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I have no idea how to proceed with the following question. Please help!

"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 $."

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closed as off-topic by José Carlos Santos, Sil, John Omielan, TheSimpliFire, RRL Apr 8 at 1:39

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Remember that $$a-b\mid P(a)-P(b)$$

so $$P(x)-x\mid P(P(x))-P(x)$$ 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 $$P(x)-x\mid P(P(P(x)))-x$$

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    $\begingroup$ Modular arithmetic was invented to clarify proofs like this where the divisibilty relation greatly obfuscates the algebraic (operational) essence of the matter - here the simple notion of a fixed point - see my answer. $\endgroup$ – Bill Dubuque Apr 7 at 18:36
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$\bmod P(x)\!-\!x\!:\,\ \color{#c00}{P(x)\equiv x}\,\Rightarrow\, P(P(\color{#c00}{P(x)}))\equiv P(P(\color{#c00}{x)}))\equiv P(x)\equiv x$

Remark $ $ The proof is a special case of: fixed points stay fixed on iteration by induction,

namely: $\, $ if $\ \color{#c00}{f(x) = x}\ $ then $\, {f^{\large n}(x) = x}\,\Rightarrow\, f^{\large n+1}(x) = f^n(\color{#c00}{f(x)})=f^n(\color{#c00}x)=x$

Corollary $\ P(x)\!-\!x\,$ divides $\, P^n(x)\!-\!x\,$ for all $\,n\in\Bbb N,\,$ and all polynomials $\,P(x)$

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