Proof for divisibility of polynomials. [on hold] The 2019 Stack Overflow Developer Survey Results Are InShow 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

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Proof for divisibility of polynomials. [on hold]



The 2019 Stack Overflow Developer Survey Results Are InShow 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










2












$begingroup$


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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HeetGorakhiya is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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put on hold as off-topic by José Carlos Santos, Sil, John Omielan, TheSimpliFire, RRL 2 days ago


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – José Carlos Santos, Sil, John Omielan, TheSimpliFire, RRL
If this question can be reworded to fit the rules in the help center, please edit the question.




















    2












    $begingroup$


    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







    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.







    $endgroup$



    put on hold as off-topic by José Carlos Santos, Sil, John Omielan, TheSimpliFire, RRL 2 days ago


    This question appears to be off-topic. The users who voted to close gave this specific reason:


    • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – José Carlos Santos, Sil, John Omielan, TheSimpliFire, RRL
    If this question can be reworded to fit the rules in the help center, please edit the question.


















      2












      2








      2





      $begingroup$


      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







      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.







      $endgroup$




      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




      polynomials divisibility






      share|cite|improve this question







      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







      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




      share|cite|improve this question






      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 2 days ago









      HeetGorakhiyaHeetGorakhiya

      283




      283




      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.





      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.






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




      put on hold as off-topic by José Carlos Santos, Sil, John Omielan, TheSimpliFire, RRL 2 days ago


      This question appears to be off-topic. The users who voted to close gave this specific reason:


      • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – José Carlos Santos, Sil, John Omielan, TheSimpliFire, RRL
      If this question can be reworded to fit the rules in the help center, please edit the question.







      put on hold as off-topic by José Carlos Santos, Sil, John Omielan, TheSimpliFire, RRL 2 days ago


      This question appears to be off-topic. The users who voted to close gave this specific reason:


      • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – José Carlos Santos, Sil, John Omielan, TheSimpliFire, RRL
      If this question can be reworded to fit the rules in the help center, please edit the question.




















          2 Answers
          2






          active

          oldest

          votes


















          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











          $endgroup$








          • 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
            2 days ago



















          4












          $begingroup$

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



          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











          $endgroup$



















            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            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











            $endgroup$








            • 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
              2 days ago
















            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











            $endgroup$








            • 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
              2 days ago














            4












            4








            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











            $endgroup$



            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














            share|cite|improve this answer



            share|cite|improve this answer








            edited 2 days ago

























            answered 2 days ago









            Maria MazurMaria Mazur

            49.9k1361125




            49.9k1361125







            • 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
              2 days ago













            • 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
              2 days ago








            1




            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
            2 days ago





            $begingroup$
            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
            2 days ago












            4












            $begingroup$

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



            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











            $endgroup$

















              4












              $begingroup$

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



              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











              $endgroup$















                4












                4








                4





                $begingroup$

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



                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











                $endgroup$



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



                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














                share|cite|improve this answer



                share|cite|improve this answer








                edited 2 days ago

























                answered 2 days ago









                Bill DubuqueBill Dubuque

                214k29196656




                214k29196656













                    -divisibility, polynomials
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