Problem 1500

1. Assume Without Loss of Generality (WLOG):
Let us assume $P(0) > Q(0)$. This assumption is made without loss of generality because if $P(0) < Q(0)$, we can simply swap $P$ and $Q$ in the argument.

2. **Define the Function $f(x)$:**
Define $f(x) = P(x) - Q(x)$. Since $P(x) = Q(x)$ has no real solutions, $f(x) \neq 0$ for all $x \in \mathbb{R}$.

3. **Continuity and Sign of $f(x)$:**
Since $P$ and $Q$ are polynomials, they are continuous functions. Therefore, $f(x)$ is also a continuous function. Given $f(0) > 0$ (since $P(0) > Q(0)$), and $f(x)$ is continuous, $f(x)$ must be positive for all $x \in \mathbb{R}$. If $f(x)$ were to change sign, there would exist some $a$ such that $f(a) < 0$, and by the Intermediate Value Theorem, there would be some $b$ such that $f(b) = 0$, which contradicts the assumption that $P(x) = Q(x)$ has no real solutions.

4. **Inequality for All $x$:**
Therefore, $f(x) > 0$ for all $x \in \mathbb{R}$, which implies $P(x) > Q(x)$ for all $x \in \mathbb{R}$.

5. **Evaluate $P(P(x))$ and $Q(Q(x))$:**
Consider the expressions $P(P(x))$ and $Q(Q(x))$. We need to show that $P(P(x)) = Q(Q(x))$ has no real solutions.

6. Use the Commutativity Condition:
Given $P(Q(x)) = Q(P(x))$ for all $x \in \mathbb{R}$, we can use this to compare $P(P(x))$ and $Q(Q(x))$:
$$Q(Q(x)) < P(Q(x)) = Q(P(x)) < P(P(x))$$
This chain of inequalities follows from the fact that $P(x) > Q(x)$ for all $x \in \mathbb{R}$.

7. Conclusion:
Since $Q(Q(x)) < P(P(x))$ for all $x \in \mathbb{R}$, it follows that $P(P(x)) = Q(Q(x))$ has no real solutions.

$\blacksquare$