System of equations $a(1 - b^2) = b(1 -c^2) = c(1 -d^2) = d(1 - a^2)$











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Given a positive real number $t$, find the number of real solutions $a, b, c, d$ of the system
$$a(1 - b^2) = b(1 -c^2) = c(1 -d^2) = d(1 - a^2) = t$$



I have a solution



Let $f(x)=frac t{1-x^2}$ and we get $f(f(f(f(a))))=a$ and so we are looking for the number of fixed points of $f(f(f(f(x))))$



1) fixed points of $f(x)$
Any fixed point $u$ of $f(x)$ gives a solution $(u,u,u,u)$.
These fixed points are solution of $x^3-x+t=0$ and so :
If $tin(0,frac 2{3sqrt 3})$ : three such solutions
If $t=frac 2{3sqrt 3})$ : two such solutions
If $tin(frac 2{3sqrt 3},+infty)$ : one such solution



2) fixed points of $f(f(x))$ which are not fixed points of $f(x)$
Any fixed point $u$ of $f(f(x))$ which is not fixed point of $f(x)$ gives a solution $(u,f(u),u,f(u))$
These fixed points are solution of $x^2-tx-1=0$ and so :
Always two solutions to this quadratic giving the same solutions to the original equation (just a circular permutation)



3) fixed points of $f(f(f(f(x))))$ which are neither fixed points of $f(x)$, neither fixed points of $f(f(x))$
These are solutions of a degree $12$ or degree $11$ ugly polynomial :
$(t^4-3t^2+1)x^{12}-t^3x^{11}+(-7t^4+18t^2-6)x^{10}$
$+(-t^5+6t^3)x^{9}+(26t^4-48t^2+15)x^{8}+(6t^5-14t^3)x^{7}$
$+(t^6-54t^4+72t^2-20)x^{6}+(-14t^5+16t^3)x^{5}+(-7t^6+64t^4-63t^2+15)x^{4}$
$+(-t^7+14t^5-9t^3)x^{3}+(11t^6-41t^4+30t^2-6)x^{2}+(2t^7-5t^5+2t^3)x^{1}$
$+(t^8-6t^6+11t^4-6t^2+1)$



And I dont know how to determine the number of real roots of this polynomial :



For certain values (for example $t=frac 38$), this polynomial has $12$ real roots and so $3$ new solutions of original equation (leading to $3+1+3=7$ solutions to the equation (counting as one all circular permutations))



For certain values (for example $t=10$), this polynomial has no real roots and so only $2$ solutions for the equation.



There is certainly a clever transformation for the equation $f(f(f(f(x))))=x$ allowing to count the roots depending on $t$ easily.



But I did not see it up to now.










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    up vote
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    Given a positive real number $t$, find the number of real solutions $a, b, c, d$ of the system
    $$a(1 - b^2) = b(1 -c^2) = c(1 -d^2) = d(1 - a^2) = t$$



    I have a solution



    Let $f(x)=frac t{1-x^2}$ and we get $f(f(f(f(a))))=a$ and so we are looking for the number of fixed points of $f(f(f(f(x))))$



    1) fixed points of $f(x)$
    Any fixed point $u$ of $f(x)$ gives a solution $(u,u,u,u)$.
    These fixed points are solution of $x^3-x+t=0$ and so :
    If $tin(0,frac 2{3sqrt 3})$ : three such solutions
    If $t=frac 2{3sqrt 3})$ : two such solutions
    If $tin(frac 2{3sqrt 3},+infty)$ : one such solution



    2) fixed points of $f(f(x))$ which are not fixed points of $f(x)$
    Any fixed point $u$ of $f(f(x))$ which is not fixed point of $f(x)$ gives a solution $(u,f(u),u,f(u))$
    These fixed points are solution of $x^2-tx-1=0$ and so :
    Always two solutions to this quadratic giving the same solutions to the original equation (just a circular permutation)



    3) fixed points of $f(f(f(f(x))))$ which are neither fixed points of $f(x)$, neither fixed points of $f(f(x))$
    These are solutions of a degree $12$ or degree $11$ ugly polynomial :
    $(t^4-3t^2+1)x^{12}-t^3x^{11}+(-7t^4+18t^2-6)x^{10}$
    $+(-t^5+6t^3)x^{9}+(26t^4-48t^2+15)x^{8}+(6t^5-14t^3)x^{7}$
    $+(t^6-54t^4+72t^2-20)x^{6}+(-14t^5+16t^3)x^{5}+(-7t^6+64t^4-63t^2+15)x^{4}$
    $+(-t^7+14t^5-9t^3)x^{3}+(11t^6-41t^4+30t^2-6)x^{2}+(2t^7-5t^5+2t^3)x^{1}$
    $+(t^8-6t^6+11t^4-6t^2+1)$



    And I dont know how to determine the number of real roots of this polynomial :



    For certain values (for example $t=frac 38$), this polynomial has $12$ real roots and so $3$ new solutions of original equation (leading to $3+1+3=7$ solutions to the equation (counting as one all circular permutations))



    For certain values (for example $t=10$), this polynomial has no real roots and so only $2$ solutions for the equation.



    There is certainly a clever transformation for the equation $f(f(f(f(x))))=x$ allowing to count the roots depending on $t$ easily.



    But I did not see it up to now.










    share|cite|improve this question
























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      Given a positive real number $t$, find the number of real solutions $a, b, c, d$ of the system
      $$a(1 - b^2) = b(1 -c^2) = c(1 -d^2) = d(1 - a^2) = t$$



      I have a solution



      Let $f(x)=frac t{1-x^2}$ and we get $f(f(f(f(a))))=a$ and so we are looking for the number of fixed points of $f(f(f(f(x))))$



      1) fixed points of $f(x)$
      Any fixed point $u$ of $f(x)$ gives a solution $(u,u,u,u)$.
      These fixed points are solution of $x^3-x+t=0$ and so :
      If $tin(0,frac 2{3sqrt 3})$ : three such solutions
      If $t=frac 2{3sqrt 3})$ : two such solutions
      If $tin(frac 2{3sqrt 3},+infty)$ : one such solution



      2) fixed points of $f(f(x))$ which are not fixed points of $f(x)$
      Any fixed point $u$ of $f(f(x))$ which is not fixed point of $f(x)$ gives a solution $(u,f(u),u,f(u))$
      These fixed points are solution of $x^2-tx-1=0$ and so :
      Always two solutions to this quadratic giving the same solutions to the original equation (just a circular permutation)



      3) fixed points of $f(f(f(f(x))))$ which are neither fixed points of $f(x)$, neither fixed points of $f(f(x))$
      These are solutions of a degree $12$ or degree $11$ ugly polynomial :
      $(t^4-3t^2+1)x^{12}-t^3x^{11}+(-7t^4+18t^2-6)x^{10}$
      $+(-t^5+6t^3)x^{9}+(26t^4-48t^2+15)x^{8}+(6t^5-14t^3)x^{7}$
      $+(t^6-54t^4+72t^2-20)x^{6}+(-14t^5+16t^3)x^{5}+(-7t^6+64t^4-63t^2+15)x^{4}$
      $+(-t^7+14t^5-9t^3)x^{3}+(11t^6-41t^4+30t^2-6)x^{2}+(2t^7-5t^5+2t^3)x^{1}$
      $+(t^8-6t^6+11t^4-6t^2+1)$



      And I dont know how to determine the number of real roots of this polynomial :



      For certain values (for example $t=frac 38$), this polynomial has $12$ real roots and so $3$ new solutions of original equation (leading to $3+1+3=7$ solutions to the equation (counting as one all circular permutations))



      For certain values (for example $t=10$), this polynomial has no real roots and so only $2$ solutions for the equation.



      There is certainly a clever transformation for the equation $f(f(f(f(x))))=x$ allowing to count the roots depending on $t$ easily.



      But I did not see it up to now.










      share|cite|improve this question













      Given a positive real number $t$, find the number of real solutions $a, b, c, d$ of the system
      $$a(1 - b^2) = b(1 -c^2) = c(1 -d^2) = d(1 - a^2) = t$$



      I have a solution



      Let $f(x)=frac t{1-x^2}$ and we get $f(f(f(f(a))))=a$ and so we are looking for the number of fixed points of $f(f(f(f(x))))$



      1) fixed points of $f(x)$
      Any fixed point $u$ of $f(x)$ gives a solution $(u,u,u,u)$.
      These fixed points are solution of $x^3-x+t=0$ and so :
      If $tin(0,frac 2{3sqrt 3})$ : three such solutions
      If $t=frac 2{3sqrt 3})$ : two such solutions
      If $tin(frac 2{3sqrt 3},+infty)$ : one such solution



      2) fixed points of $f(f(x))$ which are not fixed points of $f(x)$
      Any fixed point $u$ of $f(f(x))$ which is not fixed point of $f(x)$ gives a solution $(u,f(u),u,f(u))$
      These fixed points are solution of $x^2-tx-1=0$ and so :
      Always two solutions to this quadratic giving the same solutions to the original equation (just a circular permutation)



      3) fixed points of $f(f(f(f(x))))$ which are neither fixed points of $f(x)$, neither fixed points of $f(f(x))$
      These are solutions of a degree $12$ or degree $11$ ugly polynomial :
      $(t^4-3t^2+1)x^{12}-t^3x^{11}+(-7t^4+18t^2-6)x^{10}$
      $+(-t^5+6t^3)x^{9}+(26t^4-48t^2+15)x^{8}+(6t^5-14t^3)x^{7}$
      $+(t^6-54t^4+72t^2-20)x^{6}+(-14t^5+16t^3)x^{5}+(-7t^6+64t^4-63t^2+15)x^{4}$
      $+(-t^7+14t^5-9t^3)x^{3}+(11t^6-41t^4+30t^2-6)x^{2}+(2t^7-5t^5+2t^3)x^{1}$
      $+(t^8-6t^6+11t^4-6t^2+1)$



      And I dont know how to determine the number of real roots of this polynomial :



      For certain values (for example $t=frac 38$), this polynomial has $12$ real roots and so $3$ new solutions of original equation (leading to $3+1+3=7$ solutions to the equation (counting as one all circular permutations))



      For certain values (for example $t=10$), this polynomial has no real roots and so only $2$ solutions for the equation.



      There is certainly a clever transformation for the equation $f(f(f(f(x))))=x$ allowing to count the roots depending on $t$ easily.



      But I did not see it up to now.







      calculus systems-of-equations fixed-point-theorems fixedpoints






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      asked Nov 13 at 4:07









      Trần Văn Lâm

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          2 Answers
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          I set your polynomial to $0$ and graphed it in Desmos with $t$ along the x-axis (and $x$ along the y-axis... oops!) For each value of $t$, I drew a vertical line and counted the number of intersections:



          begin{align*}
          12 quad &text{when}; 0<t<frac{sqrt{5}-1}{2}\
          11 quad &text{when}; t=frac{sqrt{5}-1}{2}\
          12 quad &text{when}; frac{sqrt{5}-1}{2}<t<0.6432\
          8 quad &text{when}; t approx 0.6432\
          4 quad &text{when}; 0.6432<t<frac{sqrt{5}+1}{2}\
          3 quad &text{when}; t=frac{sqrt{5}+1}{2}\
          4 quad &text{when}; frac{sqrt{5}+1}{2}<t<2\
          2 quad &text{when}; t=2\
          0 quad &text{when}; t>2\
          end{align*}



          When $tapprox0.6432$, it appears to be tangent at 4 points simultaneously, but I'm not sure.



          When $t=2$, Wolfram Alpha can factor your polynomial into the form shown below, so it's clear that this transition occurs at exactly $t=2$. These two roots are already accounted for by $f(f(x))$.



          $$ (x^2 - 2 x - 1)^2 (5 x^8 + 12 x^7 - 8 x^6 - 60 x^5 - 38 x^4 + 68 x^3 + 64 x^2 + 12 x + 25) $$



          There are two vertical asymptotes at $t=frac{sqrt{5}pm 1}{2}$. I'm guessing 11 intersections means 2 circular permutations, and 3 means 0? Anyway, the way I got the asymptotes was using this equation (sorry about the confusing variables):



          $$
          frac{y-x}{y}=left(frac{xleft(1-left(frac{x}{1-y^2}right)^2right)^2}{left(1-left(frac{x}{1-y^2}right)^2right)^2-x^2}right)^2
          $$



          As $y toinfty$, the left-hand side goes to 1 and $frac{x}{1-y^2} to 0$. Therefore:



          $$
          pm xleft(1-0^2right)^2=left(1-0^2right)^2-x^2
          $$






          share|cite|improve this answer






























            up vote
            0
            down vote













            If you graph the function $x(1-y^2)=t$, you get a clear picture of what the solutions are.



            Curva_Iter_1



            Assume $1<t$, then you can have infinite negative solutions as per the path
            shown. You cannot have a positive value for any one of the unknowns.



            Only if $0<t<<1$ you may find infinite positive solutions starting with a high value for $a$.






            share|cite|improve this answer





















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              2 Answers
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              2 Answers
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              active

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              up vote
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              down vote













              I set your polynomial to $0$ and graphed it in Desmos with $t$ along the x-axis (and $x$ along the y-axis... oops!) For each value of $t$, I drew a vertical line and counted the number of intersections:



              begin{align*}
              12 quad &text{when}; 0<t<frac{sqrt{5}-1}{2}\
              11 quad &text{when}; t=frac{sqrt{5}-1}{2}\
              12 quad &text{when}; frac{sqrt{5}-1}{2}<t<0.6432\
              8 quad &text{when}; t approx 0.6432\
              4 quad &text{when}; 0.6432<t<frac{sqrt{5}+1}{2}\
              3 quad &text{when}; t=frac{sqrt{5}+1}{2}\
              4 quad &text{when}; frac{sqrt{5}+1}{2}<t<2\
              2 quad &text{when}; t=2\
              0 quad &text{when}; t>2\
              end{align*}



              When $tapprox0.6432$, it appears to be tangent at 4 points simultaneously, but I'm not sure.



              When $t=2$, Wolfram Alpha can factor your polynomial into the form shown below, so it's clear that this transition occurs at exactly $t=2$. These two roots are already accounted for by $f(f(x))$.



              $$ (x^2 - 2 x - 1)^2 (5 x^8 + 12 x^7 - 8 x^6 - 60 x^5 - 38 x^4 + 68 x^3 + 64 x^2 + 12 x + 25) $$



              There are two vertical asymptotes at $t=frac{sqrt{5}pm 1}{2}$. I'm guessing 11 intersections means 2 circular permutations, and 3 means 0? Anyway, the way I got the asymptotes was using this equation (sorry about the confusing variables):



              $$
              frac{y-x}{y}=left(frac{xleft(1-left(frac{x}{1-y^2}right)^2right)^2}{left(1-left(frac{x}{1-y^2}right)^2right)^2-x^2}right)^2
              $$



              As $y toinfty$, the left-hand side goes to 1 and $frac{x}{1-y^2} to 0$. Therefore:



              $$
              pm xleft(1-0^2right)^2=left(1-0^2right)^2-x^2
              $$






              share|cite|improve this answer



























                up vote
                2
                down vote













                I set your polynomial to $0$ and graphed it in Desmos with $t$ along the x-axis (and $x$ along the y-axis... oops!) For each value of $t$, I drew a vertical line and counted the number of intersections:



                begin{align*}
                12 quad &text{when}; 0<t<frac{sqrt{5}-1}{2}\
                11 quad &text{when}; t=frac{sqrt{5}-1}{2}\
                12 quad &text{when}; frac{sqrt{5}-1}{2}<t<0.6432\
                8 quad &text{when}; t approx 0.6432\
                4 quad &text{when}; 0.6432<t<frac{sqrt{5}+1}{2}\
                3 quad &text{when}; t=frac{sqrt{5}+1}{2}\
                4 quad &text{when}; frac{sqrt{5}+1}{2}<t<2\
                2 quad &text{when}; t=2\
                0 quad &text{when}; t>2\
                end{align*}



                When $tapprox0.6432$, it appears to be tangent at 4 points simultaneously, but I'm not sure.



                When $t=2$, Wolfram Alpha can factor your polynomial into the form shown below, so it's clear that this transition occurs at exactly $t=2$. These two roots are already accounted for by $f(f(x))$.



                $$ (x^2 - 2 x - 1)^2 (5 x^8 + 12 x^7 - 8 x^6 - 60 x^5 - 38 x^4 + 68 x^3 + 64 x^2 + 12 x + 25) $$



                There are two vertical asymptotes at $t=frac{sqrt{5}pm 1}{2}$. I'm guessing 11 intersections means 2 circular permutations, and 3 means 0? Anyway, the way I got the asymptotes was using this equation (sorry about the confusing variables):



                $$
                frac{y-x}{y}=left(frac{xleft(1-left(frac{x}{1-y^2}right)^2right)^2}{left(1-left(frac{x}{1-y^2}right)^2right)^2-x^2}right)^2
                $$



                As $y toinfty$, the left-hand side goes to 1 and $frac{x}{1-y^2} to 0$. Therefore:



                $$
                pm xleft(1-0^2right)^2=left(1-0^2right)^2-x^2
                $$






                share|cite|improve this answer

























                  up vote
                  2
                  down vote










                  up vote
                  2
                  down vote









                  I set your polynomial to $0$ and graphed it in Desmos with $t$ along the x-axis (and $x$ along the y-axis... oops!) For each value of $t$, I drew a vertical line and counted the number of intersections:



                  begin{align*}
                  12 quad &text{when}; 0<t<frac{sqrt{5}-1}{2}\
                  11 quad &text{when}; t=frac{sqrt{5}-1}{2}\
                  12 quad &text{when}; frac{sqrt{5}-1}{2}<t<0.6432\
                  8 quad &text{when}; t approx 0.6432\
                  4 quad &text{when}; 0.6432<t<frac{sqrt{5}+1}{2}\
                  3 quad &text{when}; t=frac{sqrt{5}+1}{2}\
                  4 quad &text{when}; frac{sqrt{5}+1}{2}<t<2\
                  2 quad &text{when}; t=2\
                  0 quad &text{when}; t>2\
                  end{align*}



                  When $tapprox0.6432$, it appears to be tangent at 4 points simultaneously, but I'm not sure.



                  When $t=2$, Wolfram Alpha can factor your polynomial into the form shown below, so it's clear that this transition occurs at exactly $t=2$. These two roots are already accounted for by $f(f(x))$.



                  $$ (x^2 - 2 x - 1)^2 (5 x^8 + 12 x^7 - 8 x^6 - 60 x^5 - 38 x^4 + 68 x^3 + 64 x^2 + 12 x + 25) $$



                  There are two vertical asymptotes at $t=frac{sqrt{5}pm 1}{2}$. I'm guessing 11 intersections means 2 circular permutations, and 3 means 0? Anyway, the way I got the asymptotes was using this equation (sorry about the confusing variables):



                  $$
                  frac{y-x}{y}=left(frac{xleft(1-left(frac{x}{1-y^2}right)^2right)^2}{left(1-left(frac{x}{1-y^2}right)^2right)^2-x^2}right)^2
                  $$



                  As $y toinfty$, the left-hand side goes to 1 and $frac{x}{1-y^2} to 0$. Therefore:



                  $$
                  pm xleft(1-0^2right)^2=left(1-0^2right)^2-x^2
                  $$






                  share|cite|improve this answer














                  I set your polynomial to $0$ and graphed it in Desmos with $t$ along the x-axis (and $x$ along the y-axis... oops!) For each value of $t$, I drew a vertical line and counted the number of intersections:



                  begin{align*}
                  12 quad &text{when}; 0<t<frac{sqrt{5}-1}{2}\
                  11 quad &text{when}; t=frac{sqrt{5}-1}{2}\
                  12 quad &text{when}; frac{sqrt{5}-1}{2}<t<0.6432\
                  8 quad &text{when}; t approx 0.6432\
                  4 quad &text{when}; 0.6432<t<frac{sqrt{5}+1}{2}\
                  3 quad &text{when}; t=frac{sqrt{5}+1}{2}\
                  4 quad &text{when}; frac{sqrt{5}+1}{2}<t<2\
                  2 quad &text{when}; t=2\
                  0 quad &text{when}; t>2\
                  end{align*}



                  When $tapprox0.6432$, it appears to be tangent at 4 points simultaneously, but I'm not sure.



                  When $t=2$, Wolfram Alpha can factor your polynomial into the form shown below, so it's clear that this transition occurs at exactly $t=2$. These two roots are already accounted for by $f(f(x))$.



                  $$ (x^2 - 2 x - 1)^2 (5 x^8 + 12 x^7 - 8 x^6 - 60 x^5 - 38 x^4 + 68 x^3 + 64 x^2 + 12 x + 25) $$



                  There are two vertical asymptotes at $t=frac{sqrt{5}pm 1}{2}$. I'm guessing 11 intersections means 2 circular permutations, and 3 means 0? Anyway, the way I got the asymptotes was using this equation (sorry about the confusing variables):



                  $$
                  frac{y-x}{y}=left(frac{xleft(1-left(frac{x}{1-y^2}right)^2right)^2}{left(1-left(frac{x}{1-y^2}right)^2right)^2-x^2}right)^2
                  $$



                  As $y toinfty$, the left-hand side goes to 1 and $frac{x}{1-y^2} to 0$. Therefore:



                  $$
                  pm xleft(1-0^2right)^2=left(1-0^2right)^2-x^2
                  $$







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Nov 14 at 21:17

























                  answered Nov 13 at 22:15









                  ignorantFid

                  21916




                  21916






















                      up vote
                      0
                      down vote













                      If you graph the function $x(1-y^2)=t$, you get a clear picture of what the solutions are.



                      Curva_Iter_1



                      Assume $1<t$, then you can have infinite negative solutions as per the path
                      shown. You cannot have a positive value for any one of the unknowns.



                      Only if $0<t<<1$ you may find infinite positive solutions starting with a high value for $a$.






                      share|cite|improve this answer

























                        up vote
                        0
                        down vote













                        If you graph the function $x(1-y^2)=t$, you get a clear picture of what the solutions are.



                        Curva_Iter_1



                        Assume $1<t$, then you can have infinite negative solutions as per the path
                        shown. You cannot have a positive value for any one of the unknowns.



                        Only if $0<t<<1$ you may find infinite positive solutions starting with a high value for $a$.






                        share|cite|improve this answer























                          up vote
                          0
                          down vote










                          up vote
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                          If you graph the function $x(1-y^2)=t$, you get a clear picture of what the solutions are.



                          Curva_Iter_1



                          Assume $1<t$, then you can have infinite negative solutions as per the path
                          shown. You cannot have a positive value for any one of the unknowns.



                          Only if $0<t<<1$ you may find infinite positive solutions starting with a high value for $a$.






                          share|cite|improve this answer












                          If you graph the function $x(1-y^2)=t$, you get a clear picture of what the solutions are.



                          Curva_Iter_1



                          Assume $1<t$, then you can have infinite negative solutions as per the path
                          shown. You cannot have a positive value for any one of the unknowns.



                          Only if $0<t<<1$ you may find infinite positive solutions starting with a high value for $a$.







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Nov 13 at 23:46









                          G Cab

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