Find the number of different positive integer triples $(x, y,z)$ satisfying the equations
$x^2 + y -z = 100$ and $x + y^2 - z = 124$:
1. We start with the given equations:
   \[
   x^2 + y - z = 100
   \]
   \[
   x + y^2 - z = 124
   \]

2. Subtract the first equation from the second equation to eliminate \(z\):
   \[
   (x + y^2 - z) - (x^2 + y - z) = 124 - 100
   \]
   Simplifying, we get:
   \[
   x + y^2 - x^2 - y = 24
   \]
   Rearrange the terms:
   \[
   y^2 - y - x^2 + x = 24
   \]

3. Notice that the left-hand side can be rewritten using the difference of squares:
   \[
   y^2 - y - x^2 + x = 24
   \]
   \[
   (y^2 - y + \frac{1}{4}) - (x^2 - x + \frac{1}{4}) = 24 + \frac{1}{4} - \frac{1}{4}
   \]
   \[
   \left(y - \frac{1}{2}\right)^2 - \left(x - \frac{1}{2}\right)^2 = 24
   \]
   Multiply both sides by 4 to clear the fractions:
   \[
   (2y - 1)^2 - (2x - 1)^2 = 96
   \]

4. Use the difference of squares formula \(a^2 - b^2 = (a - b)(a + b)\):
   \[
   (2y - 1)^2 - (2x - 1)^2 = 96
   \]
   \[
   (2y - 1 - (2x - 1))(2y - 1 + (2x - 1)) = 96
   \]
   Simplify the factors:
   \[
   (2y - 2x)(2y + 2x - 2) = 96
   \]
   \[
   2(y - x) \cdot 2(y + x - 1) = 96
   \]
   \[
   (y - x)(y + x - 1) = 24
   \]

5. Solve for the integer pairs \((y - x)\) and \((y + x - 1)\) that multiply to 24. The factor pairs of 24 are:
   \[
   (1, 24), (2, 12), (3, 8), (4, 6), (-1, -24), (-2, -12), (-3, -8), (-4, -6)
   \]

6. Check each pair to find valid integer solutions for \(x\) and \(y\):
   - For \((y - x, y + x - 1) = (1, 24)\):
     \[
     y - x = 1
     \]
     \[
     y + x - 1 = 24
     \]
     Solving these equations:
     \[
     y = x + 1
     \]
     \[
     (x + 1) + x - 1 = 24
     \]
     \[
     2x = 24
     \]
     \[
     x = 12, \quad y = 13
     \]

   - For \((y - x, y + x - 1) = (3, 8)\):
     \[
     y - x = 3
     \]
     \[
     y + x - 1 = 8
     \]
     Solving these equations:
     \[
     y = x + 3
     \]
     \[
     (x + 3) + x - 1 = 8
     \]
     \[
     2x + 2 = 8
     \]
     \[
     2x = 6
     \]
     \[
     x = 3, \quad y = 6
     \]

7. Substitute the valid pairs \((x, y)\) back into the original equations to find \(z\):
   - For \((x, y) = (12, 13)\):
     \[
     x^2 + y - z = 100
     \]
     \[
     12^2 + 13 - z = 100
     \]
     \[
     144 + 13 - z = 100
     \]
     \[
     157 - z = 100
     \]
     \[
     z = 57
     \]

   - For \((x, y) = (3, 6)\):
     \[
     x^2 + y - z = 100
     \]
     \[
     3^2 + 6 - z = 100
     \]
     \[
     9 + 6 - z = 100
     \]
     \[
     15 - z = 100
     \]
     \[
     z = -85
     \]
     Since \(z\) must be a positive integer, this solution is invalid.

8. The only valid solution is \((x, y, z) = (12, 13, 57)\).

The final answer is \(\boxed{(12, 13, 57)}\).