Find all polynomials $P(x)$ with real coefficients that satisfy \[P(x\sqrt{2})=P(x+\sqrt{1-x^2})\]for all real $x$ with $|x|\le 1$.
1. **Initial Condition and Symmetry**:
   Given the polynomial \( P(x) \) satisfies:
   \[
   P(x\sqrt{2}) = P(x + \sqrt{1 - x^2})
   \]
   for all \( |x| \leq 1 \). Notice that if \( x^2 + y^2 = 2 \), then \( P(x) = P(y) \). This implies that \( P \) is symmetric with respect to the transformation \( x \to \sqrt{2 - x^2} \).

2. **Even Function**:
   Since \( (-x)^2 = x^2 \), \( P \) must be an even function. Therefore, we can write \( P(x) \) as a function of \( x^2 \). Let \( Q(x) \) be a polynomial such that \( P(x) = Q(x^2) \).

3. **Transforming the Equation**:
   Substitute \( P(x) = Q(x^2) \) into the original equation:
   \[
   Q(2x^2) = Q(1 + 2x\sqrt{1 - x^2})
   \]
   Notice that \( Q(2x^2) = Q(2(1 - x^2)) \).

4. **New Polynomial \( R \)**:
   Let \( R(x) = Q(x + 1) \). Then the equation becomes:
   \[
   R(2x^2 - 1) = R(2(1 - x^2) - 1)
   \]
   This implies \( R(a) = R(-a) \) for any \( a \in [-1, 1] \), indicating that \( R \) is an even function.

5. **Further Transformation**:
   Let \( S(x) \) be a polynomial such that \( R(x) = S(x^2) \). The equation now becomes:
   \[
   S((2x^2 - 1)^2) = S(4x^2(1 - x^2))
   \]
   for \( |x| \leq 1 \). For any \( 0 \leq a \leq 1 \), let \( a = x^2 \), then:
   \[
   S((2a - 1)^2) = S(4a(1 - a))
   \]

6. **Infinitely Many Values**:
   Let \( b = 4a(1 - a) \), which can take on infinitely many values in the interval \( [0, 1] \). We have:
   \[
   S(b + 1) = S(-b)
   \]
   Since this is true for infinitely many \( b \), \( S(x + 1) = S(-x) \) for all \( x \).

7. **Final Polynomial \( T \)**:
   Let \( T(x) = S\left( x + \frac{1}{2} \right) \). The above equation gives \( T(x) = T(-x) \) for all \( x \), indicating that \( T \) is an even function. Let \( U(x) \) be a polynomial such that \( T(x) = U(x^2) \).

8. **Expression for \( P(x) \)**:
   Notice:
   \[
   P(x) = Q(x^2) = R(x^2 - 1) = S((x^2 - 1)^2) = T\left( (x^2 - 1)^2 - \frac{1}{2} \right) = U\left( \left( (x^2 - 1)^2 - \frac{1}{2} \right)^2 \right)
   \]
   For the original equation to be satisfied, we need:
   \[
   P(\sqrt{2}x) = P(x + \sqrt{1 - x^2}) \Leftrightarrow T\left( (2x^2 - 1)^2 - \frac{1}{2} \right) = T\left( ((x + \sqrt{1 - x^2})^2 - 1)^2 - \frac{1}{2} \right)
   \]
   This is a direct consequence of \( T \) being even.

9. **Final Form**:
   Notice:
   \[
   \left( (x^2 - 1)^2 - \frac{1}{2} \right)^2 = x^8 - 4x^6 + 5x^4 - 2x^2 + \frac{1}{4}
   \]
   Therefore, the answer is:
   \[
   P(x) = U\left( x^8 - 4x^6 + 5x^4 - 2x^2 + \frac{1}{4} \right)
   \]
   for any \( U \in \mathbb{R}[X] \).

The final answer is \( \boxed{ P(x) = U\left( x^8 - 4x^6 + 5x^4 - 2x^2 + \frac{1}{4} \right) } \) for any \( U \in \mathbb{R}[X] \).