How many solutions of the equation $x^2-y^2=1$ are there with $x,y\in\mathbb F_{p^n}$?

Question

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?

Answer 1

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.

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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 $x\in F$.

Answer 2

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.

Answer 3

This question most likely is arising from an algebraic geometry context. In order to better visualize it, suppose for a moment that you are working in $\mathbf{R}$. Then you are considering the conic $x^2 - y^2 = 1$. All the points on such a curve of degree 2 has an easy parameterization via Bezout's Theorem, informally saying that the number of points of intersection is equal to the product of the degree. Therefore, we can consider using a family of lines (degree 1) to intersect this degree 2 curve which by Bezout's Theorem should give us two intersection points. Now, if we choose this line more carefully, in particular by fixing a predetermined point on the curve and considering lines going through that point, it will intersect the conic at exactly one other point (some care needs to be taken here).

Now, there is nothing special about $\mathbf{R}$ other than giving you an easy way to visualize the situation. Bezout's theorem holds for general fields, including those of positive characteristic as you have here. So apply the above discussion to the problem at hand. You should be able to easily find one point on the curve and then you can get all of the other points as a parameterization. Don't forget about the parenthetical warning of what may go wrong (draw a picture in the real case with such a family of lines). At this point, the theory mostly stops and the computation kicks in.