<s>[INST] 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?
 [/INST] Here is a partial answer for $n = 1$. Is it possible to extend this proof generally? I have having trouble doing this since we can no longer use the legendre symbol. 
Anyway, denote $N(x^2-y^2 = 1)$ to be the number of solutions. Then
$$N(x^2-y^2 = 1) = \sum_{a+b = 1} N(x^2 = a)N(y^2 = -b).$$
Now $N(x^2 = a) = 1+ \left( \frac{a}p \right)$ so
$$N(x^2-y^2 = 1) = \sum_{a+b= 1}1 + \sum_a \left( \frac{a}p \right)+\sum_{-b} \left( \frac{-b}p \right)+\sum_{a+b= 1}\left( \frac{a}p \right) \left( \frac{-b}p \right).$$
The first sum is obviously $p$ and it is a well known result that the next two sums are $0$. Now using Jacobi sums (see $\star$) we can arrive at 
$$ \sum_{a+b= 1}\left( \frac{a}p \right) \left( \frac{b}p \right) = - (-1)^{\frac{p-1}2}.$$
So in the sum that we are interested in, we can factor out $\left( \frac{-1}p \right)$ and depending on if $p \equiv \pm 1 \pmod 4$, we have $\left( \frac{-1}p \right) = \pm 1$ and we can find the value of our sum using the result from $\star$. In anycase, we have $$N(x^2-y^2 = 1) = p-1.$$
$\star$ A Classical Introduction to Modern Number Theory by Ireland and Rosen.
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