This question is from a rings and modules course I'm doing:
Let $p>2$ be prime and $f:\mathbb{F}_{p}^{*}\rightarrow \mathbb{F}_{p}^{*}$ be given by $x \mapsto x^\frac{p-1}{2}$. Prove that $ker(f)$ has at most $\frac{p-1}{2}$ elements and $im(f)=\left \{ \pm1 \right \}$.
Attempt: Let $\mathbb{F}_{p}^{*}={1,x_{1},y_{1},...,x_{n},y_{n}}$ where $x_{i}y_{i}=1$
$x_{i} \in ker(f) \Rightarrow x_{i}^\frac{p-1}{2}=1$ Therefore $(x_{i}y_{i})^\frac{p-1}{2}=1^\frac{p-1}{2} \Rightarrow x_{i}^\frac{p-1}{2}y_{i}^\frac{p-1}{2}=1 \Rightarrow y_{i}^\frac{p-1}{2}=1 \Rightarrow y_{i} \in Ker(f)$
This is as far as I've got since it seems to say that if $\mathbb{F}_{p}^{*}=\mathbb{F}_{p}\backslash\left \{ 0 \right \}$ then $ker(f)$ can be as large as $p-1$?
I had assumed that what I wanted to prove was that if $x_{i} \in ker(f)$ then $y_{i} \notin ker(f)$ but that doesn't seem to work?
Thanks in advance for your help.
Hints:
$$\begin{align*}(1)&\;\;\forall x\in\Bbb F_p^*\;,\;\;x^{p-1}=1\\ (2)&\;\;\text{For}\;\;x\in\Bbb F_p^*\;,\;\;x^2=1\iff x=\pm 1\\(3)&\;\;\forall\,n\in\Bbb N\;,\;\;\text{a polynomial of degree $\,n\,$ over}\;\;\Bbb F_p^n\;\;\text{has at most $\,n\,$ different roots}\end{align*}$$