Proof for divisibility of polynomials. [on hold]












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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 $."






polynomials divisibility







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put on hold as off-topic by José Carlos Santos, Sil, John Omielan, TheSimpliFire, RRL 23 hours 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 23 hours 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






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









      asked yesterday









      HeetGorakhiyaHeetGorakhiya

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      303




      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 23 hours 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 23 hours 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

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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











          $endgroup$









          • 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
            yesterday





















          4












          $begingroup$

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



          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 $,ninBbb N,,$ 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
              yesterday


















            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
              yesterday
















            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 yesterday

























            answered yesterday









            Maria MazurMaria Mazur

            50k1361125




            50k1361125








            • 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
              yesterday
















            • 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
              yesterday










            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
            yesterday






            $begingroup$
            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
            yesterday













            4












            $begingroup$

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



            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 $,ninBbb N,,$ and all polynomials $,P(x)$






            share|cite|improve this answer











            $endgroup$


















              4












              $begingroup$

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



              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 $,ninBbb N,,$ and all polynomials $,P(x)$






              share|cite|improve this answer











              $endgroup$
















                4












                4








                4





                $begingroup$

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



                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 $,ninBbb N,,$ and all polynomials $,P(x)$






                share|cite|improve this answer











                $endgroup$



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



                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 $,ninBbb N,,$ and all polynomials $,P(x)$







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited yesterday

























                answered yesterday









                Bill DubuqueBill Dubuque

                214k29196655




                214k29196655















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