Proof for divisibility of polynomials. [closed]












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


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.







    $endgroup$



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


    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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      asked Apr 7 at 18:12









      HeetGorakhiyaHeetGorakhiya

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      263




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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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      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.




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


      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.







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


      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
            Apr 7 at 18:36





















          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
              Apr 7 at 18:36


















            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
              Apr 7 at 18:36
















            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 Apr 7 at 18:25

























            answered Apr 7 at 18:16









            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
              Apr 7 at 18:36
















            • 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
              Apr 7 at 18:36










            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
            Apr 7 at 18:36






            $begingroup$
            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













            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 Apr 7 at 18:45

























                answered Apr 7 at 18:18









                Bill DubuqueBill Dubuque

                214k29197658




                214k29197658















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