$P(x)$ is Polynomial with real coefficients & $P^n(x)$ is equal to $P(P(...(P(x))...)$ that the number of $P$ is $n$

Question

Now prove that $Q(x)$ is divisible by polynomials $P(x)-x$

$Q(x) $=$ P^{2018} (x) $- $2P^{1439} (x)$ -$ P^{1397} (x)$

At the first i tried to Invoicing the $Q(x)$ but i got nowhere , could you help guys?

Answer 1

Hint: Prove that $$P(x)-x\mid P^n(x) -x$$ for all positive integers $n$.

Answer 2

I think your question is wrong, because if $P(x)=x^2$ then $P^n(x)=x^{2^n}$ and we need $x^2-x=x(x-1)$ to divide $Q(x)=x^{2^{2018}}-2x^{2^{1439}}-x^{2^{1397}}$. But for this $Q(1)$ must be $0$, but actually we have $Q(1)=1-2-1=-2$. But if $Q(x)=P^{2018}(x)-2P^{1439}(x)+P^{1397}(x)$ then $P(x)-x|Q(x)$. To see this let $P(x)-x=(x-a_1)(x-a_2)(x-a_3)\cdots (x-a_m)$. Then for $x=a_i, 1\le i \le m$, $Q(x)=P^{2018}(x)-2P^{1439}(x)+P^{1397}(x)=x-2x+x=0$. Here we used the fact that for $x=a_i, P(x)=x$. Now factor theorem finishes the problem.