Prove that $P(x)-x$ divides $ P(P(P(x))) − x $ 
Prove that for any polynomial $ P(x) $ with real coefficients, other than polynomial $x$, the polynomial $ P(P(P(x))) − x $ is divisible by $ P(x) − x $.

Since we have $a-b\mid P(a)-P(b)$ for each $a,b$, we have also $$P(x)-x\mid P(P(x))-P(x)$$ and thus $$P(x)-x\mid (P(P(x))-P(x))+ (P(x)-x)$$
so $$P(x)-x\mid P(P(x))-x\mid P(P(P(x)))-P(x)$$
and thus $$P(x)-x\mid (P(P(P(x)))-P(x))+ (P(x)-x)$$
and finaly we have $$P(x)-x\mid P(P(P(x)))-x$$
I'm not satisfied with the lenght of the proof. Any shorter way?
 A: Yes, $\bmod P(x)\!-\!x\!:\ \color{#c00}{P(x)\equiv x}\,$ so $\,P(P(\color{#c00}{P(x)}))\equiv P(P(\color{#c00}{x}))\equiv P(x)\equiv x\ $ 
We iteratively applied the Polynomial Congruence Rule thrice to replace $\,P(x)\,$ by the congruent $\,x\,$ (the linked proof of the rule is given for integers but works in any commutative ring since it uses only ring axioms). An obvious induction extends this to $\,P^n(x)\equiv x\,$ for all $\,n\ge 0\,$ yielding 
Corollary $\ P(x)\!-\!x\,$ divides $\, P^n(x)\!-\!x\,$ for all $\,n\in\Bbb N,\,$ and all polynomials $\,P(x)$
Hence - from a modular perspective - we can view it as a special case of the obvious


*

*fixed points stay fixed under iteration, which has an obvious inductive step,


$\quad\ $ namely: $\, $ if $\ \color{#c00}{f(x) = x}\ $ then $\, {f^{\large n}(x) = x}\,\Rightarrow\, f^{\large n+1}(x) = f^n(\color{#c00}{f(x)})=f^n(\color{#c00}x)=x$
As usual, modular arithmetic simplifies the matter by replacing divisibility relations by more intuitive arithmetic (ring) operations, enabling us to appeal to our well-honed algebraic (ring) intuition from common number systems.
