Pingzhi Yuan and Yongzhong Hu: On the Diophantine Equation $ax^2+(3a+1)^m=(4a+1)^n$, p.51-59

Abstract:

Let $a$ be a positive integer, and let $p$ be an odd prime such that $p$ does not divide $a(3a+1)$ and $4a+1$ is a power of $p$. In this paper, by the deep result of Bilu, Hanrot and Voutier, i.e. the existence of primitive prime factors of Lucas and Lehmer sequences, by the computation of Jacobi's symbol and by elementary arguments, we prove that: if $a\neq1, \,2$, then the Diophantine equation of the title has at most two positive integer solutions $(x, \,m, \,n)$. Moreover, the diophantine equations $x^2+4^m=5^n$ and $2x^2+7^m=9^n$ have precisely three positive integer solutions $(x, \,m, \,n)$.

Key Words: Generalized Ramanujan-Nagell equations, primitive prime factors,Lucas and Lehmer sequences.

2000 Mathematics Subject Classification: Primary: 11D25,
Secondary: 11D61.

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