The Number of Solutions of Pell Equations $x^2–k{y}^2=N$ And $x^2+2y–k{y}^2=N$ Over ${\mathbb{F}}_p$

Abstract

Let $p$ be a prime number such that $p\equiv 1,3(\mathrm{mod}4)$, let ${\mathbb{F}}_p$ be a finite field, and let $N\in {\mathbb{F}}_p^{\ast }={\mathbb{F}}_p–\{0\}$ be a fixed element. Let $P_p^k(N):x^2–k{y}^2=N$ and ${\tilde{P}}_p^k(N):x^2+2y–k{y}^2=N$ be two Pell equations over ${\mathbb{F}}_p$, where $k=\frac{p-1}{4}$ or $k=\frac{p-3}{4}$, respectively. Let $P_p^k(N)({\mathbb{F}}_p)$ and ${\tilde{P}}_p^k(N)({\mathbb{F}}_p)$ denote the set of integer solutions of the Pell equations $P_p^k(N)$ and ${\tilde{P}}_p^k(N)$, respectively. In the first section, we give some preliminaries from the general Pell equation $x^2–k{y}^2=\pm N$. In the second section, we determine the number of integer solutions of $P_p^k(N)$. We prove that $P_p^k(N)({\mathbb{F}}_p)=p+1$ if $p\equiv 1(\mathrm{mod}4)$ or $p\equiv 7(\mathrm{mod}12)$ and $P_p^k(N)({\mathbb{F}}_p)=p-1$ if $p\equiv 11(\mathrm{mod}12)$. In the third section, we consider the Pell equation ${\tilde{P}}_p^k(N)$. We prove that ${\tilde{P}}_p^k(N)({\mathbb{F}}_p)=2p$ if $p\equiv 1(\mathrm{mod}4)$ and $N\in Q_p$; ${\tilde{P}}_p^k(N)({\mathbb{F}}_p)=0$ if $p\equiv 1(\mathrm{mod}4)$ and $N\notin Q_p$; ${\tilde{P}}_p^k(N)({\mathbb{F}}_p)=p+1$ if $p\equiv 3(\mathrm{mod}4)$.