12・105 Find the positive integer solutions $(x, y, z)$ that satisfy the equation $8^{x}+15^{y}=17^{z}$.
(Canadian Mathematical Olympiad Training Problem, 1992)
[Solution] Consider the remainders of $8^{x}, 15^{y}, 17^{z}$ when divided by 4,
for all $x$,
$8^{x} \equiv 0 \quad(\bmod 4)$.

for all $z$,
$17^{z} \equiv 1 \quad(\bmod 4)$.

Thus, from the given equation we have
$$
15^{y} \equiv 1(\bmod 4) .
$$

Therefore, $y$ must be even.
Next, consider the remainders of $8^{x}, 15^{y}, 17^{z}$ when divided by 7.
for all $x$,
$8^{x} \equiv 1 \quad(\bmod 7)$.

for all $y$,
$15^{y} \equiv 1(\bmod 7)$.

Thus, from the given equation we get
$$
17^{z} \equiv 2(\bmod 7) .
$$

Therefore, $z$ must be even.
Consider the remainders of $8^{x}, 15^{y}, 17^{z}$ when divided by 3,
for all $y$,
$$
15^{y} \equiv 0 \quad(\bmod 3) \text {. }
$$

for even $z$,
Thus, from the given equation we get
$$
8^{x} \equiv 1 \quad(\bmod 3)
$$

Therefore, $x$ is also even.
Hence, $x, y, z$ are all even.
Let $x=2 k, y=2 m, z=2 n$, then we have
$$
2^{6 k}=\left(17^{n}-15^{m}\right)\left(17^{n}+15^{m}\right) \text {. }
$$

Thus, we have
$$
\begin{array}{c}
17^{n}-15^{m}=2^{t}, \\
17^{n}+15^{m}=2^{6 k-t}
\end{array}
$$

From (3) we know
So $2^{6 k-2 t}+1$ is odd.
Therefore, from (4) we get
$$
t=1 .
$$

From (1) we have
$$
17^{n}-15^{m}=2
$$

Consider the remainders of $17^{n}, 15^{m}$ when divided by 9,
When $m \geqslant 2$,

When $n$ is odd,
$$
\begin{array}{ll}
15^{m} \equiv 0 & (\bmod 9) \\
17^{n} \equiv 8 & (\bmod 9) \\
17^{n} \equiv 1 & (\bmod 9)
\end{array}
$$

When $n$ is even,
So when $m \geqslant 2$, (5) does not hold.
Therefore, (5) has a solution only when $m=1, n=1$.
At this point, from (2) we know $k=1$.
Thus, the given equation has a unique solution:
$$
(x, y, z)=(2,2,2) .
$$