Problem 1012

[Proof] For $n \in N$ using induction. Since the polynomial $Q_{1}(x)$ divides itself, the conclusion holds when $n=1$. Now assume that the identity $Q_{n}(x)=R_{n}(x) Q_{1}(x)$ has been proven for some $n$, where $R_{n}(x)$ is a polynomial, then we have
$$\begin{aligned} Q_{n+1}(x) & \equiv P\left[Q_{n}(x)+x\right]-x \\ & \equiv P\left[R_{n}(x) Q_{1}(x)+x\right]-x \\ & \equiv\left\{P\left[R_{n}(x) \cdot Q_{1}(x)+x\right\}-P(x)+[P(x)-x]\right. \end{aligned}$$

Assume

then
$$\begin{aligned} Q_{n+1}(x) & \equiv \sum_{k=0}^{m} a_{k}\left(\left(R_{n}(x) Q_{1}(x)+x\right)^{k}-x^{k}\right)+Q_{1}(x) \\ & \equiv \sum_{k=0}^{m} a_{k} R_{n}(x) Q_{1}(x) S_{k}(x)+Q_{1}(x) \\ & \equiv Q_{1}(x)\left(1+\sum_{k=0}^{m} a_{k} R_{n}(x) S_{k}(x)\right) \end{aligned}$$

where
$$S_{k}(x)=\sum_{j=0}^{k-1}\left(R_{n}(x) Q_{1}(x)\right)^{j} x^{k-j-1},$$

Here, we applied the identity
$$a^{k}-b^{k}=(a-b) \sum_{j=0}^{k-1} a^{j} b^{k-j-1},$$

and took $\quad a=R_{n}(x) Q_{1}(x)+x, b=x$.
This proves that the conclusion holds for $n+1$.