On the exponential Diophantine equation $ \left(\frac{q^{2l}-p^{2k}}{2}n\right)^x+(p^kq^ln)^y = \left(\frac{q^{2l}+p^{2k}}{2}n\right)^z $

On the exponential Diophantine equation $ \left(\frac{q^{2l}-p^{2k}}{2}n\right)^x+(p^kq^ln)^y = \left(\frac{q^{2l}+p^{2k}}{2}n\right)^z $

Let $ k, l, m_1, m_2 $ be positive integers and let both $ p $ and $ q $ be odd primes such that $ p^k = 2^{m_1}-a^{m_2} $ and $ q^l = 2^{m_1}+a^{m_2} $ where $ a $ is odd prime with $ a\equiv 5\pmod 8 $ and $ a\not\equiv 1\pmod 5 $. In this paper, using only the elementary methods of factorization,...

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Bibliographic Details
Main Authors: Cheng Feng, Jiagui Luo
Format: Article
Language:English
Published: AIMS Press 2022-03-01
Series:AIMS Mathematics
Subjects:
exponential diophantine equations
positive integer solution
quadratic residue
Online Access:http://www.aimspress.com/article/doi/10.3934/math.2022481?viewType=HTML

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http://www.aimspress.com/article/doi/10.3934/math.2022481?viewType=HTML

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