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

Author(s): Pingzhi Yuan
Journal: Proc. Amer. Math. Soc. 132 (2004), 1561-1566.
MSC (2000): Primary 11D09; Secondary 11D25
Posted: January 20, 2004
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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.

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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: 10.1090/S0002-9939-04-07418-0
PII: S 0002-9939(04)07418-0
Keywords: Simultaneous Diophantine equations, Pell equations, Lucas sequences
Received by editor(s): September 3, 2002
Posted: January 20, 2004
Communicated by: David E. Rohrlich
Copyright of article: Copyright 2004, American Mathematical Society

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