# Proof for divisibility of polynomials. [closed]

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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## 2 Answers

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

• 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. – Bill Dubuque Apr 7 at 18:36

$$\bmod P(x)\!-\!x\!:\,\ \color{#c00}{P(x)\equiv x}\,\Rightarrow\, P(P(\color{#c00}{P(x)}))\equiv P(P(\color{#c00}{x)}))\equiv P(x)\equiv x$$

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 $$\,P(x)$$