Show that $\psi_{b}=\psi_{c}\Leftrightarrow b=c$.

Question

Let $K$ finite field with $|K|=p^{n}$,where $p$ is prime number.Let the transformation (Trace) : $$Tr:K\to K,\ Tr(a)=a+a^{p}+a^{p^{2}}+\cdots a^{p^{n-1}}.$$ For every $a\in K$ we khow it's true that $Tr(a)\in \mathbb{Z}_{p}$ and $Tr$ is $\mathbb{Z}_{p}$- linear transformation.
Let $b\in L$.We define the following transformation :
$$\psi_{b}:K\to K,\ \psi_{b}=Tr(ba).$$ We have that $\psi_{b}$ is $\mathbb{Z}_{p}$- linear tranformation, for every $b\in K$. $\quad$ But how can we prove that $\psi_{b}=\psi_{c} \Leftrightarrow b=c\ ?$
It's obvius that, if $b=c\ $ then $\ \psi_{b}=\psi_{c}$. Let we assume now, that for $b,c\in K$ we have that $$\psi_{b}=\psi_{c}\Leftrightarrow Tr(ba)=Tr(ca)\Leftrightarrow Tr((b-c)a)=0,\ \forall\ a\in K.$$ I suppose that we must find the right choice of $a\in K$ to give us the required result but i couldn't something more.
How can we prove that if $\psi_{b}=\psi_{c}\ $ then $\ b=c$ ?

Answer 1

Hint: $\psi_b-\psi_c=\psi_{b-c}$. And if $b-c\neq0$ then $\psi_{b-c}$ is a polynomial of degree $p^{n-1}$. Therefore its number of zeros ____?