WSEAS Transactions on Mathematics
Print ISSN: 1109-2769, E-ISSN: 2224-2880
Volume 19, 2020
On the Diophantine Equation (4^n)^x − P^y = Z^2
Authors: ,
Abstract: In this paper, the Diophantine equation $$(4^n)^x − p^y = z^2.$$, where $$p$$ is an odd prime, $$n ∈ Z^+$$ and $$x,y,z$$ are non-negative integers, has been investigated to show that the solutions are given by $${(x, y, z, p)} = {(k, 1, 2^{nk} − 1, 2^{nk+1} − 1)} ∪ {(0, 0, 0, p)}.$$