$p(x)$ and $q(x)$ are polynomials which satisfy the identity $p(q(x)) = q(p(x))$ for all real $x$ . If the equation $p(x) = q(x)$ has no real solutions, prove that the equation $p(p(x)) = q(q(x))$ has no real solutions.
What I Tried :- I just assumed $p(x) = a_nx^n + \dots + a_1x + a_0$ and $q(x) = b_nx^n + \dots + b_1x + b_0$.
It was also given that $p(x) = q(x)$ has no real solutions, so in the context of real numbers, I can claim $p(x) \neq q(x)$ for all $x$. But I am not sure how that is going to help, and I have not used this information yet.
After this, we have $p(b_nx^n + \dots + b_1x + b_0) = q(a_nx^n + \dots + a_1x + a_0)$.
But now, I am stuck. Expanding more is going to make it complicated.
I am also thinking of showing this by contradiction somehow, that I am assuming first there exists a real root to $p(p(x)) = q(q(x))$, but I am not sure how to do it.
Can anyone help me? Thank You.
$p-q$ is a continuous function. If it has no real zeros then it is always positive or always negative. Suppose $p-q>0$. Then $p(p(x)) >q(p(x))=p(q(x))> q(q(x))$ so $p(p(x)) \neq q(q(x))$. The other case is similar.