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Proof for divisibility of polynomials.

Show that $a^p^n=amod p$Divisibility problemPolynomial divisibility proofPolynomials and Divisibility Rule.Induction proof, divisibilityDivisibility of a polynomial by another polynomialDoes there exist a polynomial $f(x)$ with real coefficients such that $f(x)^2$ has fewer nonzero coefficients than $f(x)$?Polynomials - Relation of DivisibilityProof using concept of polynomials.Proof of Existence of A Rational Polynomial which has Irrational Root for an EquationTricky problem of infinite harmonic sum of polynomials

0

$begingroup$

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 $."

polynomials divisibility

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asked 5 hours ago

HeetGorakhiyaHeetGorakhiya

203

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0

$begingroup$

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 $."

polynomials divisibility

share|cite|improve this question

asked 5 hours ago

HeetGorakhiyaHeetGorakhiya

203

New contributor

HeetGorakhiya is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

add a comment | 

$begingroup$

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 $."

polynomials divisibility

share|cite|improve this question

asked 5 hours ago

HeetGorakhiyaHeetGorakhiya

203

New contributor

HeetGorakhiya is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

share|cite|improve this question

asked 5 hours ago

HeetGorakhiyaHeetGorakhiya

203

New contributor

HeetGorakhiya is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|cite|improve this question

asked 5 hours ago

HeetGorakhiyaHeetGorakhiya

203

New contributor

HeetGorakhiya is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked 5 hours ago

HeetGorakhiyaHeetGorakhiya

203

New contributor

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asked 5 hours ago

HeetGorakhiyaHeetGorakhiya

203

2 Answers
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4

$begingroup$

Remember that $$a-bmid P(a)-P(b)$$

so $$P(x)-xmid P(P(x))-P(x)$$ and thus $$P(x)-xmid (P(P(x))-P(x))+ (P(x)-x)$$

so $$P(x)-xmid P(P(x))-xmid P(P(P(x)))-P(x)$$

and thus $$P(x)-xmid (P(P(P(x)))-P(x))+ (P(x)-x)$$

and finaly we have $$P(x)-xmid P(P(P(x)))-x$$

share|cite|improve this answer

edited 5 hours ago

answered 5 hours ago

Maria MazurMaria Mazur

50k1361124

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  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
  5 hours ago
  
  

  
  
  
  

add a comment | 

3

$begingroup$

$bmod P(x)!-!x!:, color#c00P(x)equiv x,Rightarrow, P(P(color#c00P(x)))equiv P(P(color#c00x)))equiv P(x)equiv x$

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

namely: $, $ if $ color#c00f(x) = x $ then $, f^large n(x) = x,Rightarrow, f^large n+1(x) = f^n(color#c00f(x))=f^n(color#c00x)=x$

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

share|cite|improve this answer

edited 5 hours ago

answered 5 hours ago

Bill DubuqueBill Dubuque

214k29196654

$endgroup$

add a comment | 

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4

$begingroup$

Remember that $$a-bmid P(a)-P(b)$$

so $$P(x)-xmid P(P(x))-P(x)$$ and thus $$P(x)-xmid (P(P(x))-P(x))+ (P(x)-x)$$

so $$P(x)-xmid P(P(x))-xmid P(P(P(x)))-P(x)$$

and thus $$P(x)-xmid (P(P(P(x)))-P(x))+ (P(x)-x)$$

and finaly we have $$P(x)-xmid P(P(P(x)))-x$$

share|cite|improve this answer

edited 5 hours ago

answered 5 hours ago

Maria MazurMaria Mazur

50k1361124

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  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
  5 hours ago
  
  

  
  
  
  

add a comment | 

4

$begingroup$

Remember that $$a-bmid P(a)-P(b)$$

so $$P(x)-xmid P(P(x))-P(x)$$ and thus $$P(x)-xmid (P(P(x))-P(x))+ (P(x)-x)$$

so $$P(x)-xmid P(P(x))-xmid P(P(P(x)))-P(x)$$

and thus $$P(x)-xmid (P(P(P(x)))-P(x))+ (P(x)-x)$$

and finaly we have $$P(x)-xmid P(P(P(x)))-x$$

share|cite|improve this answer

edited 5 hours ago

answered 5 hours ago

Maria MazurMaria Mazur

50k1361124

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  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
  5 hours ago
  
  

  
  
  
  

add a comment | 

$begingroup$

Remember that $$a-bmid P(a)-P(b)$$

so $$P(x)-xmid P(P(x))-P(x)$$ and thus $$P(x)-xmid (P(P(x))-P(x))+ (P(x)-x)$$

so $$P(x)-xmid P(P(x))-xmid P(P(P(x)))-P(x)$$

and thus $$P(x)-xmid (P(P(P(x)))-P(x))+ (P(x)-x)$$

and finaly we have $$P(x)-xmid P(P(P(x)))-x$$

share|cite|improve this answer

edited 5 hours ago

answered 5 hours ago

Maria MazurMaria Mazur

50k1361124

$endgroup$

share|cite|improve this answer

edited 5 hours ago

answered 5 hours ago

Maria MazurMaria Mazur

50k1361124

answered 5 hours ago

Maria MazurMaria Mazur

50k1361124

answered 5 hours ago

Maria MazurMaria Mazur

50k1361124

3

$begingroup$

$bmod P(x)!-!x!:, color#c00P(x)equiv x,Rightarrow, P(P(color#c00P(x)))equiv P(P(color#c00x)))equiv P(x)equiv x$

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

namely: $, $ if $ color#c00f(x) = x $ then $, f^large n(x) = x,Rightarrow, f^large n+1(x) = f^n(color#c00f(x))=f^n(color#c00x)=x$

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

share|cite|improve this answer

edited 5 hours ago

answered 5 hours ago

Bill DubuqueBill Dubuque

214k29196654

$endgroup$

add a comment | 

3

$begingroup$

$bmod P(x)!-!x!:, color#c00P(x)equiv x,Rightarrow, P(P(color#c00P(x)))equiv P(P(color#c00x)))equiv P(x)equiv x$

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

namely: $, $ if $ color#c00f(x) = x $ then $, f^large n(x) = x,Rightarrow, f^large n+1(x) = f^n(color#c00f(x))=f^n(color#c00x)=x$

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

share|cite|improve this answer

edited 5 hours ago

answered 5 hours ago

Bill DubuqueBill Dubuque

214k29196654

$endgroup$

add a comment | 

$begingroup$

$bmod P(x)!-!x!:, color#c00P(x)equiv x,Rightarrow, P(P(color#c00P(x)))equiv P(P(color#c00x)))equiv P(x)equiv x$

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

namely: $, $ if $ color#c00f(x) = x $ then $, f^large n(x) = x,Rightarrow, f^large n+1(x) = f^n(color#c00f(x))=f^n(color#c00x)=x$

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

share|cite|improve this answer

edited 5 hours ago

answered 5 hours ago

Bill DubuqueBill Dubuque

214k29196654

$endgroup$

share|cite|improve this answer

edited 5 hours ago

answered 5 hours ago

Bill DubuqueBill Dubuque

214k29196654

answered 5 hours ago

Bill DubuqueBill Dubuque

214k29196654

answered 5 hours ago

Bill DubuqueBill Dubuque

214k29196654