# Find all solutions of the equation 2^x+3^y=z^2

by .d9n.
Tags: 3yz2, equation, solutions
 Sci Advisor HW Helper P: 2,020 Sorry, I'm too tired to read all that, so let me indicate what I'd meant with my hint. Looking at the equation mod 8 and mod 3, and using the fact that ##2\nmid z## and ##3\nmid z##, we can conclude that x and y must be even. Thus the equation $$(2^{x/2})^2 + (3^{y/2})^2 = z^2$$ is primitive Pythagorean, and the solutions to such equations are fully known (google it). If you write down a parametrization for these Pythagorean triples, and use the fact that the only integer solutions to $$2^a = 3^b + 1$$ are ##(a,b) \in \{(1,0), (2,1)\}## (easy to prove), you should be able to obtain all positive integral solutions (x,y,z) to the original equation. (I got (3,0,3) and (4,2,5) as the only solutions.)