Solve the following system of equations in real numbers:
$$\left\{\begin{array}{l}x^2=y^2+z^2,\\x^{2024}=y^{2024}+z^{2024},\\x^{2025}=y^{2025}+z^{2025}.\end{array}\right.$$
[i]Proposed by Mykhailo Shtandenko, Anton Trygub, Bogdan Rublov[/i]
1. We start with the given system of equations:
   \[
   \left\{
   \begin{array}{l}
   x^2 = y^2 + z^2 \\
   x^{2024} = y^{2024} + z^{2024} \\
   x^{2025} = y^{2025} + z^{2025}
   \end{array}
   \right.
   \]

2. From the first equation, we have:
   \[
   x^2 = y^2 + z^2
   \]

3. From the second equation, we have:
   \[
   x^{2024} = y^{2024} + z^{2024}
   \]

4. From the third equation, we have:
   \[
   x^{2025} = y^{2025} + z^{2025}
   \]

5. Consider the second equation \(x^{2024} = y^{2024} + z^{2024}\). We can rewrite it using the first equation \(x^2 = y^2 + z^2\):
   \[
   (y^2 + z^2)^{1012} = y^{2024} + z^{2024}
   \]

6. Notice that \((y^2)^{1012} = y^{2024}\) and \((z^2)^{1012} = z^{2024}\). Therefore, we have:
   \[
   (y^2 + z^2)^{1012} = y^{2024} + z^{2024}
   \]

7. For the equality \((y^2 + z^2)^{1012} = y^{2024} + z^{2024}\) to hold, it must be that either \(y = 0\) or \(z = 0\). This is because the sum of two non-negative terms raised to a power can only equal the sum of the individual terms raised to the same power if one of the terms is zero.

8. Without loss of generality, assume \(y = 0\). Then the equations reduce to:
   \[
   x^2 = z^2
   \]
   \[
   x^{2024} = z^{2024}
   \]
   \[
   x^{2025} = z^{2025}
   \]

9. From \(x^2 = z^2\), we get \(x = z\) or \(x = -z\).

10. If \(x = z\), then the solutions are of the form \((x, 0, x)\).

11. If \(x = -z\), then the solutions are of the form \((x, 0, -x)\).

12. Similarly, if \(z = 0\), then the equations reduce to:
    \[
    x^2 = y^2
    \]
    \[
    x^{2024} = y^{2024}
    \]
    \[
    x^{2025} = y^{2025}
    \]

13. From \(x^2 = y^2\), we get \(x = y\) or \(x = -y\).

14. If \(x = y\), then the solutions are of the form \((x, x, 0)\).

15. If \(x = -y\), then the solutions are of the form \((x, -x, 0)\).

16. Therefore, the complete set of solutions is:
    \[
    (x, x, 0), (x, -x, 0), (x, 0, x), (x, 0, -x)
    \]

The final answer is \( \boxed{ (x, x, 0), (x, -x, 0), (x, 0, x), (x, 0, -x) } \).