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 offtopic by Martin Sleziak, user44191, Sean Lawton, Emil Jeřábek, Jeremy Rickard Mar 14 at 14:09
This question appears to be offtopic. 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

5$\begingroup$ How about $x=y=z=1$? $\endgroup$ – Joel Reyes Noche Mar 14 at 13:26

2$\begingroup$ Is there any motivation for this? $\endgroup$ – Noah Schweber Mar 14 at 13:33

2$\begingroup$ 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. $\endgroup$ – quarague Mar 14 at 13:34

2$\begingroup$ It's trivial to produce a solution for each odd value $\ge 3$ of $z$, with $x,y$ both odd or both even. Not researchlevel. $\endgroup$ – YCor Mar 14 at 14:14

1$\begingroup$ $x=4893263478963457849263789516578165183$, $z=9657385467358467868634867$, $y=\frac12(5^{9657385467358467868634867}3^{4893263478963457849263789516578165183})$. Do you understand now what quarague wrote above? $\endgroup$ – Emil Jeřábek Mar 14 at 14:50