We have from the given that: P(Q(Q(x)))=Q(Q(Q(x))) , so polYnomials P and Q take for a sufficiet big nr of values(as many as Q(Q(x))) take the same value, so they must be identical! I hope I've explained it well enough ,at least the idee should be clear!

I think a problem from a recent Romanian TST stated that $P=Q$ if $P\circ P=Q\circ Q$, but I'm not sure right now. I'd have to check.

What about $P(x)=x$ and $Q(x)=-x$?

Yes, sorry about that, $P$ and $Q$ were monic complex polynomials in that problem. It's from a TST in $2000$.

Myth, your polynomials do not satisfy the second condition.

grobber wrote: $P$ and $Q$ were monic complex polynomials in that problem. It is obvious statement for real polynomials, but complex... I need additional reasoning.

Xixas wrote: Myth, your polynomials do not satisfy the second condition. It was counterexample for grobber's post.

Myth wrote: grobber wrote: $P$ and $Q$ were monic complex polynomials in that problem. It is obvious statement for real polynomials, but complex... I need additional reasoning. Yes, for real polynomials you don't need any messy coefficient identification: you can do it by some analysis, studying what happens when $x\to\infty$. I think that's what I did when I first saw the problem, because I didn't read the text properly and missed "complex polynomials". On the other hand, I think we have to resort to identifying the coefficients if $P,Q$ are complex, and I don't like such problems, so I won't try to solve it .

Actually, at this time I am writing solution, which uses behavour when $x\to\infty$.

Xixas wrote: P(x) and Q(x) are polynomials of positive degree such that for all x P(P(x))=Q(Q(x)) and P(P(P(x)))=Q(Q(Q(x))). Does this necessarily mean that P(x)=Q(x)? From $P(P(P(x)))=Q(Q(Q(x)))$ we immediately obtain $P$ and $Q$ have the same leading coefficient. Suppose $P(x)\neq Q(x)$, then $P(x)=a_nx^n+...+a_kx^k+p_{k-1}x^{k-1}+...+p_0$ and $P(x)=a_nx^n+...+a_kx^k+q_{k-1}x^{k-1}+...+q_0$, where $p_{k-1}>q_{k-1}$ (for example). 0) $n=1$. Any questions? 1) $a>0$. Then there is $M$ s.t. $P(x)$ and $Q(x)$ increase on $[M,+\infty]$, $P(M)>M$, $Q(M)>M$, $P(x)>Q(x)$ on $[M,+\infty]$. We see that $P(P(x))>Q(P(x))>Q(Q(x))$ on $[M,+\infty]$. 2) $a<0$. Almost the same argumentation, but a bit more boring.

Xixas wrote: P(x) and Q(x) are polynomials of positive degree such that for all x P(P(x))=Q(Q(x)) and P(P(P(x)))=Q(Q(Q(x))). Does this necessarily mean that P(x)=Q(x)? Let $S$ be the image of $P(P(x))$. Since $\deg P > 0$, $S$ has infinitely many elements. Since $P(P(P(x)))=Q(Q(Q(x)))$, we see that if $y \in S$, then $P(y) = Q(y)$. Two polynomials agree on infinitely many values imply that they're identical. Hence $P = Q$.

It is strange. What if we remove condition $P(P(x))=Q(Q(x))$.

grobber wrote: Yes, sorry about that, $P$ and $Q$ were monic complex polynomials in that problem. It's from a TST in $2000$. Yes I remember that problem. Nice one, solved it quickly, with equalizing coefficients