17. (Hong Kong) Find all positive integer solutions \(x, y, z\) for the equation \(3^{x} + 4^{y} = 5^{x}\).
Solving: $x=y=z=2$ is a solution. The following proves that this is the only solution. Let positive integers $x, y, z$ satisfy $3 j$ product, then
$$
1 \equiv 5^{\circ}(\bmod 3)=2^{\circ}(\bmod 3),
$$

thus $z$ is even. Let $z=2 z_{1}$, then the original equation becomes
$$
\left(5^{x}+2^{y}\right)\left(5^{x}-2^{y}\right)=3^{x} \text {. }
$$

From this, we deduce $\left\{\begin{array}{l}5^{x}+2^{y}=3^{x}, \\ 5^{x}-2^{y}=1 .\end{array}\right.$
Therefore,
$$
\left\{\begin{array}{l}
(-1)^{7}+(-1)^{y} \equiv 0(\bmod 3), \\
(-1)^{x}-(-1)^{y} \equiv 1(\bmod 3) .
\end{array}\right.
$$

Thus $z_{1}$ is odd, $y$ is even, and from (1) we get
$$
1=(-1)^{x}(\bmod 4),
$$
$$
5 \text { (mod8), }
$$

which is a contradiction. Therefore, $y=2$. From (2) and (1), we know $z_{1}=1$, $x=2$. So $x=2, y=2, z=2 z_{1}=2$ is the only solution to the equation.