Find all polynomials $ P(x,y)$ such that for all reals $ x$ and $y$,
\[P(x^{2},y^{2}) =P\left(\frac {(x + y)^{2}}{2},\frac {(x - y)^{2}}{2}\right).\]
1. **Verification of the given solution:**

   We need to verify that the polynomial \( P(x, y) \equiv Q(x + y, xy(x - y)^2) \) satisfies the given functional equation:
   \[
   P(x^2, y^2) = P\left(\frac{(x + y)^2}{2}, \frac{(x - y)^2}{2}\right).
   \]

   Let \( P(x, y) = Q(x + y, xy(x - y)^2) \). Then:
   \[
   P(x^2, y^2) = Q(x^2 + y^2, x^2 y^2 (x^2 - y^2)^2).
   \]

   On the other hand:
   \[
   P\left(\frac{(x + y)^2}{2}, \frac{(x - y)^2}{2}\right) = Q\left(\frac{(x + y)^2}{2} + \frac{(x - y)^2}{2}, \left(\frac{(x + y)^2}{2}\right) \left(\frac{(x - y)^2}{2}\right) \left(\frac{(x + y)^2}{2} - \frac{(x - y)^2}{2}\right)^2\right).
   \]

   Simplifying the arguments of \( Q \):
   \[
   \frac{(x + y)^2}{2} + \frac{(x - y)^2}{2} = \frac{x^2 + 2xy + y^2 + x^2 - 2xy + y^2}{2} = x^2 + y^2,
   \]
   and
   \[
   \left(\frac{(x + y)^2}{2}\right) \left(\frac{(x - y)^2}{2}\right) \left(\frac{(x + y)^2}{2} - \frac{(x - y)^2}{2}\right)^2 = \frac{(x^2 + 2xy + y^2)(x^2 - 2xy + y^2)}{4} \left(\frac{4xy}{2}\right)^2 = x^2 y^2 (x^2 - y^2)^2.
   \]

   Therefore:
   \[
   P\left(\frac{(x + y)^2}{2}, \frac{(x - y)^2}{2}\right) = Q(x^2 + y^2, x^2 y^2 (x^2 - y^2)^2).
   \]

   This shows that \( P(x, y) \equiv Q(x + y, xy(x - y)^2) \) satisfies the given functional equation.

2. **Proving the converse:**

   We need to show that any polynomial \( P(x, y) \) satisfying the given functional equation must be of the form \( Q(x + y, xy(x - y)^2) \).

   - **Symmetry:**

     By changing \( y \) to \( -y \) in the given equation, we get:
     \[
     P(x^2, y^2) = P\left(\frac{(x + y)^2}{2}, \frac{(x - y)^2}{2}\right) = P\left(\frac{(x - y)^2}{2}, \frac{(x + y)^2}{2}\right).
     \]
     This implies that \( P \) is symmetric, i.e., \( P(x, y) = P(y, x) \) for all \( x, y \in \mathbb{R} \).

   - **Divisibility by \( xy \):**

     Note that if we subtract from \( P \) any polynomial \( A(x + y) \), then \( P \) would still satisfy the assertion. By doing this, we may assume without loss of generality that \( P(x, 0) \equiv 0 \). Since \( P \) is symmetric, it follows that \( xy \) divides \( P(x, y) \).

   - **Divisibility by \( (x - y)^2 \):**

     Putting \( y = x \) in the given equation, we get:
     \[
     P(x^2, x^2) = P(2x^2, 0) = 0.
     \]
     This implies that \( P \) is divisible by \( (x - y) \). Since \( P \) is symmetric, it follows that \( P \) is divisible by \( (x - y)^2 \).

   - **Form of \( P \):**

     Therefore, \( P \) is divisible by \( xy(x - y)^2 \). We can write \( P(x, y) = xy(x - y)^2 R(x, y) \) for some polynomial \( R(x, y) \). Since \( P \) satisfies the given functional equation, it follows that \( R(x, y) \) must be of the form \( Q(x + y) \) for some polynomial \( Q \).

     Hence, \( P(x, y) \equiv Q(x + y, xy(x - y)^2) \).

The final answer is \( \boxed{ P(x, y) \equiv Q(x + y, xy(x - y)^2) } \) for any \( Q(x, y) \in \mathbb{R}[x, y] \).