On solutions of the simultaneous Pell equations $$ x^{2}-\left( a^{2}-1\right) y^{2}=1$$ x 2 - a 2 - 1 y 2 = 1 and $$y^{2}-pz^{2}=1$$ y 2 - p z 2 = 1

Abstract

Let $$a\ge 2$$ be an integer and p prime number. It is well-known that the solutions of the Pell equation have recurrence relations. For the simultaneous Pell equations $$\begin{aligned}&x^{2}-\left( a^{2}-1\right) y^{2} =1 \\&y^{2}-pz^{2} =1 \end{aligned}$$ assume that $$x=x_{m}$$ and $$y=y_{m}$$ . In this paper, we show that if $$m\ge 3$$ is an odd integer, then there is no positive solution to the system. Moreover, we find the solutions completely for $$5\le a\le 14$$ in the cases when $$m\ge 2$$ is even integer and $$m=1$$ .