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On the number of solutions of $x^2-4m(m+1)y^2=y^2-bz^2=1$


Author: Pingzhi Yuan
Translated by:
Journal: Proc. Amer. Math. Soc. 132 (2004), 1561-1566
MSC (2000): Primary 11D09; Secondary 11D25
DOI: https://doi.org/10.1090/S0002-9939-04-07418-0
Published electronically: January 20, 2004
MathSciNet review: 2051114
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Abstract | References | Similar Articles | Additional Information

Abstract: In this paper, using a result of Ljunggren and some results on primitive prime factors of Lucas sequences of the first kind, we prove the following results by an elementary argument: if $m$ and $b$ are positive integers, then the simultaneous Pell equations

\begin{displaymath}x^2-4m(m+1)y^2=y^2-bz^2=1\end{displaymath}

possesses at most one solution $(x,y,z)$ in positive integers.


References [Enhancements On Off] (What's this?)

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Additional Information

Pingzhi Yuan
Affiliation: Department of Mathematics, Zhongshan University, Guangzhou 510275, P.R. China
Email: yuanpz@csru.edu.cn, mcsypz@zsu.edu.cn, yuanpz@mail.csu.edu.cn

DOI: https://doi.org/10.1090/S0002-9939-04-07418-0
Keywords: Simultaneous Diophantine equations, Pell equations, Lucas sequences
Received by editor(s): September 3, 2002
Published electronically: January 20, 2004
Communicated by: David E. Rohrlich
Article copyright: © Copyright 2004 American Mathematical Society

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