# Find all polynomials P(x) which have the property P[F(x)]=F[P(x)], P(0)=0 where F(x) is a given function with the property F(x)>x for all x>=0.

Find all polynomials $P\left(x\right)$ which have the property
$P\left[F\left(x\right)\right]=F\left[P\left(x\right)\right],\phantom{\rule{1em}{0ex}}P\left(0\right)=0$
where $F\left(x\right)$ is a given function with the property $F\left(x\right)>x$ for all $x\ge 0$
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veirenca77
Denote ${x}_{0}=0$ and ${x}_{n+1}=F\left({x}_{n}\right)$ for every $n\ge 0$. Since $F\left(x\right)>x$ for every $x\ge 0$${x}_{n+1}>{x}_{n}$ for every $n\ge 0$. Note that $P\left({x}_{0}\right)={x}_{0}$ and
$P\left({x}_{n+1}\right)=P\left(F\left({x}_{n}\right)\right)=F\left(P\left({x}_{n}\right)\right),\phantom{\rule{1em}{0ex}}\mathrm{\forall }n\ge 0.$
Then by induction, it is easy to see that
$\begin{array}{}\text{(1)}& P\left({x}_{n}\right)={x}_{n},\phantom{\rule{1em}{0ex}}\mathrm{\forall }n\ge 0.\end{array}$
(1) implies that the polynomial $P\left(x\right)-x$ has infinitely many roots, i.e. $P\left(x\right)\equiv x$