Find all polynomials $P(x)$, so that $P[F(x)] = F[P(x)],P(0)=0$ where $F$ is a given function with the property $F(x)>x$ for all $x\ge0$

Question

Find all polynomials $P(x)$ so that $$P[F(x)] = F[P(x)]\text{ and } P(0)=0,$$ where $F$ is a given function with the property $F(x)>x$ for all $x≥0$.

This question is from the book Problem Solving Strategies by A. Engel.

Here is the solution Transcribed from this image

Let $F(0)=a_0>0$. Then $P(F(0))=F(P(0))\Leftrightarrow P(a_0)=a_0$. Similarly, we get $F(a_n)=a_{n+1}$, $P(a_n)=a_n$, and $a_{n+1}>a_n$. We must find all polynomials with infinitely many points on $y=x$. Then $P(x)-x$ has infinitely many zeros, i.e., $P(x)=x$.

Can anyone please explain the solution in detail (from second line)?

Thanks in advance.

Answer 1

$a_n$'s are defined inductively by $a_{n+1}=F(a_n)$. Note that $P(F(a_0))=F(P(a_0))$ gives $P(a_1)=F(a_0)=a_1$ since $P(a_0)=a_0$. Now I will let you use induction to prove that $P(a_n)=a_n$ for all $n$. Let $Q(x)=P(x)-x$. Then $Q(a_n)=0$ for all $n$. Also $a_{n+1}=F(a_n) >a_n$ which makes the numbers $a_n$ distinct. Hence the polynomial $Q$ has infinitely many zeros. This implies that $Q(x)=0$ for all $x$ or $P(x)=x$ for all $x$.