Find all functions $f: \mathbb{R}\to [0;+\infty)$ such that: 
\[f(x^2+y^2)=f(x^2-y^2)+f(2xy)\]
for all real numbers $x$ and $y$.

[i]Laurentiu Panaitopol[/i]
1. Let \( P(x, y) \) be the assertion \( f(x^2 + y^2) = f(x^2 - y^2) + f(2xy) \).

2. First, consider \( P(0, 0) \):
   \[
   P(0, 0) \implies f(0^2 + 0^2) = f(0^2 - 0^2) + f(2 \cdot 0 \cdot 0) \implies f(0) = f(0) + f(0) \implies f(0) = 0.
   \]

3. Next, consider \( P(0, x) \):
   \[
   P(0, x) \implies f(0^2 + x^2) = f(0^2 - x^2) + f(2 \cdot 0 \cdot x) \implies f(x^2) = f(-x^2) + f(0) \implies f(x^2) = f(-x^2).
   \]
   This shows that \( f \) is an even function.

4. Now, let \( x \ge y \ge z \ge 0 \). We will use the following substitutions:
   - (a) \( P\left(\sqrt{\frac{x+y}{2}}, \sqrt{\frac{x-y}{2}}\right) \):
     \[
     f\left(\left(\sqrt{\frac{x+y}{2}}\right)^2 + \left(\sqrt{\frac{x-y}{2}}\right)^2\right) = f\left(\left(\sqrt{\frac{x+y}{2}}\right)^2 - \left(\sqrt{\frac{x-y}{2}}\right)^2\right) + f\left(2 \cdot \sqrt{\frac{x+y}{2}} \cdot \sqrt{\frac{x-y}{2}}\right).
     \]
     Simplifying, we get:
     \[
     f\left(\frac{x+y}{2} + \frac{x-y}{2}\right) = f\left(\frac{x+y}{2} - \frac{x-y}{2}\right) + f\left(2 \sqrt{\frac{x+y}{2}} \cdot \sqrt{\frac{x-y}{2}}\right) \implies f(x) = f(y) + f(\sqrt{x^2 - y^2}).
     \]

   - (b) \( P\left(\sqrt{\frac{y+z}{2}}, \sqrt{\frac{y-z}{2}}\right) \):
     \[
     f\left(\left(\sqrt{\frac{y+z}{2}}\right)^2 + \left(\sqrt{\frac{y-z}{2}}\right)^2\right) = f\left(\left(\sqrt{\frac{y+z}{2}}\right)^2 - \left(\sqrt{\frac{y-z}{2}}\right)^2\right) + f\left(2 \cdot \sqrt{\frac{y+z}{2}} \cdot \sqrt{\frac{y-z}{2}}\right).
     \]
     Simplifying, we get:
     \[
     f\left(\frac{y+z}{2} + \frac{y-z}{2}\right) = f\left(\frac{y+z}{2} - \frac{y-z}{2}\right) + f\left(2 \sqrt{\frac{y+z}{2}} \cdot \sqrt{\frac{y-z}{2}}\right) \implies f(y) = f(z) + f(\sqrt{y^2 - z^2}).
     \]

   - (c) \( P\left(\sqrt{\frac{x+z}{2}}, \sqrt{\frac{x-z}{2}}\right) \):
     \[
     f\left(\left(\sqrt{\frac{x+z}{2}}\right)^2 + \left(\sqrt{\frac{x-z}{2}}\right)^2\right) = f\left(\left(\sqrt{\frac{x+z}{2}}\right)^2 - \left(\sqrt{\frac{x-z}{2}}\right)^2\right) + f\left(2 \cdot \sqrt{\frac{x+z}{2}} \cdot \sqrt{\frac{x-z}{2}}\right).
     \]
     Simplifying, we get:
     \[
     f\left(\frac{x+z}{2} + \frac{x-z}{2}\right) = f\left(\frac{x+z}{2} - \frac{x-z}{2}\right) + f\left(2 \sqrt{\frac{x+z}{2}} \cdot \sqrt{\frac{x-z}{2}}\right) \implies f(x) = f(z) + f(\sqrt{x^2 - z^2}).
     \]

5. Combining (a), (b), and (c):
   \[
   f(\sqrt{x^2 - z^2}) = f(\sqrt{x^2 - y^2}) + f(\sqrt{y^2 - z^2}).
   \]

6. Let \( f(x) = g(x^2) \). Then the equation becomes:
   \[
   g(x^2 + y^2) = g(x^2) + g(y^2).
   \]
   This is Cauchy's functional equation for non-negative arguments. Since \( g(x) \ge 0 \), the only solution is \( g(x) = ax \) for some \( a \ge 0 \).

7. Therefore, \( f(x) = ax^2 \) for all \( x \ge 0 \). Since \( f \) is even, we have:
   \[
   f(x) = ax^2 \quad \forall x \in \mathbb{R}.
   \]

Conclusion:
\[
\boxed{f(x) = ax^2 \text{ for some } a \ge 0}
\]