Find all monic polynomials $P,Q$ which are non-constant, have real coefficients and they satisfy $2P(x)=Q(\frac{(x+1)^2}{2})-Q(\frac{(x-1)^2}{2})$ and $P(1)=1$ for all real $x$.
1. **Initial Conditions and Degree Analysis:**
   - Given the equation \(2P(x) = Q\left(\frac{(x+1)^2}{2}\right) - Q\left(\frac{(x-1)^2}{2}\right)\) and \(P(1) = 1\).
   - Plugging \(x = 0\) into the equation, we get:
     \[
     2P(0) = Q\left(\frac{1}{2}\right) - Q\left(\frac{1}{2}\right) = 0 \implies P(0) = 0.
     \]
   - Plugging \(x = 1\) and \(x = -1\) into the equation, we get:
     \[
     2P(1) = Q(2) - Q(0) \quad \text{and} \quad 2P(-1) = Q(0) - Q(2).
     \]
     Since \(P(1) = 1\), we have:
     \[
     2 \cdot 1 = Q(2) - Q(0) \implies Q(2) - Q(0) = 2.
     \]
     Similarly, for \(x = -1\):
     \[
     2P(-1) = Q(0) - Q(2) \implies 2P(-1) = -2 \implies P(-1) = -1.
     \]

2. **Degree Comparison:**
   - Let \(P(x) = x^n + p_{n-1}x^{n-1} + \cdots + p_1x + p_0\) and \(Q(x) = x^m + q_{m-1}x^{m-1} + \cdots + q_1x + q_0\).
   - The degree of \(P(x)\) is \(n\) and the degree of \(Q(x)\) is \(m\).
   - On the LHS of the given equation, the degree is \(n\).
   - On the RHS, we have:
     \[
     Q\left(\frac{(x+1)^2}{2}\right) - Q\left(\frac{(x-1)^2}{2}\right).
     \]
     The highest degree term in \(Q\left(\frac{(x+1)^2}{2}\right)\) and \(Q\left(\frac{(x-1)^2}{2}\right)\) is \(\left(\frac{(x+1)^2}{2}\right)^m\) and \(\left(\frac{(x-1)^2}{2}\right)^m\), respectively.
   - The degree of \(\left(\frac{(x+1)^2}{2}\right)^m\) and \(\left(\frac{(x-1)^2}{2}\right)^m\) is \(2m\), but these terms cancel out, leaving the highest degree term as \(2m-1\).
   - Therefore, \(n = 2m - 1\).

3. **Leading Coefficient Comparison:**
   - The leading coefficient on the LHS is \(2\).
   - On the RHS, consider the \(k\)-th term without the coefficient \(q_k\):
     \[
     \left(\frac{(x+1)^2}{2}\right)^k - \left(\frac{(x-1)^2}{2}\right)^k.
     \]
     Using the binomial expansion and symmetry, the leading term in the difference is:
     \[
     2x \left(\left(\frac{(x+1)^2}{2}\right)^{k-1} + \left(\frac{(x+1)^2}{2}\right)^{k-2}\left(\frac{(x-1)^2}{2}\right) + \cdots + \left(\frac{(x-1)^2}{2}\right)^{k-1}\right).
     \]
   - For \(k = m\), the coefficient of \(x^{2m-1}\) is:
     \[
     \frac{2m}{2^{m-1}} = \frac{m}{2^{m-2}}.
     \]
   - Equating this to the leading coefficient on the LHS, we get:
     \[
     2 = \frac{m}{2^{m-2}} \implies m = 1 \text{ or } m = 2.
     \]

4. **Case Analysis:**
   - **Case 1: \(m = 1\)**
     - Then \(n = 2m - 1 = 1\).
     - Since \(P(0) = 0\), we have \(P(x) = x\).
     - Plugging \(P(x) = x\) into the original equation, we get:
       \[
       2x = Q\left(\frac{(x+1)^2}{2}\right) - Q\left(\frac{(x-1)^2}{2}\right).
       \]
       This implies \(Q(x) = x + c\) for some constant \(c\).

   - **Case 2: \(m = 2\)**
     - Then \(n = 2m - 1 = 3\).
     - Given \(P(1) = 1\), \(P(0) = 0\), and \(P(-1) = -1\), we can deduce \(P(x) = x^3\).
     - Plugging \(P(x) = x^3\) into the original equation, we get:
       \[
       2x^3 = Q\left(\frac{(x+1)^2}{2}\right) - Q\left(\frac{(x-1)^2}{2}\right).
       \]
       After algebraic manipulation, we find \(Q(x) = x^2 - x + c\) for some constant \(c\).

The final answer is \( \boxed{ P(x) = x } \) and \(Q(x) = x + c\) for \(m = 1\), or \(P(x) = x^3\) and \(Q(x) = x^2 - x + c\) for \(m = 2\).