Find all number triples $(x,y,z)$ such that when any of these numbers is added to the product of the other two, the result is 2.
We are given the following system of equations:
\[
\begin{cases}
x + yz = 2 \\
y + xz = 2 \\
z + xy = 2
\end{cases}
\]

1. **Subtract the first equation from the second:**
\[
(y + xz) - (x + yz) = 2 - 2
\]
\[
y + xz - x - yz = 0
\]
\[
y - x + xz - yz = 0
\]
\[
y - x + z(x - y) = 0
\]
\[
(y - x)(1 - z) = 0
\]
This implies:
\[
y = x \quad \text{or} \quad z = 1
\]

2. **Case 1: \( z = 1 \)**
\[
x + y \cdot 1 = 2 \implies x + y = 2
\]
\[
y + x \cdot 1 = 2 \implies y + x = 2
\]
\[
1 + xy = 2 \implies xy = 1
\]
We now have the system:
\[
\begin{cases}
x + y = 2 \\
xy = 1
\end{cases}
\]
Solving this system, we use the quadratic equation:
\[
t^2 - (x+y)t + xy = 0
\]
\[
t^2 - 2t + 1 = 0
\]
\[
(t-1)^2 = 0
\]
\[
t = 1
\]
Thus, \( x = 1 \) and \( y = 1 \). Therefore, \( (x, y, z) = (1, 1, 1) \).

3. **Case 2: \( x = y \)**
\[
x + xz = 2 \implies x(1 + z) = 2 \implies z = \frac{2}{x} - 1
\]
\[
z + x^2 = 2 \implies \frac{2}{x} - 1 + x^2 = 2
\]
\[
\frac{2}{x} + x^2 - 1 = 2
\]
\[
\frac{2}{x} + x^2 = 3
\]
Multiplying through by \( x \):
\[
2 + x^3 = 3x
\]
\[
x^3 - 3x + 2 = 0
\]
Factoring the cubic equation:
\[
(x-1)(x^2 + x - 2) = 0
\]
\[
(x-1)(x-1)(x+2) = 0
\]
Thus, \( x = 1 \) or \( x = -2 \).

- If \( x = 1 \), then \( y = 1 \) and \( z = 1 \). Thus, \( (x, y, z) = (1, 1, 1) \).
- If \( x = -2 \), then \( y = -2 \) and \( z = -2 \). Thus, \( (x, y, z) = (-2, -2, -2) \).

Therefore, the solutions are:
\[
(x, y, z) = (1, 1, 1) \quad \text{or} \quad (x, y, z) = (-2, -2, -2)
\]

The final answer is \(\boxed{(1, 1, 1), (-2, -2, -2)}\).