On the Solutions of the Diophantine Equation $p^x +(p+4k)^y=z^2$ for Prime Pairs $p$ and $p+4k$

Abstract

In this paper, we solve the Diophantine equation px + (p + 4k)y = z2 in N0 for prime pairs (p, p+ 4k). First, we consider cousin primes p and p+ 4. Then we extend the study to solving px + (p + 4)y = z 2n, where n ∈ N\{1}. Furthermore, we solve the equation px + (p + 4k)y = z2 for k ≥ 2. As a result, we show that this equation has a unique solution (p, p + 4k, x, y, z) =(3, 11, 5, 2, 122) whenever x 1 and y 1. Finally, we show the finiteness of number of solutions in N.