Polynomials without real solutions

[tanmoy]

$P(x),Q(x)$ are two polynomials such that $P(x) = Q(x)$ has no real solution,and $P(Q(x)) \equiv Q(P(x))$ $\forall x \in \mathbb{R}$.
Prove that $P(P(x)) = Q(Q(x))$ has no real solution.

[joydip]

My solution:

Let $P$ and $Q$ be the graph of $P(X),Q(X)$.
$P,Q$ doesn't intersect (as $P(X)$ can't be equal to $Q(x)$).Assume $P(P(X))=Q(Q(X))$.Consider the points $A(P(X),P(P(X))),B(P(X),Q(P(X))),C(Q(X),P(Q(X))),D(Q(X),Q(Q(X)))$. $AD \parallel BC \parallel XX^{'}$ and $AB \parallel YY^{'} \parallel CD$.So $ABCD$ makes a rectangle.
$P,Q$ can't intersect any line parallel to $YY^{'}$ twice (obviously,they are functions)..............(1)
$P$ goes though $A,C$.$Q$ goes though $B,D$.$P$ divides the plane into two regions one containing $B$,other containing $D$(because of (1)).So $P,Q$ must intersect,contradiction.

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