Solutions of $ 0 = x^2 -ay^2 -1$ in $\mathbb F_q$ where $a$ is not a square.

Question

Assume $F = \mathbb F_q$ where $q = p^r$ for $p$ prime and $r > 0$. I have to count $$ \{(x,y) \in F^2 \mid x^2 -ay^2 -1 = 0\} $$ where $a$ is not a square in $F^*$. The equation is equivalent to $ay^2 = x^2-1 = (x-1)(x+1)$ so $y^2 = a^{-1} (x-1)(x+1)$ which means that $a^{-1}(x-1)(x+1)$ is a square. But $a$ is not so $a^{-1}$ is not thus $(x-1)(x+1)$ must be not a square. This means

• $(x-1)$ is a square and $(x+1)$ not or
• $(x+1)$ is a square and $(x-1)$ not

I use here several times the fact that $F^*$ a cyclic group is. Assume now that $(x-1)$ is a square. Is it then possible that $x+1$ is a square, too ? If this is not true I would get $(q-1)/2$ solutions.

Answer 1

Hint: $F(\sqrt{a})$ is an extension of degree $2$ of $F$, and you have to count the number of its elements of norm $1$. This can be done using the famous Theorem

Hilbert 90. If $L/K$ is a finite Galois extension whose Galois group is cyclic with generator $\sigma$, then $L^*/K^* \cong \ker(N_{L/K} : L^* \to K^*)$ via $[x] \mapsto \sigma(x)/x$