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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.
Mina et al. (Tue,) studied this question.