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











    $endgroup$













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      2 Answers
      2






      active

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      2 Answers
      2






      active

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      active

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      active

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      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






























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