[b]Q.[/b] Find all polynomials $P: \mathbb{R \times R}\to\mathbb{R\times R}$ with real coefficients, such that $$P(x,y) = P(x+y,x-y), \ \forall\ x,y \in \mathbb{R}.$$

[i]Proposed by TuZo[/i]
1. **Claim:** \( P(x,y) = (a,b) \) where \( a, b \in \mathbb{R} \) are arbitrary constants.

2. **Step 1:** Observe the given functional equation:
   \[
   P(x,y) = P(x+y, x-y) \quad \forall x, y \in \mathbb{R}.
   \]

3. **Step 2:** Consider the transformation \( (x, y) \to (x+y, x-y) \). We need to show that this transformation implies that \( P(x, y) \) is constant.

4. **Step 3:** Apply the transformation iteratively:
   \[
   P(2x, 2y) = P((x+y) + (x-y), (x+y) - (x-y)) = P(x+y, x-y) = P(x, y).
   \]
   This implies:
   \[
   P(x, y) = P\left(\frac{x}{2}, \frac{y}{2}\right).
   \]

5. **Step 4:** Continue applying the transformation:
   \[
   P(x, y) = P\left(\frac{x}{2}, \frac{y}{2}\right) = P\left(\frac{x}{4}, \frac{y}{4}\right) = \dots = P\left(\frac{x}{2^n}, \frac{y}{2^n}\right).
   \]

6. **Step 5:** As \( n \to \infty \), \(\frac{x}{2^n} \to 0\) and \(\frac{y}{2^n} \to 0\). Since any polynomial function is continuous, we have:
   \[
   P(x, y) = \lim_{n \to \infty} P\left(\frac{x}{2^n}, \frac{y}{2^n}\right) = P\left(\lim_{n \to \infty} \frac{x}{2^n}, \lim_{n \to \infty} \frac{y}{2^n}\right) = P(0, 0).
   \]

7. **Step 6:** Let \( P(0, 0) = (a, b) \) where \( a, b \in \mathbb{R} \). Therefore, we conclude:
   \[
   P(x, y) = (a, b) \quad \forall x, y \in \mathbb{R}.
   \]

8. **Conclusion:** The polynomial \( P(x, y) \) is a constant polynomial.

\(\blacksquare\)

The final answer is \( \boxed{ P(x, y) = (a, b) } \) where \( a, b \in \mathbb{R} \).