On the determination of solutions of simultaneous Pell equations \(x^2 - (a^2 - 1) y^2 = y^2 - pz^2 = 1\).(English)Zbl 1413.11064
Summary: In this paper, we consider the simultaneous Pell equations \[x^2-(a^2-1)y^2 = 1,\]
\[y^2 - pz^2 = 1,\] where \(p\) is prime and \(a > 1\). Assuming the solutions of the Pell equation \(x^2-(a^2-1)y^2 = 1\) are \(x = x_m\) and \(y = y_m\) with \(m \geq 2\), we prove that the system has solutions only when \(m = 2\) or \(m = 3\). In the case of \(m = 3\), we show that \(p = 2\) and give the solutions of the system in terms of Pell and Pell-Lucas sequences. When \(m = 2\) and \(p \equiv 3 (\bmod 4)\), we determine the values of \(a, x, y\), and \(z\). Lastly, we show that the system has no solutions when \(p \equiv 1(\bmod 4)\).
\[y^2 - pz^2 = 1,\] where \(p\) is prime and \(a > 1\). Assuming the solutions of the Pell equation \(x^2-(a^2-1)y^2 = 1\) are \(x = x_m\) and \(y = y_m\) with \(m \geq 2\), we prove that the system has solutions only when \(m = 2\) or \(m = 3\). In the case of \(m = 3\), we show that \(p = 2\) and give the solutions of the system in terms of Pell and Pell-Lucas sequences. When \(m = 2\) and \(p \equiv 3 (\bmod 4)\), we determine the values of \(a, x, y\), and \(z\). Lastly, we show that the system has no solutions when \(p \equiv 1(\bmod 4)\).
MSC:
| 11D25 | Cubic and quartic Diophantine equations |
| 11B37 | Recurrences |
| 11B39 | Fibonacci and Lucas numbers and polynomials and generalizations |
Software:
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Full Text:DOI
References:
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