solution of the Diophantine equation of the form $(2^n)^x + p^y = z^2 $

Question

Can we find solutions of Diophantine equations of the form :

$$(2^n)^x + p^y = z^2 $$

where $k, x, y, z$ and $n$ are positive integers.

-Richard Simson

Answer 1

Some easily-spotted solutions:

1. $p = 0$, $nx$ is even, $z$ is a square root of $(2^{nx})$.
2. $x = y = 2$, $(2^{n},p,z)$ is a Pythagorean triple, e.g. $(4,3,5)$.
3. In 2., adding $k$ to $n$ and multiplying $p$ and $z$ by $2^{k}$ will produce a new solution for any positive integer $k$.

Answer 2

Another simple solution:

x = 1, p = 2, n even, y = n+3

Then, putting m = n/2:

$(2^n)^x + p^y = 2^n + 2^3.2^n = (1 + 2^3).2^n = 9.2^n = 9.4^m = (3.2^m)^2$