17. (HKG 4) Find all positive integer solutions \(x, y, z\) of the equation \(3^{x} + 4^{y} = 5^{z}\).
17. Taking the equation $3^{x}+4^{y}=5^{z}(x, y, z>0)$ modulo 3, we get that $5^{z} \equiv 1(\bmod 3)$, hence $z$ is even, say $z=2 z_{1}$. The equation then becomes $3^{x}=5^{2 z_{1}}-4^{y}=\left(5^{z_{1}}-2^{y}\right)\left(5^{z_{1}}+2^{y}\right)$. Each factor $5^{z_{1}}-2^{y}$ and $5^{z_{1}}+2^{y}$ is a power of 3, for which the only possibility is $5^{z_{1}}+2^{y}=3^{x}$ and $5^{z_{1}}-2^{y}=$ 1. Again modulo 3 these equations reduce to $(-1)^{z_{1}}+(-1)^{y}=0$ and $(-1)^{z_{1}}-(-1)^{y}=1$, implying that $z_{1}$ is odd and $y$ is even. Particularly, $y \geq 2$. Reducing the equation $5^{z_{1}}+2^{y}=3^{x}$ modulo 4 we get that $3^{x} \equiv 1$, hence $x$ is even. Now if $y>2$, modulo 8 this equation yields $5 \equiv 5^{z_{1}} \equiv$ $3^{x} \equiv 1$, a contradiction. Hence $y=2, z_{1}=1$. The only solution of the original equation is $x=y=z=2$.