@article{On the Solutions of the Diophantine Equation $p^x +(p+4k)^y=z^2$ for Prime Pairs $p$ and $p+4k$_2021, place={Maryland, USA}, volume={14}, url={https://www.ejpam.com/index.php/ejpam/article/view/3947}, DOI={10.29020/nybg.ejpam.v14i2.3947}, abstractNote={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 âˆˆ N\{1}. Furthermore, we solve the equation px + (p + 4k)y = z2 for k â‰¥ 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 &gt; 1 and y &gt; 1. Finally, we show the finiteness of number of solutions in N.}, number={2}, journal={European Journal of Pure and Applied Mathematics}, year={2021}, month={May}, pages={471–479} }