Exercise 5.7

Proof. Let $S=\{(\bar{x},\bar{y})\in {𝔽}_p^2|{\bar{x}}^2-{\bar{y}}^2=\bar{a}\}$, and $T=\{(\bar{u},\bar{v})\in {𝔽}_p^2|\bar{u}\bar{v}=\bar{a}\}$. Then $f:\{\begin{array}{ccc}\hfill S\hfill & \hfill \to \hfill & \hfill T\hfill \\ \hfill (\bar{x},\bar{y})\hfill & \hfill \mapsto \hfill & \hfill (\bar{x}+\bar{y},\bar{x}-\bar{y})\hfill \end{array}$ is well defined (if $(\bar{x},\bar{y})\in S,(\bar{x}-\bar{y})(\bar{x}+\bar{y})=a$, so $(\bar{x}+\bar{y},\bar{x}-\bar{y})\in T$). Moreover $f$ is a bijection, with inverse $(\bar{u},\bar{v})\mapsto ((\bar{u}+\bar{v})∕2,(\bar{u}-\bar{v})∕2)$, so $|S|=|T|$.

We compute $|T|$.

$\text{∙}$ Suppose that $p\nmid a$, so $\bar{a}\ne \bar{0}$. For $\bar{v}=0$, there is no solution, and for each $\bar{v}\ne 0$, we obtain the unique solution $(\bar{a}{\bar{v}}^{-1},\bar{v})$, so there exist $p-1$ solutions.

$\text{∙}$ Suppose that $p\mid a$. The solutions of $\bar{u}\bar{v}=\bar{0}$ are $(\bar{0},\bar{0})$ if $\bar{u}=\bar{v}=\bar{0}$, $(\bar{0},\bar{v})$ for each $\bar{v}\ne \bar{0}$, and $(\bar{u},\bar{0})$ for each $\bar{v}=\bar{0}$, that is to say $N=1+(p-1)+(p-1)=2p-1$ solutions.

Conclusion :