How many solutions of the equation $x^2-y^2=1$ are there with $x,y\in\mathbb F_{p^n}$? 
Let $p>2$ be an odd prime. Let $\mathbb F_{p^n}$ be the field with $p^n$ elements. How many solutions of the equation $x^2-y^2=1$ are there with $x,y\in\mathbb F_{p^n}$?

My work:
Char $F=p$. 
$x^2-y^2=(x+y)(x-y)=1$. Since we are looking for $x,y\in F$, $x+y,x-y\in F$ and if $x+y=\alpha$, $x-y=\alpha^{-1}$. Hence, $\displaystyle x=\frac{\alpha+\alpha^{-1}}{2}, y=\frac{\alpha-\alpha^{-1}}{2}$. I was stuck afterwards since I cannot find the distinct sets of $x,y$ from these relations. Can anyone please give me a hint?

If you want something even more explicit than André's mention of the fact, note that we have
$$\begin{pmatrix} 1 & -1 \\ 1 & 1\end{pmatrix}\begin{pmatrix} x \\ y\end{pmatrix}=\begin{pmatrix} \alpha \\ \alpha^{-1}\end{pmatrix}$$
since the characteristic is not $2$ and the determinant of that matrix is $2$, we see that
$$2^{-1}\begin{pmatrix} 1 & 1 \\ -1 & 1\end{pmatrix}\begin{pmatrix} \alpha \\ \alpha^{-1}\end{pmatrix}=\begin{pmatrix} x \\ y\end{pmatrix}$$
So that establishes the bijection, in particular since there are $p^n-1$ elements $\alpha\in F$ so that $\alpha^{-1}$ exists, we have there are $p^n-1$ total solutions.

Incidentally, if the characteristic is $2$, this is even easier:  $a^2-b^2=(a-b)^2$ so it's easy to see that all you need to do is solve
$$(x-y)^2-1^2=(x-y-1)^2=0.$$
Of course in this case this clearly has $p^n$ total solutions, all given by $(x, x-1)$ and $x\in F$.
