Let us consider a polynomial $P(x)$ with integers coefficients satisfying
$$P(-1)=-4,\ P(-3)=-40,\text{ and } P(-5)=-156.$$
What is the largest possible number of integers $x$ satisfying
$$P(P(x))=x^2?$$

To solve the problem, we need to determine the largest possible number of integers \( x \) satisfying \( P(P(x)) = x^2 \) for a polynomial \( P(x) \) with integer coefficients, given the conditions:
\[ P(-1) = -4, \quad P(-3) = -40, \quad P(-5) = -156. \]

We will use modular arithmetic to analyze the problem.

1. **Consider the polynomial modulo 3:**
   Since \( P(x) \) has integer coefficients, we know that \( P(x+3) \equiv P(x) \pmod{3} \). This property will help us analyze the polynomial under modulo 3.

2. **Case 1: \( u \equiv 0 \pmod{3} \)**
   \[
   \begin{align*}
   P(P(u)) &\equiv P(P(0)) \pmod{3} \\
   u^2 &\equiv 0 \pmod{3} \\
   P(P(0)) &\equiv 0 \pmod{3}
   \end{align*}
   \]
   However, we need to check the given values:
   \[
   P(-3) = -40 \implies P(-3) \equiv -1 \pmod{3} \\
   P(-1) = -4 \implies P(-1) \equiv 2 \pmod{3}
   \]
   Since \( P(-3) \equiv -1 \pmod{3} \) and \( P(-1) \equiv 2 \pmod{3} \), we have a contradiction because \( P(P(0)) \equiv 0 \pmod{3} \) does not match \( u^2 \equiv 0 \pmod{3} \).

3. **Case 2: \( u \equiv 1 \pmod{3} \)**
   \[
   \begin{align*}
   P(P(u)) &\equiv P(P(1)) \pmod{3} \\
   u^2 &\equiv 1 \pmod{3} \\
   P(P(1)) &\equiv 1 \pmod{3}
   \end{align*}
   \]
   Given values:
   \[
   P(-5) = -156 \implies P(-5) \equiv 0 \pmod{3} \\
   P(-3) = -40 \implies P(-3) \equiv -1 \pmod{3}
   \]
   Since \( P(-5) \equiv 0 \pmod{3} \) and \( P(-3) \equiv -1 \pmod{3} \), we have a contradiction because \( P(P(1)) \equiv 1 \pmod{3} \) does not match \( u^2 \equiv 1 \pmod{3} \).

4. **Case 3: \( u \equiv 2 \pmod{3} \)**
   \[
   \begin{align*}
   P(P(u)) &\equiv P(P(2)) \pmod{3} \\
   u^2 &\equiv 4 \equiv 1 \pmod{3} \\
   P(P(2)) &\equiv 1 \pmod{3}
   \end{align*}
   \]
   Given values:
   \[
   P(-1) = -4 \implies P(-1) \equiv 2 \pmod{3} \\
   P(-3) = -40 \implies P(-3) \equiv -1 \pmod{3}
   \]
   Since \( P(-1) \equiv 2 \pmod{3} \) and \( P(-3) \equiv -1 \pmod{3} \), we have a contradiction because \( P(P(2)) \equiv 1 \pmod{3} \) does not match \( u^2 \equiv 1 \pmod{3} \).

Since all cases lead to contradictions, there are no integers \( x \) that satisfy \( P(P(x)) = x^2 \).

The final answer is \( \boxed{0} \)