A* Find all integer triples $(x, y, z)$ that satisfy $8^{x}+15^{y}=17^{z}$.
Parse modulo $4, 8^{x} \equiv 0, 17^{z} \equiv 1$. Therefore $15^{y}=1(\bmod 4)$. Thus $y$ must be even. Consider modulo 7, then $8^{x} \equiv 1, 15^{y} \equiv 1$, so $17^{z} \equiv 2(\bmod 7)$ holds only when $z$ is even. Modulo 3, similarly, we get $x$ is even. Thus $x, y, z$ are all even.

Let $x=2 k, y=2 m, z=2 n$, then $2^{6 k}=\left(17^{n}-15^{m}\right)\left(17^{n}+15^{m}\right)$. Therefore $17^{n}-15^{m}=2^{t}, 17^{n}+15^{m}=2^{6 k-t}, 1 \leqslant t \leqslant 3 k$. Adding these two equations gives $2 \cdot 17^{n}-2^{t}\left(2^{6 k-2 t}+1\right)$. Thus, $t=1$, so $17^{n}-15^{m}=2$. Modulo 9, when $n$ is odd, $17^{n} \equiv 8$, and when $n$ is even, $17^{n}=1$. On the other hand, $15=6$ and when $m \geqslant 2$, $15^{m} \equiv 0$. Therefore $m=1$, which implies $n=1$, $k=1$. Hence the unique solution to the original equation is $(x, y, z)=(2,2,2)$.