Determine and justify all solutions $(x,y, z)$ of the system of equations:

$x^2 = y + z$
$y^2 = x + z$
$z^2 = x + y$
1. **Finding Obvious Solutions:**
   Suppose \( x = y = z \). Then the system of equations reduces to:
   \[
   x^2 = x + x \implies x^2 = 2x
   \]
   Solving this equation:
   \[
   x^2 - 2x = 0 \implies x(x - 2) = 0
   \]
   Thus, \( x = 0 \) or \( x = 2 \). Therefore, the solutions are:
   \[
   (0, 0, 0) \quad \text{and} \quad (2, 2, 2)
   \]

2. **Exploring Other Solutions:**
   Rearrange the second equation to express \( z \) in terms of \( x \) and \( y \):
   \[
   y^2 = x + z \implies z = y^2 - x
   \]
   Substitute \( z = y^2 - x \) into the first equation:
   \[
   x^2 = y + (y^2 - x) \implies x^2 + x = y^2 + y
   \]
   Similarly, from the third equation:
   \[
   z^2 = x + y \implies (y^2 - x)^2 = x + y
   \]
   This implies:
   \[
   y^4 - 2xy^2 + x^2 = x + y
   \]

3. **Equating the Quantities:**
   Let \( c = x^2 + x = y^2 + y = z^2 + z \). Then:
   \[
   t^2 + t = c
   \]
   This quadratic equation can have at most two solutions for \( t \). Therefore, at least two of \( x, y, z \) must be equal.

4. **Case 1: \( x = y \):**
   The system reduces to:
   \[
   x^2 = x + z \quad \text{and} \quad z^2 = 2x
   \]
   Substitute \( z^2 = 2x \) into the first equation:
   \[
   x^2 = x + \sqrt{2x}
   \]
   Solving for \( x \):
   \[
   x^2 - x - \sqrt{2x} = 0
   \]
   This equation is complex to solve directly, so we consider the polynomial form:
   \[
   z^4 - 2z^2 - 4z = 0
   \]
   Factoring:
   \[
   z(z - 2)(z^2 + 2z + 2) = 0
   \]
   The solutions are:
   \[
   z = 0, \quad z = 2, \quad z = -1 + i, \quad z = -1 - i
   \]

5. **Finding Corresponding \( x \) and \( y \):**
   For \( z = 0 \) and \( z = 2 \):
   \[
   z^2 = 2x \implies x = 0 \quad \text{or} \quad x = 2
   \]
   For \( z = -1 + i \):
   \[
   z^2 = 2x \implies (-1 + i)^2 = 2x \implies 1 - 2i - 1 = 2x \implies x = -i
   \]
   For \( z = -1 - i \):
   \[
   z^2 = 2x \implies (-1 - i)^2 = 2x \implies 1 + 2i - 1 = 2x \implies x = i
   \]

6. **Permutations of Solutions:**
   The solutions are:
   \[
   (0, 0, 0), \quad (2, 2, 2), \quad (-1+i, -i, -i), \quad (-i, -1+i, -i), \quad (-i, -i, -1+i), \quad (-1-i, i, i), \quad (i, -1-i, i), \quad (i, i, -1-i)
   \]

The final answer is \( \boxed{ (0, 0, 0), (2, 2, 2), (-1+i, -i, -i), (-i, -1+i, -i), (-i, -i, -1+i), (-1-i, i, i), (i, -1-i, i), (i, i, -1-i) } \)