prove it. By definition, every $\overline{A} \in \operatorname{PSL}_2(\mathbb{F}_p)$ has two possible representants: $A= A \cdot I$ and $-A= A \cdot (-I_2) \in \operatorname{SL}_2(\mathbb{F}_p)$.
Let study $A= \left(\begin{matrix} a & b \\ c & d \end{matrix}\right)$. Then $ad-bc=det(A)=1$ since it lives in $SL_2$.
Passing to algebraic closure $\overline{\mathbb{F}_p}$ in corresponding $\operatorname{SL}_2(\overline{\mathbb{F}_p})$ matix $A$ may have two possible characteristic polynomials $P_A(X)$ (if we exclude the tedious case that $A$ is a scalar matrix):
$$ P_A(X)= (X-r)^2 \text{ or } (X-s) \cdot (X-t) $$
with $r, s, t\in \overline{\mathbb{F}_p}$ such that $s \neq t$. If $P_A(X)= (X-r)^2$, we conclude $r^2= det(A)=1$, so $r= \pm 1$. If $r=1$ we are done, if $r=-1$, we pass to other representant $-A$ and are also done.
I cannot show that case $P_A(X)=(X-s) \cdot (X-t) $ never happens here. Since $A^p= \pm I_2$ we get $s^p= \pm 1$ and $t^p= \pm 1$. Why this leads to a contradiction? That means, why case $P_A(X)=(X-s) \cdot (X-t) $ can't happen in this context?