TITLE: Prove that for any polynomial $P(x)$ there exist polynomials $F(x)$ and $G(x)$ such that $F\left(G(x) \right)-G\left(F(x) \right)=P(x)$
QUESTION [6 upvotes]: Prove that for any polynomial $P(x)$ there exist polynomials $F(x)$ and $G(x)$ such that $\forall x \in \mathbb R$,
$$F\big(G(x) \big)-G\big(F(x) \big)=P(x)\,.$$

My work so far:
Let $G(x)=x+1$. Then $$F(x+1)-F(x)=P(x)+1\,.$$
I need help here.

REPLY [2 votes]: Hint $\ $ Just like the derivative, the linear operator $\, D f(x) = f(x\!+\!1) - f(x) \,$ acts on polynomials by decreasing the degree by $1,\,$ since $\,D x^n = (x+1)^n-x^n = c_n x^{n-1} +g(x)\,$ with $\,c_n\neq 0,\,\deg g < n-1$. For any such linear operator one can solve equations of the form $\, D f(x) = g(x)\,$ for given polynomial $g$ by using undetermined coefficients and induction, e.g. 
$$ g_n x^n+\cdots = D (f_{n+1} x^{n+1} + \cdots) = c_{n+1} f_{n+1} x^n + \cdots\,\Rightarrow\, f_{n+1} = g_n/c_{n+1}$$
Substituting that value for $\,f_{n+1}\,$ we reduce to a problem with smaller degree $\,g$