12・104 Find all positive integer solutions \(x, y, z\) to the equation \(3^{x}+4^{y}=5^{z}\).
(32nd International Mathematical Olympiad Preliminary Question, 1991)
[Solution] Clearly, $x=y=z=2$ is a positive integer solution to the equation.
Next, we prove that the equation
$$
3^{x}+4^{y}=5^{z}
$$

has no other positive integer solutions besides $x=y=z=2$.
Suppose positive integers $x, y, z$ satisfy equation (1), then
$$
5^{z} \equiv 1 \pmod{3}.
$$

Since $5^{z}=(3+2)^{z}$, then
$$
2^{z} \equiv 1 \pmod{3}.
$$

Thus, $z$ is even. Let $z=2z_{1}$, then (1) becomes
$$
\begin{array}{c}
5^{2z_{1}}-2^{2y}=3^{x}, \\
\left(5^{z_{1}}+2^{y}\right)\left(5^{z_{1}}-2^{y}\right)=3^{x}.
\end{array}
$$

From this, we can derive

By $5=6-1, 2=3-1$, we get
$$
\left\{\begin{array}{ll}
(-1)^{z_{1}}+(-1)^{y} \equiv 0 & \pmod{3}, \\
(-1)^{z_{1}}-(-1)^{y} \equiv 1 & \pmod{3}.
\end{array}\right.
$$

Thus, $z_{1}$ is odd, and $y$ is even. Let $y=2y_{1}$,
From (2), we get
$$
\begin{array}{l}
(4+1)^{z_{1}}+4^{y_{1}}=(4-1)^{x}, \\
1 \equiv(-1)^{x} \pmod{4}.
\end{array}
$$

Therefore, $x$ is also even.
If $y>2$, then from (2) we get
$$
5 \equiv 1 \pmod{8}
$$

which is impossible.
Therefore, we can only have
From (3), we get
$$
\begin{array}{c}
y=2. \\
5^{z_{1}}=1+2^{2}=5, \\
z_{1}=1, \quad z=2. \\
x=2.
\end{array}
$$

Thus, from (2)
Therefore, $x=y=z=2$ is the only positive integer solution to (1).