Find all polynomials $P\in \mathbb{R}[x]$, for which $P(P(x))=\lfloor P^2 (x)\rfloor$ is true for $\forall x\in \mathbb{Z}$.
1. **Given Problem:**
   We need to find all polynomials \( P \in \mathbb{R}[x] \) such that \( P(P(x)) = \lfloor P^2(x) \rfloor \) for all \( x \in \mathbb{Z} \).

2. **Analyzing \( P(P(x)) \):**
   Since \( P(P(x)) \) is a polynomial and polynomials are continuous functions, \( P(P(x)) \) must be continuous. However, \( \lfloor P^2(x) \rfloor \) is an integer-valued function for all \( x \in \mathbb{Z} \), which implies that \( P(P(x)) \) must also be an integer for all \( x \in \mathbb{Z} \).

3. **Considering the Continuity and Integer Values:**
   Since \( P(P(x)) \) is both continuous and integer-valued for all \( x \in \mathbb{Z} \), it must be constant. Let \( P(P(x)) = c \) for some constant \( c \in \mathbb{Z} \).

4. **Implication for \( P^2(x) \):**
   Given \( P(P(x)) = c \), we have:
   \[
   c = \lfloor P^2(x) \rfloor
   \]
   Since \( \lfloor P^2(x) \rfloor \) is an integer, \( P^2(x) \) must be very close to an integer. However, since \( P^2(x) \) is a polynomial and must be constant (as it is bounded and polynomial), we conclude that \( P^2(x) \) itself must be an integer.

5. **Solving the Equation \( c = \lfloor c^2 \rfloor \):**
   We need to solve:
   \[
   c = \lfloor c^2 \rfloor
   \]
   This implies \( c \) must be an integer such that \( c = c^2 \). The only integer solutions to this equation are \( c = 0 \) and \( c = 1 \).

6. **Finding Corresponding Polynomials:**
   - For \( c = 0 \):
     \[
     P(x) = 0 \quad \forall x
     \]
   - For \( c = 1 \):
     \[
     P(x) = 1 \quad \forall x
     \]

Thus, the only polynomials \( P \) that satisfy the given condition are \( P(x) = 0 \) and \( P(x) = 1 \).

The final answer is \( \boxed{ P(x) = 0 } \) and \( P(x) = 1 \).