BALKAN 2009 Problem 1

We start by observing that $z$ must be even, so $z^{2}=3^{x}-5^{y} \equiv(-1)^{x}-1(\bmod 4)$ is divisible by 4 , which implies that $x$ is even, say $x=2 t$. Then our equation can be rewritten as $\left(3^{t}-z\right)\left(3^{t}+z\right)=5^{y}$, which means that both $3^{t}-z=5^{k}$ and $3^{t}+z=5^{y-k}$ for some nonnegative integer $k$. Since $5^{k}+5^{y-k}=2 \cdot 3^{t}$ is not divisible by 5 , it follows that $k=0$ and

$$2 \cdot 3^{t}=5^{y}+1$$

Suppose that $t \geq 2$. Then $5^{y}+1$ is divisible by 9 , which is only possible if $y \equiv 3$ $(\bmod 6)$. However, in this case $5^{y}+1 \equiv 5^{3}+1 \equiv 0(\bmod 7)$, so $5^{y}+1$ is also divisible by 7 , which is impossible. Therefore we must have $t \leq 1$, which yields a (unique) solution $(x, y, z)=(2,1,2)$.