# A Diophatine equation question [closed]

The question is the following. Consider the Diophantine equation $$3^x + 2y = 5^z$$. Then $$x, y , z$$ must necessarily be even numbers ?

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## closed as off-topic by Martin Sleziak, user44191, Sean Lawton, Emil Jeřábek, Jeremy RickardMar 14 at 14:09

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question does not appear to be about research level mathematics within the scope defined in the help center." – Martin Sleziak, user44191, Sean Lawton, Jeremy Rickard
If this question can be reworded to fit the rules in the help center, please edit the question.

• How about $x=y=z=1$? – Joel Reyes Noche Mar 14 at 13:26
• Is there any motivation for this? – Noah Schweber Mar 14 at 13:33
• Do you mean $2^y$ or actually $2\cdot y$ as currently written? Otherwise you can pick any natural numbers for $x$ and $z$ you want and then compute the $y$ that solves the equation. – quarague Mar 14 at 13:34
• It's trivial to produce a solution for each odd value $\ge 3$ of $z$, with $x,y$ both odd or both even. Not research-level. – YCor Mar 14 at 14:14
• $x=4893263478963457849263789516578165183$, $z=9657385467358467868634867$, $y=\frac12(5^{9657385467358467868634867}-3^{4893263478963457849263789516578165183})$. Do you understand now what quarague wrote above? – Emil Jeřábek Mar 14 at 14:50