4. 139 Find all triples $(x, y, z)$ such that adding any one of the numbers to the product of the other two results in 2.
[Solution] From the given conditions, we have
$$
\left\{\begin{array}{l}
x+y z=2, \\
y+z x=2, \\
z+x y=2 .
\end{array}\right.
$$
(1) - (2) gives $\quad(x-y)(1-z)=0$.
(2) - (3) gives $(y-z)(1-x)=0$.

Thus, according to (4) and (5), we can discuss in four cases.
Case $1 \quad x-y=0$ and $y-z=0$. In this case, we have $x=y=z$. Substituting into the original equations, we get

Therefore,
$$
\begin{array}{l}
x^{2}+x-2=0, \\
x=1 \text { or }-2 .
\end{array}
$$

Thus, we have $x=y=z=1$ or $x=y=z=-2$.
Case $2 \quad x-y=0$ and $1-x=0$. Thus, we get $x=y=z=1$.
Case $3 \quad 1-z=0$ and $y-z=0$. Thus, we get $x=y=z=1$.
Case $4 \quad 1-z=0$ and $1-x=0$. Thus, we get $x=y=z=1$.
In summary, the required triplets are $(1,1,1)$ and $(-2,-2,-2)$.