Find all polynomials $P(x, y)$ with real coefficients which for all real numbers $x$ and $y$ satisfy $P(x + y, x - y) = 2P(x, y)$.
1. We start with the given functional equation for the polynomial \( P(x, y) \):
   \[
   P(x + y, x - y) = 2P(x, y)
   \]
   We need to find all polynomials \( P(x, y) \) that satisfy this equation for all real numbers \( x \) and \( y \).

2. Let's make a substitution to simplify the problem. Set \( x = \frac{a + b}{2} \) and \( y = \frac{a - b}{2} \). Then:
   \[
   x + y = \frac{a + b}{2} + \frac{a - b}{2} = a
   \]
   \[
   x - y = \frac{a + b}{2} - \frac{a - b}{2} = b
   \]
   Substituting these into the original equation, we get:
   \[
   P(a, b) = 2P\left(\frac{a + b}{2}, \frac{a - b}{2}\right)
   \]

3. Now, let's consider another substitution. Set \( x = \frac{a}{2} \) and \( y = \frac{b}{2} \). Then:
   \[
   x + y = \frac{a}{2} + \frac{b}{2} = \frac{a + b}{2}
   \]
   \[
   x - y = \frac{a}{2} - \frac{b}{2} = \frac{a - b}{2}
   \]
   Substituting these into the original equation, we get:
   \[
   P\left(\frac{a + b}{2}, \frac{a - b}{2}\right) = 2P\left(\frac{a}{2}, \frac{b}{2}\right)
   \]

4. Combining the results from steps 2 and 3, we have:
   \[
   P(a, b) = 2P\left(\frac{a + b}{2}, \frac{a - b}{2}\right) = 2 \cdot 2P\left(\frac{a}{2}, \frac{b}{2}\right) = 4P\left(\frac{a}{2}, \frac{b}{2}\right)
   \]
   Therefore, we have:
   \[
   P(x, y) = 4P\left(\frac{x}{2}, \frac{y}{2}\right)
   \]

5. To find the form of \( P(x, y) \), consider a general polynomial \( P(x, y) = \sum_{m, n} a_{mn} x^m y^n \). Using the equation \( P(x, y) = 4P\left(\frac{x}{2}, \frac{y}{2}\right) \), we substitute and compare coefficients:
   \[
   \sum_{m, n} a_{mn} x^m y^n = 4 \sum_{m, n} a_{mn} \left(\frac{x}{2}\right)^m \left(\frac{y}{2}\right)^n
   \]
   \[
   \sum_{m, n} a_{mn} x^m y^n = 4 \sum_{m, n} a_{mn} \frac{x^m}{2^m} \frac{y^n}{2^n}
   \]
   \[
   \sum_{m, n} a_{mn} x^m y^n = \sum_{m, n} a_{mn} \frac{x^m y^n}{2^{m+n-2}}
   \]
   For the equality to hold for all \( x \) and \( y \), the coefficients must satisfy:
   \[
   a_{mn} = 4 \cdot \frac{a_{mn}}{2^{m+n}}
   \]
   Simplifying, we get:
   \[
   2^{m+n} = 4 \implies m + n = 2
   \]

6. Therefore, the polynomial \( P(x, y) \) must be of the form:
   \[
   P(x, y) = a_{20} x^2 + a_{11} xy + a_{02} y^2
   \]
   Letting \( b = a_{20} \), \( c = a_{11} \), and \( d = a_{02} \), we have:
   \[
   P(x, y) = bx^2 + cxy + dy^2
   \]

7. Finally, we verify that this form satisfies the original equation:
   \[
   P(x + y, x - y) = b(x + y)^2 + c(x + y)(x - y) + d(x - y)^2
   \]
   \[
   = b(x^2 + 2xy + y^2) + c(x^2 - y^2) + d(x^2 - 2xy + y^2)
   \]
   \[
   = bx^2 + 2bxy + by^2 + cx^2 - cy^2 + dx^2 - 2dxy + dy^2
   \]
   \[
   = (b + c + d)x^2 + (2b - 2d)xy + (b - c + d)y^2
   \]
   \[
   2P(x, y) = 2(bx^2 + cxy + dy^2) = 2bx^2 + 2cxy + 2dy^2
   \]
   For these to be equal, we must have:
   \[
   b + c + d = 2b, \quad 2b - 2d = 2c, \quad b - c + d = 2d
   \]
   Solving these, we find that the conditions are satisfied for arbitrary \( b \), \( c \), and \( d \).

The final answer is \( \boxed{ P(x, y) = bx^2 + cxy + dy^2 } \) for arbitrary real constants \( b \), \( c \), and \( d \).