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 $

Abstract

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, congruence methods and the quadratic reciprocity law, we show that 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 $ has only the positive integer solution $ (x, y, z) = (2, 2, 2) $.

Keywords

• exponential diophantine equations
• positive integer solution
• quadratic residue