Find all polynomials $p(x)$ such that $(p(x))^2 = 1 + x cdot p(x + 1)$ for all $xin mathbb{R}$












6












$begingroup$



Find all polynomials $p(x)$ such that $(p(x))^2 = 1 + x cdot p(x + 1)$ for all $xin mathbb{R}$.




I assume that $O(p(x)=n)$ (Where O(p) denotes the order of p)
let $p(x)=a_nx^n+a_{n-1}x^{n-1}+a_{n-2}x^{n-2} dots dots +a_{1}x^{1}+a_{n-1}$ also $a_n neq 0$



Then we have $a_n^2x^{2n} +dots +a_0^2=1+a_nx^{n+1}+ dots a_0x$
since $a_n neq 0$ we must have $2n=n+1$ and hence $n=1$ hence the polynomial must be linear and hence there doesn't exist any solution(I proved assuming $p(x)=ax+b$)



Is it correct?






real-analysis abstract-algebra polynomials







share|cite|improve this question









$endgroup$








  • 2




    $begingroup$
    What about $p(x)=x+1$?
    $endgroup$
    – Michael Burr
    Jan 10 at 0:46






  • 4




    $begingroup$
    $p$ cannot be zero. If $p$ has degree $r,$ the left hand side has degree $2r$ or $1$ and the right hand side has degree $r + 1,$ hence $r = 1.$ Now propose $p(x)=ax+b$ and solve.
    $endgroup$
    – Will M.
    Jan 10 at 0:48








  • 5




    $begingroup$
    What you wrote looks good, but you should check the $ax+b$ case more carefully.
    $endgroup$
    – lulu
    Jan 10 at 0:51










  • $begingroup$
    @WillM. That's an ideal hint, I think---perhaps write it up as an answer?
    $endgroup$
    – Travis
    Jan 10 at 2:57
















6












$begingroup$



Find all polynomials $p(x)$ such that $(p(x))^2 = 1 + x cdot p(x + 1)$ for all $xin mathbb{R}$.




I assume that $O(p(x)=n)$ (Where O(p) denotes the order of p)
let $p(x)=a_nx^n+a_{n-1}x^{n-1}+a_{n-2}x^{n-2} dots dots +a_{1}x^{1}+a_{n-1}$ also $a_n neq 0$



Then we have $a_n^2x^{2n} +dots +a_0^2=1+a_nx^{n+1}+ dots a_0x$
since $a_n neq 0$ we must have $2n=n+1$ and hence $n=1$ hence the polynomial must be linear and hence there doesn't exist any solution(I proved assuming $p(x)=ax+b$)



Is it correct?






real-analysis abstract-algebra polynomials







share|cite|improve this question









$endgroup$








  • 2




    $begingroup$
    What about $p(x)=x+1$?
    $endgroup$
    – Michael Burr
    Jan 10 at 0:46






  • 4




    $begingroup$
    $p$ cannot be zero. If $p$ has degree $r,$ the left hand side has degree $2r$ or $1$ and the right hand side has degree $r + 1,$ hence $r = 1.$ Now propose $p(x)=ax+b$ and solve.
    $endgroup$
    – Will M.
    Jan 10 at 0:48








  • 5




    $begingroup$
    What you wrote looks good, but you should check the $ax+b$ case more carefully.
    $endgroup$
    – lulu
    Jan 10 at 0:51










  • $begingroup$
    @WillM. That's an ideal hint, I think---perhaps write it up as an answer?
    $endgroup$
    – Travis
    Jan 10 at 2:57














6












6








6


1



$begingroup$



Find all polynomials $p(x)$ such that $(p(x))^2 = 1 + x cdot p(x + 1)$ for all $xin mathbb{R}$.




I assume that $O(p(x)=n)$ (Where O(p) denotes the order of p)
let $p(x)=a_nx^n+a_{n-1}x^{n-1}+a_{n-2}x^{n-2} dots dots +a_{1}x^{1}+a_{n-1}$ also $a_n neq 0$



Then we have $a_n^2x^{2n} +dots +a_0^2=1+a_nx^{n+1}+ dots a_0x$
since $a_n neq 0$ we must have $2n=n+1$ and hence $n=1$ hence the polynomial must be linear and hence there doesn't exist any solution(I proved assuming $p(x)=ax+b$)



Is it correct?






real-analysis abstract-algebra polynomials







share|cite|improve this question









$endgroup$





Find all polynomials $p(x)$ such that $(p(x))^2 = 1 + x cdot p(x + 1)$ for all $xin mathbb{R}$.




I assume that $O(p(x)=n)$ (Where O(p) denotes the order of p)
let $p(x)=a_nx^n+a_{n-1}x^{n-1}+a_{n-2}x^{n-2} dots dots +a_{1}x^{1}+a_{n-1}$ also $a_n neq 0$



Then we have $a_n^2x^{2n} +dots +a_0^2=1+a_nx^{n+1}+ dots a_0x$
since $a_n neq 0$ we must have $2n=n+1$ and hence $n=1$ hence the polynomial must be linear and hence there doesn't exist any solution(I proved assuming $p(x)=ax+b$)



Is it correct?






real-analysis abstract-algebra polynomials




real-analysis abstract-algebra polynomials






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 10 at 0:37









SunShineSunShine

1384




1384








  • 2




    $begingroup$
    What about $p(x)=x+1$?
    $endgroup$
    – Michael Burr
    Jan 10 at 0:46






  • 4




    $begingroup$
    $p$ cannot be zero. If $p$ has degree $r,$ the left hand side has degree $2r$ or $1$ and the right hand side has degree $r + 1,$ hence $r = 1.$ Now propose $p(x)=ax+b$ and solve.
    $endgroup$
    – Will M.
    Jan 10 at 0:48








  • 5




    $begingroup$
    What you wrote looks good, but you should check the $ax+b$ case more carefully.
    $endgroup$
    – lulu
    Jan 10 at 0:51










  • $begingroup$
    @WillM. That's an ideal hint, I think---perhaps write it up as an answer?
    $endgroup$
    – Travis
    Jan 10 at 2:57














  • 2




    $begingroup$
    What about $p(x)=x+1$?
    $endgroup$
    – Michael Burr
    Jan 10 at 0:46






  • 4




    $begingroup$
    $p$ cannot be zero. If $p$ has degree $r,$ the left hand side has degree $2r$ or $1$ and the right hand side has degree $r + 1,$ hence $r = 1.$ Now propose $p(x)=ax+b$ and solve.
    $endgroup$
    – Will M.
    Jan 10 at 0:48








  • 5




    $begingroup$
    What you wrote looks good, but you should check the $ax+b$ case more carefully.
    $endgroup$
    – lulu
    Jan 10 at 0:51










  • $begingroup$
    @WillM. That's an ideal hint, I think---perhaps write it up as an answer?
    $endgroup$
    – Travis
    Jan 10 at 2:57








2




2




$begingroup$
What about $p(x)=x+1$?
$endgroup$
– Michael Burr
Jan 10 at 0:46




$begingroup$
What about $p(x)=x+1$?
$endgroup$
– Michael Burr
Jan 10 at 0:46




4




4




$begingroup$
$p$ cannot be zero. If $p$ has degree $r,$ the left hand side has degree $2r$ or $1$ and the right hand side has degree $r + 1,$ hence $r = 1.$ Now propose $p(x)=ax+b$ and solve.
$endgroup$
– Will M.
Jan 10 at 0:48






$begingroup$
$p$ cannot be zero. If $p$ has degree $r,$ the left hand side has degree $2r$ or $1$ and the right hand side has degree $r + 1,$ hence $r = 1.$ Now propose $p(x)=ax+b$ and solve.
$endgroup$
– Will M.
Jan 10 at 0:48






5




5




$begingroup$
What you wrote looks good, but you should check the $ax+b$ case more carefully.
$endgroup$
– lulu
Jan 10 at 0:51




$begingroup$
What you wrote looks good, but you should check the $ax+b$ case more carefully.
$endgroup$
– lulu
Jan 10 at 0:51












$begingroup$
@WillM. That's an ideal hint, I think---perhaps write it up as an answer?
$endgroup$
– Travis
Jan 10 at 2:57




$begingroup$
@WillM. That's an ideal hint, I think---perhaps write it up as an answer?
$endgroup$
– Travis
Jan 10 at 2:57










2 Answers
2






active

oldest

votes


















2












$begingroup$

After observing that the degree is at most $1$ you can simplify the rest of the argument as follows: $p(0)^{2}=1$ from the equation, so $p(0)=pm 1$. Also $p(-1)^{2}=1-p(0)$. By considering the cases $p(0)=1$ and $p(0)=-1$ separately it is quite easy to conclude that the only solution is $p(x)=x+1$.






share|cite|improve this answer









$endgroup$





















    2












    $begingroup$

    Our OP SunShine has correctly deduced that $deg p(x) = 1$, so that



    $p(x) = ax + b; tag 1$



    we compute:



    $p^2(x) = a^2x^2 + 2ab x + b^2; tag 2$



    $p(x + 1) = ax + a + b; tag 3$



    $xp(x + 1) = ax^2 + ax + bx; tag 4$



    $xp(x + 1) + 1 = ax^2 + ax + bx + 1; tag 5$



    $a^2 x^2 + 2ab x + b^2 = ax^2 + ax + bx + 1; tag 6$



    $(a^2 - a)x^2 + 2abx + b^2 = ax + bx + 1; tag 7$



    $a^2 = a, ; b^2 = 1, ; 2ab = a + b; tag 8$



    $a = 0, 1; ; b = pm 1; tag 9$



    $b = 1 Longrightarrow 2a = a + 1 Longrightarrow a = 1 in {0, 1 }; tag{10}$



    $b = -1 Longrightarrow -2a = a - 1 Longrightarrow a = dfrac{1}{3} notin {0, 1 }; tag{11}$



    thus the only solution is $a = b = 1$, so that



    $p(x) = x + 1. tag{12}$



    CHECK:



    $p^2(x) = (x + 1)^2 = x^2 + 2x + 1$
    $= x(x + 2) + 1 = x((x + 1) + 1) + 1 = xp(x + 1) + 1; tag{13}$



    it thus appears that the only solution is



    $p(x) = x + 1. tag{14}$






    share|cite|improve this answer











    $endgroup$













      Your Answer





      StackExchange.ifUsing("editor", function () {
      return StackExchange.using("mathjaxEditing", function () {
      StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
      StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
      });
      });
      }, "mathjax-editing");

      StackExchange.ready(function() {
      var channelOptions = {
      tags: "".split(" "),
      id: "69"
      };
      initTagRenderer("".split(" "), "".split(" "), channelOptions);

      StackExchange.using("externalEditor", function() {
      // Have to fire editor after snippets, if snippets enabled
      if (StackExchange.settings.snippets.snippetsEnabled) {
      StackExchange.using("snippets", function() {
      createEditor();
      });
      }
      else {
      createEditor();
      }
      });

      function createEditor() {
      StackExchange.prepareEditor({
      heartbeatType: 'answer',
      autoActivateHeartbeat: false,
      convertImagesToLinks: true,
      noModals: true,
      showLowRepImageUploadWarning: true,
      reputationToPostImages: 10,
      bindNavPrevention: true,
      postfix: "",
      imageUploader: {
      brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
      contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
      allowUrls: true
      },
      noCode: true, onDemand: true,
      discardSelector: ".discard-answer"
      ,immediatelyShowMarkdownHelp:true
      });


      }
      });














      draft saved

      draft discarded


















      StackExchange.ready(
      function () {
      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3068107%2ffind-all-polynomials-px-such-that-px2-1-x-cdot-px-1-for-all%23new-answer', 'question_page');
      }
      );

      Post as a guest















      Required, but never shown

























      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      2












      $begingroup$

      After observing that the degree is at most $1$ you can simplify the rest of the argument as follows: $p(0)^{2}=1$ from the equation, so $p(0)=pm 1$. Also $p(-1)^{2}=1-p(0)$. By considering the cases $p(0)=1$ and $p(0)=-1$ separately it is quite easy to conclude that the only solution is $p(x)=x+1$.






      share|cite|improve this answer









      $endgroup$


















        2












        $begingroup$

        After observing that the degree is at most $1$ you can simplify the rest of the argument as follows: $p(0)^{2}=1$ from the equation, so $p(0)=pm 1$. Also $p(-1)^{2}=1-p(0)$. By considering the cases $p(0)=1$ and $p(0)=-1$ separately it is quite easy to conclude that the only solution is $p(x)=x+1$.






        share|cite|improve this answer









        $endgroup$
















          2












          2








          2





          $begingroup$

          After observing that the degree is at most $1$ you can simplify the rest of the argument as follows: $p(0)^{2}=1$ from the equation, so $p(0)=pm 1$. Also $p(-1)^{2}=1-p(0)$. By considering the cases $p(0)=1$ and $p(0)=-1$ separately it is quite easy to conclude that the only solution is $p(x)=x+1$.






          share|cite|improve this answer









          $endgroup$



          After observing that the degree is at most $1$ you can simplify the rest of the argument as follows: $p(0)^{2}=1$ from the equation, so $p(0)=pm 1$. Also $p(-1)^{2}=1-p(0)$. By considering the cases $p(0)=1$ and $p(0)=-1$ separately it is quite easy to conclude that the only solution is $p(x)=x+1$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 10 at 6:26









          Kavi Rama MurthyKavi Rama Murthy

          54.5k32055




          54.5k32055























              2












              $begingroup$

              Our OP SunShine has correctly deduced that $deg p(x) = 1$, so that



              $p(x) = ax + b; tag 1$



              we compute:



              $p^2(x) = a^2x^2 + 2ab x + b^2; tag 2$



              $p(x + 1) = ax + a + b; tag 3$



              $xp(x + 1) = ax^2 + ax + bx; tag 4$



              $xp(x + 1) + 1 = ax^2 + ax + bx + 1; tag 5$



              $a^2 x^2 + 2ab x + b^2 = ax^2 + ax + bx + 1; tag 6$



              $(a^2 - a)x^2 + 2abx + b^2 = ax + bx + 1; tag 7$



              $a^2 = a, ; b^2 = 1, ; 2ab = a + b; tag 8$



              $a = 0, 1; ; b = pm 1; tag 9$



              $b = 1 Longrightarrow 2a = a + 1 Longrightarrow a = 1 in {0, 1 }; tag{10}$



              $b = -1 Longrightarrow -2a = a - 1 Longrightarrow a = dfrac{1}{3} notin {0, 1 }; tag{11}$



              thus the only solution is $a = b = 1$, so that



              $p(x) = x + 1. tag{12}$



              CHECK:



              $p^2(x) = (x + 1)^2 = x^2 + 2x + 1$
              $= x(x + 2) + 1 = x((x + 1) + 1) + 1 = xp(x + 1) + 1; tag{13}$



              it thus appears that the only solution is



              $p(x) = x + 1. tag{14}$






              share|cite|improve this answer











              $endgroup$


















                2












                $begingroup$

                Our OP SunShine has correctly deduced that $deg p(x) = 1$, so that



                $p(x) = ax + b; tag 1$



                we compute:



                $p^2(x) = a^2x^2 + 2ab x + b^2; tag 2$



                $p(x + 1) = ax + a + b; tag 3$



                $xp(x + 1) = ax^2 + ax + bx; tag 4$



                $xp(x + 1) + 1 = ax^2 + ax + bx + 1; tag 5$



                $a^2 x^2 + 2ab x + b^2 = ax^2 + ax + bx + 1; tag 6$



                $(a^2 - a)x^2 + 2abx + b^2 = ax + bx + 1; tag 7$



                $a^2 = a, ; b^2 = 1, ; 2ab = a + b; tag 8$



                $a = 0, 1; ; b = pm 1; tag 9$



                $b = 1 Longrightarrow 2a = a + 1 Longrightarrow a = 1 in {0, 1 }; tag{10}$



                $b = -1 Longrightarrow -2a = a - 1 Longrightarrow a = dfrac{1}{3} notin {0, 1 }; tag{11}$



                thus the only solution is $a = b = 1$, so that



                $p(x) = x + 1. tag{12}$



                CHECK:



                $p^2(x) = (x + 1)^2 = x^2 + 2x + 1$
                $= x(x + 2) + 1 = x((x + 1) + 1) + 1 = xp(x + 1) + 1; tag{13}$



                it thus appears that the only solution is



                $p(x) = x + 1. tag{14}$






                share|cite|improve this answer











                $endgroup$
















                  2












                  2








                  2





                  $begingroup$

                  Our OP SunShine has correctly deduced that $deg p(x) = 1$, so that



                  $p(x) = ax + b; tag 1$



                  we compute:



                  $p^2(x) = a^2x^2 + 2ab x + b^2; tag 2$



                  $p(x + 1) = ax + a + b; tag 3$



                  $xp(x + 1) = ax^2 + ax + bx; tag 4$



                  $xp(x + 1) + 1 = ax^2 + ax + bx + 1; tag 5$



                  $a^2 x^2 + 2ab x + b^2 = ax^2 + ax + bx + 1; tag 6$



                  $(a^2 - a)x^2 + 2abx + b^2 = ax + bx + 1; tag 7$



                  $a^2 = a, ; b^2 = 1, ; 2ab = a + b; tag 8$



                  $a = 0, 1; ; b = pm 1; tag 9$



                  $b = 1 Longrightarrow 2a = a + 1 Longrightarrow a = 1 in {0, 1 }; tag{10}$



                  $b = -1 Longrightarrow -2a = a - 1 Longrightarrow a = dfrac{1}{3} notin {0, 1 }; tag{11}$



                  thus the only solution is $a = b = 1$, so that



                  $p(x) = x + 1. tag{12}$



                  CHECK:



                  $p^2(x) = (x + 1)^2 = x^2 + 2x + 1$
                  $= x(x + 2) + 1 = x((x + 1) + 1) + 1 = xp(x + 1) + 1; tag{13}$



                  it thus appears that the only solution is



                  $p(x) = x + 1. tag{14}$






                  share|cite|improve this answer











                  $endgroup$



                  Our OP SunShine has correctly deduced that $deg p(x) = 1$, so that



                  $p(x) = ax + b; tag 1$



                  we compute:



                  $p^2(x) = a^2x^2 + 2ab x + b^2; tag 2$



                  $p(x + 1) = ax + a + b; tag 3$



                  $xp(x + 1) = ax^2 + ax + bx; tag 4$



                  $xp(x + 1) + 1 = ax^2 + ax + bx + 1; tag 5$



                  $a^2 x^2 + 2ab x + b^2 = ax^2 + ax + bx + 1; tag 6$



                  $(a^2 - a)x^2 + 2abx + b^2 = ax + bx + 1; tag 7$



                  $a^2 = a, ; b^2 = 1, ; 2ab = a + b; tag 8$



                  $a = 0, 1; ; b = pm 1; tag 9$



                  $b = 1 Longrightarrow 2a = a + 1 Longrightarrow a = 1 in {0, 1 }; tag{10}$



                  $b = -1 Longrightarrow -2a = a - 1 Longrightarrow a = dfrac{1}{3} notin {0, 1 }; tag{11}$



                  thus the only solution is $a = b = 1$, so that



                  $p(x) = x + 1. tag{12}$



                  CHECK:



                  $p^2(x) = (x + 1)^2 = x^2 + 2x + 1$
                  $= x(x + 2) + 1 = x((x + 1) + 1) + 1 = xp(x + 1) + 1; tag{13}$



                  it thus appears that the only solution is



                  $p(x) = x + 1. tag{14}$







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Jan 10 at 2:41

























                  answered Jan 10 at 2:15









                  Robert LewisRobert Lewis

                  44.8k22964




                  44.8k22964






























                      draft saved

                      draft discarded




















































                      Thanks for contributing an answer to Mathematics Stack Exchange!


                      • Please be sure to answer the question. Provide details and share your research!

                      But avoid …



                      • Asking for help, clarification, or responding to other answers.

                      • Making statements based on opinion; back them up with references or personal experience.


                      Use MathJax to format equations. MathJax reference.


                      To learn more, see our tips on writing great answers.




                      draft saved


                      draft discarded














                      StackExchange.ready(
                      function () {
                      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3068107%2ffind-all-polynomials-px-such-that-px2-1-x-cdot-px-1-for-all%23new-answer', 'question_page');
                      }
                      );

                      Post as a guest















                      Required, but never shown





















































                      Required, but never shown














                      Required, but never shown












                      Required, but never shown







                      Required, but never shown

































                      Required, but never shown














                      Required, but never shown












                      Required, but never shown







                      Required, but never shown







                      -

                      Popular posts from this blog

                      Mario Kart Wii

                      Image
                      Please consider supporting The Cutting Room Floor on Patreon. Thanks for all your support! Mario Kart Wii From The Cutting Room Floor Jump to: navigation, search Mario Kart Wii Developer : Nintendo Publisher : Nintendo Platform : Wii Released in JP : April 10, 2008 Released in US : April 27, 2008 Released in EU : April 11, 2008 Released in AU : April 24, 2008 Released in KR : April 30, 2009 This game has unused animations. This game has unused areas. This game has unused code. This game has unused objects. This game has unused modes / minigames. This game has unused graphics. This game has unused models. This game has unused items. This game has unused music. This game has regional differences. This game has a notes page This game has a prerelease article Mario Kart Wii is the sixth game in the series (not counting Arcade editions), which intro...
                      Read more

                      The Binding of Isaac: Rebirth/Afterbirth

                      Image
                      Please consider supporting The Cutting Room Floor on Patreon. Thanks for all your support! The Binding of Isaac: Rebirth/Afterbirth From The Cutting Room Floor < The Binding of Isaac: Rebirth Jump to: navigation, search This is a sub-page of The Binding of Isaac: Rebirth . Afterbirth is the first expansion pack for Rebirth and features a variety of improvements. Contents 1 Unused Graphics 1.1 Old Man Keeper 1.2 Skinless Hush 1.3 Miscellaneous Unused Hush Sprites 1.4 Generosity achievement 1.5 Afterbirth+ Room Editor 1.6 Bearded Ultra Greed 1.7 Early Ultra Greed 1.8 Unused Brownie Expression 1.9 Frail's Spinning Head 1.10 Mom's Dead Hand Alt 2 Unused Audio 2.1 Unused Dead Womb Layer 2.2 Unused Hush Battle Outro 2.3 The Stain Grunt Unused Graphics Old Man Keeper Image file: gfxuimain me...
                      Read more

                      What does “Dominus providebit” mean?

                      Image
                      5 Does "Dominus providebit" mean "The Lord will provide" or "The Lord provides"? I once had a bit of an argument with a guy who studied Latin over that. It's an inscription that appears on the rim of some old Swiss coins that I have. meaning latin-to-english-translation grammar-identification share | improve this question edited Jan 18 at 15:27 V2Blast 105 4 asked Jan 17 at 22:49 Donald K. Donald K. 26 1 ...
                      Read more