12.103 Find the natural number solutions to the equation $2^{x}+3^{y}=5^{z}$.
(All-Soviet Union Mathematical Winter Camp, 1991)
[Solution] $x=1, y=1, z=1$ is clearly a solution.
When $x=1, y \geqslant 2$,

So
$$
\begin{array}{l}
2^{x}+3^{y} \equiv 2 \quad(\bmod 9), \\
5^{z} \equiv 2 \quad(\bmod 9) . \\
\end{array}
$$

Since
$$
\begin{array}{l}
5^{5}=3125 \equiv 2 \quad(\bmod 9), \\
5^{6}=15625 \equiv 1 \quad(\bmod 9), \\
5^{6 k+5} \equiv 2 \quad(\bmod 9), \\
\end{array}
$$

Thus
$$
z=6 k+5 \text {. }
$$

So $\quad 3^{y}=5^{6 k+5}-2 \equiv 1(\bmod 7)$.
Thus
$$
6 \mid y \text {. }
$$

This leads to
$$
2=5^{z}-3^{y} \equiv 0 \quad(\bmod 4)
$$

which is impossible.
When $x=2$, then
$$
3^{y}=5^{z}-4 \equiv 1 \quad(\bmod 4) .
$$

So $y$ is even, let $y=2 y_{1}$, hence
$$
5^{z}=4+3^{y} \equiv 4 \quad(\bmod 9) .
$$

This implies
$$
z=6 k+4 \text {, }
$$

Therefore
$$
4=5^{z}-3^{y}=5^{6 k+4}-9^{y_{1}} \equiv 0 \quad(\bmod 8) .
$$

This is also impossible, so when $x=2$, the equation has no natural number solutions.
When $x \geqslant 3$, from
$$
3^{y}=5^{z}-2^{x} \equiv 1 \quad(\bmod 4)
$$

we know $y=2 y_{1}$ is even, thus
$$
5^{z}=2^{x}+9^{y_{1}} \equiv 1 \quad(\bmod 8),
$$

which implies $z=2 z_{1}$ is even, so
we can set
$$
2^{x}=\left(5^{z_{1}}+3^{y_{1}}\right)\left(5^{z_{1}}-3^{y_{1}}\right) .
$$
$$
\begin{array}{l}
5^{z_{1}}+3^{y_{1}}=2^{x_{1}}, \\
5^{z_{1}}-3^{y_{1}}=2^{x_{2}} .
\end{array}
$$

where $x_{1}>x_{2} \geqslant 0, x_{1}+x_{2}=x$. Eliminating $5^{z_{1}}$, we get
$$
3^{y_{1}}=2^{x_{1}-1}-2^{x_{2}-1} \text {. }
$$

This leads to
$$
\begin{array}{c}
x_{2}=1, \\
3^{y_{1}}=2^{x-2}-1 \equiv 0 \quad(\bmod 3) .
\end{array}
$$

So $x-2=2 x_{3}$ is even,
$$
3^{y_{1}}=\left(2^{x_{3}}+1\right)\left(2^{x_{3}}-1\right) \text {. }
$$

We can also set
$$
\begin{array}{l}
2^{x_{3}}+1=3^{y_{2}}, \\
2^{x_{3}}-1=3^{y_{3}},
\end{array}
$$

then
$$
3^{y_{2}}-3^{y_{3}}=2 \text {, }
$$

Thus
$$
y_{2}=1, \quad y_{3}=0 \text {. }
$$

Therefore $\quad x_{3}=1, x=4, y=2, z=2$.
In summary, there are two sets of natural number solutions $(1,1,1),(4,2,2)$.