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On the solvability of the simultaneous Pell equations \(x^2-a y^2=1\) and \(y^2-b z^2= v_1^2\). (English) Zbl 1483.11054

Summary: Let \(a\), \(b\) be fixed positive integers such that \(a\) is not a perfect square and \(b\) is squarefree, and let \(\omega(b)\) denote the number of distinct prime divisors of \(b\). Let \(( u_1, v_1)\) denote the least solution of Pell equation \(u^2-a v^2=1\). Further, for any positive integer \(n\), let \(u_n=\frac{ \alpha^n + \overline{\alpha}^n}{ 2}\) and \(v_n=\frac{ \alpha^n - \overline{\alpha}^n}{ 2 \sqrt{ a}} \), where \(\alpha= u_1+ v_1\sqrt{ a}\) and \(\overline{\alpha}= u_1- v_1\sqrt{ a} \). In this paper, using the basic properties of Pell equations and some known results on binary quartic Diophantine equations, a necessary and sufficient condition for the system of equations \((\ast) x^2-a y^2=1\) and \(y^2-b z^2= v_1^2\) to have positive integer solutions \((x,y,z)\) is obtained. By this result, we prove that if \((\ast)\) has a positive integer solution \((x,y,z)\) for \(\omega(b)\leq2\) or 3 according to \(2\nmid b\) or not, then \(4 u_1^2-\delta=b g^2\) and \((x,y,z)=( u_r, v_r,g v_{[ \frac{ r + 1}{ 2} ]})\), where \(g\) is a positive integer, \( \delta=1\) or 2 and \(r=2\) or 3 according to \(2\nmid b\) or not, \([\frac{ r + 1}{ 2}]\) is the integer part of \(\frac{ r + 1}{ 2} \), except for \((a,b,x,y,z)=(24,2134,47525,9701,210)\)

MSC:

11D09 Quadratic and bilinear Diophantine equations
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