Find all ordered triples $(x, y, z)$ such that $x, y, z \in Q_{+}$, and $x+\frac{1}{y}$, $y+\frac{1}{z}$, $z+\frac{1}{x}$ are all integers.
(2010, International Invitational Competition for Young Mathematicians in Cities)
Let $x=\frac{a}{b}, y=\frac{b}{c}, z=\frac{c}{a}\left(a, b, c \in \mathbf{N}_{+}\right)$.
(1) If $a=b=c$, then
$$
(x, y, z)=(1,1,1).
$$
(2) If $a, b, c$ are exactly two equal, without loss of generality, let $a=b$. Then
$$
\begin{array}{l}
x+\frac{1}{y}=1+\frac{c}{a}, y+\frac{1}{z}=\frac{2 a}{c}, \\
z+\frac{1}{x}=\frac{c}{a}+1 .
\end{array}
$$

Thus $a|c, c| 2 a \Rightarrow c=2 a$.
Let $a=k$. Then $(a, b, c)=(k, k, 2 k)$ and its permutations.
So, $(x, y, z)=\left(1, \frac{1}{2}, 2\right)$ and its permutations.
(3) If $a, b, c$ are all distinct, without loss of generality, let $a>b>c$.

Then $z+\frac{1}{x}=\frac{c+b}{a} \in \mathbf{N}_{+}$
$$
\begin{array}{l}
\Rightarrow a \leqslant c+b<2 a \\
\Rightarrow a=b+c . \\
\text { Also } x+\frac{1}{y}=\frac{a+c}{b}=\frac{b+2 c}{b}=1+\frac{2 c}{b} \in \mathbf{N}_{+} \\
\Rightarrow b \leqslant 2 c<2 b \\
\Rightarrow b=2 c, a=3 c .
\end{array}
$$

Therefore, $(a, b, c)=(3 k, 2 k, k)$ and its permutations.
Hence $(x, y, z)$ $=\left(\frac{3}{2}, 2, \frac{1}{3}\right),\left(3, \frac{1}{2}, \frac{2}{3}\right)$ and its permutations.
In summary,
$$
\begin{array}{l}
(x, y, z) \\
=(1,1,1),\left(1, \frac{1}{2}, 2\right),\left(\frac{3}{2}, 2, \frac{1}{3}\right),
\end{array}
$$
$\left(3, \frac{1}{2}, \frac{2}{3}\right)$ and its permutations.