Davrbek wrote: The polynomial $P (x)$ is such that the polynomials $P (P (x))$ and $P (P (P (x)))$ are strictly monotone on the whole real axis. Prove that $P (x)$ is also strictly monotone on the whole thing axis. Since $P(P(x))$ is strictly monotonous, then it has odd degree and is surjective. So $P(x)$ has odd degree and is surjective. If $P(a)=P(b)$ for some $a\ne b$, then, since surjective, $\exists u\ne v$ such that $P(u)=a$ and $P(v)=b$ And then $P(P(u))=P(P(v))$, impossible since $P(P(x))$ is monotonous. Hence $P(x)$ is injective, and so monotonous. Q.E.D. (And, btw, no need for $P(P(P(x))$ monotonous.)

I don't use assumption $P(P(P(x))$ is monotonous too. if $P(x)$ is constant everything works. Assume $P(x)$ is not constant. For all $x \in R$ the following has the same sign different than $0$. $[P(P(x))]'= P'(P(x))\cdot P'(x)$. Assume WLOG for all $x\in R$ we have $[P(P(x))]'<0$. $P(x)$ is strictly monotone iff $P'(x)$ has the same sign for all $x\in R$. Assume there exist real numbers $a\neq b\wedge P'(a)>0>P'(b)$. Then by intermediate value theorem $\exists_{c\in R} \ P'(c) =0$. Thus 0>$[P(P(c))]'= P'(P(c))\cdot P'(c)=0$ contradiction. This finishes proof.

pco wrote: If $P(a)=P(b)$ for some $a\ne b$, then, since surjective, $\exists u\ne v$ such that $P(u)=a$ and $P(v)=b$ And then $P(P(u))=P(P(v))$, impossible since $P(P(x))$ is monotonous. You could do it faster. If $P(a)=P(b)$ for some $a\ne b$, then $P(P(a))=P(P(b))$ which is contradiction with $P (P (x))$ is strictly monotone.

Davrbek wrote: The polynomial $P (x)$ is such that the polynomials $P (P (x))$ and $P (P (P (x)))$ are strictly monotone on the whole real axis. Prove that $P (x)$ is also strictly monotone on the whole thing axis. If we use the second condition as well, then it gives a one-liner, so why not! Assume on the contrary that $a>b$ but $P(a)<P(b)$. Then $P(P(P(a)))<P(P(P(b)))$ by the fact that $P(P(x))$ is stricly monotone, but this is a contradiction since $P(P(P(x)))$ is also monotone!

Addition of $P(P(P(x)))$ was one of tricky moves on the problem committee That's an easy pronlem, so why not!

$P^2$ and $P^3$ being monotonous and surjective tells us that they are bijective. We shall prove $P$ is bijective too. Since $P^2$ is surjective, it has odd degree, and the degree of $P$ is $\sqrt{\text{deg}(P^2)}$ which is also odd, hence $P$ is surjective. If $P$ was not injective, then clearly $P^2$ isnt either, contradiction. So $P$ is bijective what the heck is this

eh???