A polynomial $P$ with real coefficients is called [i]great,[/i] if for some integer $a>1$ and for all integers $x$, there exists an integer $z$ such that $aP(x)=P(z)$. Find all [i]great[/i] polynomials.

[i]Proposed by A. Golovanov[/i]
1. **Initial Assumptions and Setup:**
   - Let \( P \) be a polynomial with real coefficients.
   - Assume \( P \) is nonconstant and monic (leading coefficient is 1) with degree \( d \).
   - By the problem's condition, there exists an integer \( a > 1 \) such that for all integers \( x \), there exists an integer \( z \) satisfying \( aP(x) = P(z) \).

2. **Sequence and Limit Analysis:**
   - Let \( z_n \) be a sequence of integers such that \( aP(n) = P(z_n) \) for all \( n \geq 1 \).
   - Define \( A = \sqrt[d]{a} \).
   - Write \( P(x) = x^d + bx^{d-1} + Q(x) \) where \( \deg(Q) \leq d-2 \).
   - As \( n \to \infty \), \( |z_n| \to \infty \).

3. **Equation Transformation:**
   - Consider the equation:
     \[
     a \cdot \frac{P(n)}{n^d} = \frac{P(z_n)}{z_n^d} \cdot \left( \frac{z_n}{n} \right)^d
     \]
   - Taking the limit as \( n \to \infty \):
     \[
     \lim_{n \to \infty} \left| \frac{z_n}{n} \right| \to A
     \]
   - Define \( x_n := \frac{z_n}{An} \). Note that \( x_n \) is bounded.

4. **Limit Analysis:**
   - Rewrite the original equation in terms of \( x_n \) and take the limit as \( n \to \infty \):
     \[
     \lim_{n \to \infty} An(x_n^d - 1) + b(x_n^{d-1} - A) = 0
     \]
   - Call this equation \( \spadesuit \).

5. **Case Analysis Based on Parity of \( d \):**

   **Case 1: \( d \) is odd.**
   - As \( n \to \infty \), \( x_n \to 1 \).
   - Using this in \( \spadesuit \):
     \[
     \lim_{n \to \infty} An(x_n - 1) \to \frac{b(A - 1)}{d} \implies \lim_{n \to \infty} z_n - An \to \frac{b(A - 1)}{d} \overset{\text{def}}{:=} B
     \]
   - For large \( n \), \( z_n = An + B \) (since \( z_n \) is an integer).
   - The original equation becomes:
     \[
     A^d P(n) = P(An + B)
     \]
   - Define \( c = \frac{B}{1 - A} \) and consider \( Q(x) = P(x + c) \):
     \[
     Q(Ax) = P(Ax + c) = P(A(x + c) + B) = A^d P(x + c) = A^d Q(x)
     \]
   - By comparing coefficients, \( Q(x) = x^d \), hence \( P(x) = C(x - c)^d \).

   **Case 2: \( d \) is even.**
   - Assume \( d = 2k \).
   - \( x_n \) does not necessarily converge to 1, but \( x_n = O\left(\frac{1}{n}\right) \).
   - From \( \spadesuit \):
     \[
     \lim_{n \to \infty} Ank(x_n^2 - 1) + b(x_n - A) = 0
     \]
   - Set \( y_n = z_n + \frac{b}{2k} \):
     \[
     \lim_{n \to \infty} \frac{y_n^2}{An} - An \to \frac{bA}{k}
     \]
   - Also, \( \lim_{n \to \infty} \left| \frac{y_n}{An} \right| \to 1 \):
     \[
     \lim_{n \to \infty} \left( |y_n| - An \right) \to \frac{bA}{2k}
     \]
   - For distinct integers \( i, j \) with \( |i - j| \leq 2 \) and large \( x, y \):
     \[
     |y_i - y_j| = \pm (z_i - z_j)
     \]
   - Since \( 2A \) is an integer, for large \( n \):
     \[
     \varepsilon z_{k_n} - Ak_n \to \frac{b(A - \varepsilon)}{2k} \overset{\text{def}}{:=} C
     \]
   - The original equation translates to:
     \[
     A^d P(k_n) = (\varepsilon A)^d = P(\varepsilon Ak_n + \varepsilon C)
     \]
   - Proceeding similarly to the odd case, we conclude \( P = c(x - x_0)^d \) for some \( c \in \mathbb{R} \) and \( x_0 \in \mathbb{Q} \).

6. **Conclusion:**
   - The only great polynomials are of the form \( P(x) = c(x - x_0)^d \) where \( c \in \mathbb{R} \) and \( x_0 \in \mathbb{Q} \).

\(\blacksquare\)

The final answer is \( \boxed{ P(x) = c(x - x_0)^d } \) where \( c \in \mathbb{R} \) and \( x_0 \in \mathbb{Q} \).