Find all triples $ (x,y,z)$ of integers which satisfy

$ x(y \plus{} z) \equal{} y^2 \plus{} z^2 \minus{} 2$

$ y(z \plus{} x) \equal{} z^2 \plus{} x^2 \minus{} 2$

$ z(x \plus{} y) \equal{} x^2 \plus{} y^2 \minus{} 2$.
1. We start with the given system of equations:
   \[
   \begin{cases}
   x(y + z) = y^2 + z^2 - 2 \\
   y(z + x) = z^2 + x^2 - 2 \\
   z(x + y) = x^2 + y^2 - 2
   \end{cases}
   \]

2. Add all three equations together:
   \[
   x(y + z) + y(z + x) + z(x + y) = y^2 + z^2 - 2 + z^2 + x^2 - 2 + x^2 + y^2 - 2
   \]
   Simplify the right-hand side:
   \[
   x(y + z) + y(z + x) + z(x + y) = 2(x^2 + y^2 + z^2) - 6
   \]

3. Notice that the left-hand side can be rewritten as:
   \[
   x(y + z) + y(z + x) + z(x + y) = xy + xz + yz + yx + zx + zy = 2(xy + yz + zx)
   \]

4. Equate the simplified left-hand side and right-hand side:
   \[
   2(xy + yz + zx) = 2(x^2 + y^2 + z^2) - 6
   \]
   Divide both sides by 2:
   \[
   xy + yz + zx = x^2 + y^2 + z^2 - 3
   \]

5. Rearrange the equation to isolate the quadratic terms:
   \[
   x^2 + y^2 + z^2 - xy - yz - zx = 3
   \]

6. Recognize that the left-hand side can be rewritten using the identity for the sum of squares:
   \[
   \frac{1}{2} \left( (x - y)^2 + (y - z)^2 + (z - x)^2 \right) = 3
   \]
   Multiply both sides by 2:
   \[
   (x - y)^2 + (y - z)^2 + (z - x)^2 = 6
   \]

7. The equation \((x - y)^2 + (y - z)^2 + (z - x)^2 = 6\) implies that the differences between the pairs \((x, y, z)\) must be such that their squares sum to 6. The only integer solutions to this are permutations of \(\{1, 1, 2\}\).

8. Assume \(x \ge y \ge z\). Then we must have:
   \[
   x - z = 2, \quad x - y = 1, \quad y - z = 1
   \]

9. Solving these, we get:
   \[
   x = k + 2, \quad y = k + 1, \quad z = k
   \]

10. Substitute \(x = k + 2\), \(y = k + 1\), and \(z = k\) back into the original equations to verify:
    \[
    (k + 2)((k + 1) + k) = (k + 1)^2 + k^2 - 2
    \]
    Simplify:
    \[
    (k + 2)(2k + 1) = k^2 + 2k + 1 + k^2 - 2
    \]
    \[
    2k^2 + 5k + 2 = 2k^2 + 2k - 1
    \]
    \[
    5k + 2 = 2k - 1
    \]
    \[
    3k = -3
    \]
    \[
    k = -1
    \]

11. Therefore, the solution is:
    \[
    (x, y, z) = (-1 + 2, -1 + 1, -1) = (1, 0, -1)
    \]
    or any permutation of \((1, 0, -1)\).

The final answer is \( \boxed{ (1, 0, -1) } \) or any permutation of \((1, 0, -1)\).