# Find all polynomials with real coefficients $P(x),Q(x)$ such that for every real number $x$ we have: $P(Q(x))=P(x)^{2017}$

Find all polynomials with real coefficients $P(x),Q(x)$ such that for every real number $x$ we have: $$P(Q(x))=P(x)^{2017}$$

This problem is really hard to me!! I have no idea for the solution!
It has been given in summer camp for Iranian MO.

• Clearly $Q(x)$ must have degree $2017$ for otherwise $P(Q(x))$ and $P(x)^{2017}$ have different degrees, and therefore there are only finitely many solutions to the equation. You also get information about the leading coefficients. But, this doesn't say much. Aug 14, 2017 at 6:36
• @JyrkiLahtonen There are infinitely many; $P(x)=x^n,\,Q(x)=x^{2017}$ is a solution for all $n$. Aug 14, 2017 at 6:42
• Yes, @stewbasic. I spotted those solutions right away. $Q(x)=-x^{2017}$ also works if $n$ is even. But are they all? Aug 14, 2017 at 6:44
• @JyrkiLahtonen - Didn't you mean P(x)=x? Aug 14, 2017 at 6:52
• There are solutions $P(x)=ax^n, Q(x)=bx^{2017}$ with $a^{2016}=b^n$. In particular there is the solution $P=Q=0$. Aug 14, 2017 at 6:57

Contrary to my initial guess, we do not need that $2017$ is prime.

Theorem. Let $N>1$ and $P,Q\in\Bbb C[X]$ such that $$\tag0 P(Q(x))=P(x)^{N}\qquad\text{for infinitely many }x\in\Bbb C.$$ Then either $P$ is constant or there are $n\in\Bbb N$, $a,b,\beta\in\Bbb C$ with $a^{N-1}=b^n$ and $$\tag1 P(X)=a(X-\beta)^n,\quad Q(X)=b(X-\beta)^N+\beta.$$

Proof. First of all, note that the polynomial equation $(0)$ will also hold for all $x\in\Bbb C$. If $f$ is a polynomial and $z\in\Bbb C$, let $v_f(z)$ denote the order of $z$ as root of $f$. So $v_f(z)=0$ for almost all $z$ and $\sum_{z\in\Bbb C}v_f(z)=\deg f$.

Let $n=\deg P$, $m=\deg Q$. As we are interested only in non-constant $P$, we can assume $n>0$. The degree of $P\circ Q$ is $nm$, that of $P^{N}$ is $Nn$. We conclude that $m=N$.

From $(0)$ we see that for $\alpha\in\Bbb C$ and $\beta:=Q(\alpha)$, $$\tag2 N\cdot {v_P(\alpha)} = v_P(Q(\alpha))\cdot v_{Q-\beta}(\alpha).$$ For $\alpha$ that maximizes $v_P$ (in particular, $v_P(\alpha)>0$), $(2)$ implies $v_{Q-\beta}\ge N$. Thus $Q-\beta$ is a multiple of $(X-\alpha)^N$, i.e., $$\tag3Q(X)=b(X-\alpha)^N+\beta$$ with $\alpha,\beta,b\in\Bbb C$, $b\ne0$. Then $\alpha$ is the only root of the derivative $Q'(X)=Nb(X-\alpha)^{N-1}$, hence $Q(X)-w$ with $w\ne \beta$ has $N$ distinct, simple roots and at most one of these equals $\beta$. Accordingly, let $w^*$ be a root of $Q(X)-w$ with $w^*\ne\beta$. Adapting $(2)$, we find $Nv_P(w^*)=v_P(w)$. By infinite descent, we find that $v_P(w)=0$ for all $w\ne\beta$. In other words, $$\tag4P(X)=a(X-\beta)^n$$ for some $a\ne 0$.

Then $\beta$ is the only root of $P(X)^{N}$, hence the only root of $Q(X)-\beta$. Therefore $\beta=\alpha$. By a quick comparison of leading coefficients, we arrive at $(1)$. $\square$

Remark: If we know that $P,Q$ have real coefficients, then of course $a,b,\beta$ are real.