y divides discriminant implies y^2 divides it












2












$begingroup$


The wiki page https://en.wikipedia.org/wiki/Nagell%E2%80%93Lutz_theorem is discussing an elliptic curve



$$y^2 = x^3 + ax^2 + bx + c$$
with the cubic polynomial discriminant



$$D = -4a^3c + a^2b^2 + 18abc - 4b^3 - 27c^2.$$



Later when explaining a particular case they state:

"... $y$ divides $D$, which immediately implies that $y^2$ divides $D$."



I apologize, I'm probably missing something obvious here, but I don't see why $y$ divides $D$ implies $y^2$ divides $D$. What am I missing?










share|cite|improve this question









$endgroup$

















    2












    $begingroup$


    The wiki page https://en.wikipedia.org/wiki/Nagell%E2%80%93Lutz_theorem is discussing an elliptic curve



    $$y^2 = x^3 + ax^2 + bx + c$$
    with the cubic polynomial discriminant



    $$D = -4a^3c + a^2b^2 + 18abc - 4b^3 - 27c^2.$$



    Later when explaining a particular case they state:

    "... $y$ divides $D$, which immediately implies that $y^2$ divides $D$."



    I apologize, I'm probably missing something obvious here, but I don't see why $y$ divides $D$ implies $y^2$ divides $D$. What am I missing?










    share|cite|improve this question









    $endgroup$















      2












      2








      2





      $begingroup$


      The wiki page https://en.wikipedia.org/wiki/Nagell%E2%80%93Lutz_theorem is discussing an elliptic curve



      $$y^2 = x^3 + ax^2 + bx + c$$
      with the cubic polynomial discriminant



      $$D = -4a^3c + a^2b^2 + 18abc - 4b^3 - 27c^2.$$



      Later when explaining a particular case they state:

      "... $y$ divides $D$, which immediately implies that $y^2$ divides $D$."



      I apologize, I'm probably missing something obvious here, but I don't see why $y$ divides $D$ implies $y^2$ divides $D$. What am I missing?










      share|cite|improve this question









      $endgroup$




      The wiki page https://en.wikipedia.org/wiki/Nagell%E2%80%93Lutz_theorem is discussing an elliptic curve



      $$y^2 = x^3 + ax^2 + bx + c$$
      with the cubic polynomial discriminant



      $$D = -4a^3c + a^2b^2 + 18abc - 4b^3 - 27c^2.$$



      Later when explaining a particular case they state:

      "... $y$ divides $D$, which immediately implies that $y^2$ divides $D$."



      I apologize, I'm probably missing something obvious here, but I don't see why $y$ divides $D$ implies $y^2$ divides $D$. What am I missing?







      elementary-number-theory cubic-equations






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Jan 10 at 17:02









      PugNosePugNose

      112




      112






















          2 Answers
          2






          active

          oldest

          votes


















          1












          $begingroup$

          Let $f(x)=x^3+ax^2+bx+c$, so $f'(x) = 3x^2+2ax+b$. $D$ is the polynomial resultant
          $$
          D = Res_x(f(x), f'(x))
          $$

          and there exists some integer polynomials $g(x), h(x)$ such that
          $$
          D = g(x) f(x) + h(x)f'(x)
          $$

          Usually the proof shows that $y$ divides $f(x),f'(x)$, so that $y$ divides $D$.



          Now the part you are missing is you can also write $D$ as
          $$
          D = -(-27x^3-27ax^2-27bx+4a^3-18ab+27c)f(x) + (-3x^2-2ax+a^2-4b)f'(x)^2
          $$

          Since $y^2 = f(x)$, $y^2$ divides $f(x)$. Then $ymid f'(x)$ gives $y^2mid D$.






          share|cite|improve this answer









          $endgroup$





















            0












            $begingroup$

            In essence, all solutions will hit the same point. Here is one more solution with the danger to become "lost in notations "... we first introduce some notations with a reference, then conclude immediately. The "general" elliptic curve (over some good ring, here $Bbb Z$, considered but in the same time by abuse also over the fraction field, here $Bbb Q$) is of the shape:
            $$
            y^2+a_1xy+a_3y=x^3+a_2x^2+a_4x+a_6 .
            $$

            It is usual to denote by $kappa$ the $y$-derivative of the corresponding polynomial extracted from the equation (by moving everything on the LHS), and then, using usual notations...
            $$
            begin{aligned}
            kappa &= 2y +a_1 +a_3 ,\
            kappa^2 &= 4x^3 + b_2x^2 + 2b_4x^2 + b_6 ,\
            &qquadtext{ and one defines the following division polynomials:}\
            psi_0 &= 0 ,\
            psi_1 &= 1 ,\
            psi_2 &= kappa ,\
            psi_3 &= 3x^4+b_2x^3+3b_4x^2+3b_6x+b_8 ,\
            &qquadtext{ and so on, and one defines recursively polynomials $psi_m$,}\[2mm]
            psi_{2m+1} &= psi_{m+2}psi_m^3 - psi_{m-1}psi_{m+1}^3 ,\
            psi_{2m} &= (psi_{m+2}psi_{m-1}^2 - psi_{m-2}psi_{m+1}^2)psi_m/kappa ,\
            &qquadtext{ and then we introduce the "numerator" $phi_m$ by}\
            phi_m &= xpsi_m^2-psi_{m-1}psi_{m+1}\[2mm]
            &qquadtext{ such that one has in case of $mPne0$ the relation}\
            x(nP)
            &=
            frac{phi_m(P)}{psi_m(P)^2} ,\
            Delta
            &=
            {color{blue}{(sigma_2-x(2(x,y))tau_2)}}; kappa^2 ,\
            &qquadtext{ where}\[2mm]
            sigma_2 &=12x^3-b_2x^2-10b_4x+b_2b_4-27b_6\
            tau_2 &=48x^2+8b_2x+32b_4-b_2^2 .
            end{aligned}
            $$

            See for instance section 1.7.2, Propositions 1.7.8, 1.7.12 in the excellent book



            Ian Connel, Elliptic curve Handbook



            (this is around page 150).





            Proof of the implicit claim in the OP:




            If $P(x,y)$ is a torsion point with $2Pne O$ on an elliptic curve of equation $Y^2 = X^3 +a_2X^2+a_4X+a_6$ and discriminant $Delta$, consider $kappa=2y$, then $kappa^2$ divides $Delta$.




            More general:




            If $P,2Pne O$ have integer coordinates, then with the same notations $kappa^2$ divides $Delta$.




            The above relation $(*)$, since the blue term above ${color{blue}{(sigma_2-x(2(x,y))tau_2)}}$ is an integer. This is for the first statement the case because twice the given point $P$ is also a torsion point, not $O$, thus integral.



            $square$





            Usually it is good in such cases to have an explicit example, here we have one using sagemath:



            sage: import random
            sage: alfa = random.choice([10^4..10^5])
            sage: alfa
            40882
            sage: c = alfa^2-alfa
            sage: b = alfa*c
            sage: E = EllipticCurve( QQ, [1-c, -b, -b, 0, 0] )
            sage: E
            Elliptic Curve defined by y^2 - 1671297041*x*y - 68325965671044*y = x^3 - 68325965671044*x^2 over Rational Field
            sage: EE = EllipticCurve( QQ, [ -27*E.c4(), -54*E.c6() ] )
            sage: EE
            Elliptic Curve defined by y^2 = x^3 - 210616964878176082678138017988707694683*x + 1176489725128296463454256557863234189321847831792178736374 over Rational Field
            sage: EE.discriminant().factor()
            2^19 * 3^19 * 13627^7 * 20441^7 * 68314266509987
            sage: EE.torsion_points()
            [(0 : 1 : 0),
            (8378881486178014515 : -7379204292472752 : 1),
            (8378881486178014515 : 7379204292472752 : 1),
            (8378881546344708027 : -301661871656786182296 : 1),
            (8378881546344708027 : 301661871656786182296 : 1),
            (8381341220942172099 : -12332842306323413283199584 : 1),
            (8381341220942172099 : 12332842306323413283199584 : 1)]
            sage: P = _[-1]
            sage: P
            (8381341220942172099 : 12332842306323413283199584 : 1)
            sage: P.xy()
            (8381341220942172099, 12332842306323413283199584)
            sage: y = P.xy()[1]
            sage: y
            12332842306323413283199584
            sage: ((2*y)^2).factor()
            2^12 * 3^10 * 13627^4 * 20441^6
            sage: ((2*y)^2).divides( EE.discriminant() )
            True


            In words, we have constructed an elliptic curve with $Bbb Z/7$-torsion, then looked for a version of it of the shape $Y^2=X^ 3+aX+b$, so that $kappa$ does not involve $a_1,a_3$, then checked the OP in the given case.






            share|cite|improve this answer









            $endgroup$













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              1












              $begingroup$

              Let $f(x)=x^3+ax^2+bx+c$, so $f'(x) = 3x^2+2ax+b$. $D$ is the polynomial resultant
              $$
              D = Res_x(f(x), f'(x))
              $$

              and there exists some integer polynomials $g(x), h(x)$ such that
              $$
              D = g(x) f(x) + h(x)f'(x)
              $$

              Usually the proof shows that $y$ divides $f(x),f'(x)$, so that $y$ divides $D$.



              Now the part you are missing is you can also write $D$ as
              $$
              D = -(-27x^3-27ax^2-27bx+4a^3-18ab+27c)f(x) + (-3x^2-2ax+a^2-4b)f'(x)^2
              $$

              Since $y^2 = f(x)$, $y^2$ divides $f(x)$. Then $ymid f'(x)$ gives $y^2mid D$.






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                Let $f(x)=x^3+ax^2+bx+c$, so $f'(x) = 3x^2+2ax+b$. $D$ is the polynomial resultant
                $$
                D = Res_x(f(x), f'(x))
                $$

                and there exists some integer polynomials $g(x), h(x)$ such that
                $$
                D = g(x) f(x) + h(x)f'(x)
                $$

                Usually the proof shows that $y$ divides $f(x),f'(x)$, so that $y$ divides $D$.



                Now the part you are missing is you can also write $D$ as
                $$
                D = -(-27x^3-27ax^2-27bx+4a^3-18ab+27c)f(x) + (-3x^2-2ax+a^2-4b)f'(x)^2
                $$

                Since $y^2 = f(x)$, $y^2$ divides $f(x)$. Then $ymid f'(x)$ gives $y^2mid D$.






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  Let $f(x)=x^3+ax^2+bx+c$, so $f'(x) = 3x^2+2ax+b$. $D$ is the polynomial resultant
                  $$
                  D = Res_x(f(x), f'(x))
                  $$

                  and there exists some integer polynomials $g(x), h(x)$ such that
                  $$
                  D = g(x) f(x) + h(x)f'(x)
                  $$

                  Usually the proof shows that $y$ divides $f(x),f'(x)$, so that $y$ divides $D$.



                  Now the part you are missing is you can also write $D$ as
                  $$
                  D = -(-27x^3-27ax^2-27bx+4a^3-18ab+27c)f(x) + (-3x^2-2ax+a^2-4b)f'(x)^2
                  $$

                  Since $y^2 = f(x)$, $y^2$ divides $f(x)$. Then $ymid f'(x)$ gives $y^2mid D$.






                  share|cite|improve this answer









                  $endgroup$



                  Let $f(x)=x^3+ax^2+bx+c$, so $f'(x) = 3x^2+2ax+b$. $D$ is the polynomial resultant
                  $$
                  D = Res_x(f(x), f'(x))
                  $$

                  and there exists some integer polynomials $g(x), h(x)$ such that
                  $$
                  D = g(x) f(x) + h(x)f'(x)
                  $$

                  Usually the proof shows that $y$ divides $f(x),f'(x)$, so that $y$ divides $D$.



                  Now the part you are missing is you can also write $D$ as
                  $$
                  D = -(-27x^3-27ax^2-27bx+4a^3-18ab+27c)f(x) + (-3x^2-2ax+a^2-4b)f'(x)^2
                  $$

                  Since $y^2 = f(x)$, $y^2$ divides $f(x)$. Then $ymid f'(x)$ gives $y^2mid D$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 10 at 19:17









                  Yong Hao NgYong Hao Ng

                  3,4941220




                  3,4941220























                      0












                      $begingroup$

                      In essence, all solutions will hit the same point. Here is one more solution with the danger to become "lost in notations "... we first introduce some notations with a reference, then conclude immediately. The "general" elliptic curve (over some good ring, here $Bbb Z$, considered but in the same time by abuse also over the fraction field, here $Bbb Q$) is of the shape:
                      $$
                      y^2+a_1xy+a_3y=x^3+a_2x^2+a_4x+a_6 .
                      $$

                      It is usual to denote by $kappa$ the $y$-derivative of the corresponding polynomial extracted from the equation (by moving everything on the LHS), and then, using usual notations...
                      $$
                      begin{aligned}
                      kappa &= 2y +a_1 +a_3 ,\
                      kappa^2 &= 4x^3 + b_2x^2 + 2b_4x^2 + b_6 ,\
                      &qquadtext{ and one defines the following division polynomials:}\
                      psi_0 &= 0 ,\
                      psi_1 &= 1 ,\
                      psi_2 &= kappa ,\
                      psi_3 &= 3x^4+b_2x^3+3b_4x^2+3b_6x+b_8 ,\
                      &qquadtext{ and so on, and one defines recursively polynomials $psi_m$,}\[2mm]
                      psi_{2m+1} &= psi_{m+2}psi_m^3 - psi_{m-1}psi_{m+1}^3 ,\
                      psi_{2m} &= (psi_{m+2}psi_{m-1}^2 - psi_{m-2}psi_{m+1}^2)psi_m/kappa ,\
                      &qquadtext{ and then we introduce the "numerator" $phi_m$ by}\
                      phi_m &= xpsi_m^2-psi_{m-1}psi_{m+1}\[2mm]
                      &qquadtext{ such that one has in case of $mPne0$ the relation}\
                      x(nP)
                      &=
                      frac{phi_m(P)}{psi_m(P)^2} ,\
                      Delta
                      &=
                      {color{blue}{(sigma_2-x(2(x,y))tau_2)}}; kappa^2 ,\
                      &qquadtext{ where}\[2mm]
                      sigma_2 &=12x^3-b_2x^2-10b_4x+b_2b_4-27b_6\
                      tau_2 &=48x^2+8b_2x+32b_4-b_2^2 .
                      end{aligned}
                      $$

                      See for instance section 1.7.2, Propositions 1.7.8, 1.7.12 in the excellent book



                      Ian Connel, Elliptic curve Handbook



                      (this is around page 150).





                      Proof of the implicit claim in the OP:




                      If $P(x,y)$ is a torsion point with $2Pne O$ on an elliptic curve of equation $Y^2 = X^3 +a_2X^2+a_4X+a_6$ and discriminant $Delta$, consider $kappa=2y$, then $kappa^2$ divides $Delta$.




                      More general:




                      If $P,2Pne O$ have integer coordinates, then with the same notations $kappa^2$ divides $Delta$.




                      The above relation $(*)$, since the blue term above ${color{blue}{(sigma_2-x(2(x,y))tau_2)}}$ is an integer. This is for the first statement the case because twice the given point $P$ is also a torsion point, not $O$, thus integral.



                      $square$





                      Usually it is good in such cases to have an explicit example, here we have one using sagemath:



                      sage: import random
                      sage: alfa = random.choice([10^4..10^5])
                      sage: alfa
                      40882
                      sage: c = alfa^2-alfa
                      sage: b = alfa*c
                      sage: E = EllipticCurve( QQ, [1-c, -b, -b, 0, 0] )
                      sage: E
                      Elliptic Curve defined by y^2 - 1671297041*x*y - 68325965671044*y = x^3 - 68325965671044*x^2 over Rational Field
                      sage: EE = EllipticCurve( QQ, [ -27*E.c4(), -54*E.c6() ] )
                      sage: EE
                      Elliptic Curve defined by y^2 = x^3 - 210616964878176082678138017988707694683*x + 1176489725128296463454256557863234189321847831792178736374 over Rational Field
                      sage: EE.discriminant().factor()
                      2^19 * 3^19 * 13627^7 * 20441^7 * 68314266509987
                      sage: EE.torsion_points()
                      [(0 : 1 : 0),
                      (8378881486178014515 : -7379204292472752 : 1),
                      (8378881486178014515 : 7379204292472752 : 1),
                      (8378881546344708027 : -301661871656786182296 : 1),
                      (8378881546344708027 : 301661871656786182296 : 1),
                      (8381341220942172099 : -12332842306323413283199584 : 1),
                      (8381341220942172099 : 12332842306323413283199584 : 1)]
                      sage: P = _[-1]
                      sage: P
                      (8381341220942172099 : 12332842306323413283199584 : 1)
                      sage: P.xy()
                      (8381341220942172099, 12332842306323413283199584)
                      sage: y = P.xy()[1]
                      sage: y
                      12332842306323413283199584
                      sage: ((2*y)^2).factor()
                      2^12 * 3^10 * 13627^4 * 20441^6
                      sage: ((2*y)^2).divides( EE.discriminant() )
                      True


                      In words, we have constructed an elliptic curve with $Bbb Z/7$-torsion, then looked for a version of it of the shape $Y^2=X^ 3+aX+b$, so that $kappa$ does not involve $a_1,a_3$, then checked the OP in the given case.






                      share|cite|improve this answer









                      $endgroup$


















                        0












                        $begingroup$

                        In essence, all solutions will hit the same point. Here is one more solution with the danger to become "lost in notations "... we first introduce some notations with a reference, then conclude immediately. The "general" elliptic curve (over some good ring, here $Bbb Z$, considered but in the same time by abuse also over the fraction field, here $Bbb Q$) is of the shape:
                        $$
                        y^2+a_1xy+a_3y=x^3+a_2x^2+a_4x+a_6 .
                        $$

                        It is usual to denote by $kappa$ the $y$-derivative of the corresponding polynomial extracted from the equation (by moving everything on the LHS), and then, using usual notations...
                        $$
                        begin{aligned}
                        kappa &= 2y +a_1 +a_3 ,\
                        kappa^2 &= 4x^3 + b_2x^2 + 2b_4x^2 + b_6 ,\
                        &qquadtext{ and one defines the following division polynomials:}\
                        psi_0 &= 0 ,\
                        psi_1 &= 1 ,\
                        psi_2 &= kappa ,\
                        psi_3 &= 3x^4+b_2x^3+3b_4x^2+3b_6x+b_8 ,\
                        &qquadtext{ and so on, and one defines recursively polynomials $psi_m$,}\[2mm]
                        psi_{2m+1} &= psi_{m+2}psi_m^3 - psi_{m-1}psi_{m+1}^3 ,\
                        psi_{2m} &= (psi_{m+2}psi_{m-1}^2 - psi_{m-2}psi_{m+1}^2)psi_m/kappa ,\
                        &qquadtext{ and then we introduce the "numerator" $phi_m$ by}\
                        phi_m &= xpsi_m^2-psi_{m-1}psi_{m+1}\[2mm]
                        &qquadtext{ such that one has in case of $mPne0$ the relation}\
                        x(nP)
                        &=
                        frac{phi_m(P)}{psi_m(P)^2} ,\
                        Delta
                        &=
                        {color{blue}{(sigma_2-x(2(x,y))tau_2)}}; kappa^2 ,\
                        &qquadtext{ where}\[2mm]
                        sigma_2 &=12x^3-b_2x^2-10b_4x+b_2b_4-27b_6\
                        tau_2 &=48x^2+8b_2x+32b_4-b_2^2 .
                        end{aligned}
                        $$

                        See for instance section 1.7.2, Propositions 1.7.8, 1.7.12 in the excellent book



                        Ian Connel, Elliptic curve Handbook



                        (this is around page 150).





                        Proof of the implicit claim in the OP:




                        If $P(x,y)$ is a torsion point with $2Pne O$ on an elliptic curve of equation $Y^2 = X^3 +a_2X^2+a_4X+a_6$ and discriminant $Delta$, consider $kappa=2y$, then $kappa^2$ divides $Delta$.




                        More general:




                        If $P,2Pne O$ have integer coordinates, then with the same notations $kappa^2$ divides $Delta$.




                        The above relation $(*)$, since the blue term above ${color{blue}{(sigma_2-x(2(x,y))tau_2)}}$ is an integer. This is for the first statement the case because twice the given point $P$ is also a torsion point, not $O$, thus integral.



                        $square$





                        Usually it is good in such cases to have an explicit example, here we have one using sagemath:



                        sage: import random
                        sage: alfa = random.choice([10^4..10^5])
                        sage: alfa
                        40882
                        sage: c = alfa^2-alfa
                        sage: b = alfa*c
                        sage: E = EllipticCurve( QQ, [1-c, -b, -b, 0, 0] )
                        sage: E
                        Elliptic Curve defined by y^2 - 1671297041*x*y - 68325965671044*y = x^3 - 68325965671044*x^2 over Rational Field
                        sage: EE = EllipticCurve( QQ, [ -27*E.c4(), -54*E.c6() ] )
                        sage: EE
                        Elliptic Curve defined by y^2 = x^3 - 210616964878176082678138017988707694683*x + 1176489725128296463454256557863234189321847831792178736374 over Rational Field
                        sage: EE.discriminant().factor()
                        2^19 * 3^19 * 13627^7 * 20441^7 * 68314266509987
                        sage: EE.torsion_points()
                        [(0 : 1 : 0),
                        (8378881486178014515 : -7379204292472752 : 1),
                        (8378881486178014515 : 7379204292472752 : 1),
                        (8378881546344708027 : -301661871656786182296 : 1),
                        (8378881546344708027 : 301661871656786182296 : 1),
                        (8381341220942172099 : -12332842306323413283199584 : 1),
                        (8381341220942172099 : 12332842306323413283199584 : 1)]
                        sage: P = _[-1]
                        sage: P
                        (8381341220942172099 : 12332842306323413283199584 : 1)
                        sage: P.xy()
                        (8381341220942172099, 12332842306323413283199584)
                        sage: y = P.xy()[1]
                        sage: y
                        12332842306323413283199584
                        sage: ((2*y)^2).factor()
                        2^12 * 3^10 * 13627^4 * 20441^6
                        sage: ((2*y)^2).divides( EE.discriminant() )
                        True


                        In words, we have constructed an elliptic curve with $Bbb Z/7$-torsion, then looked for a version of it of the shape $Y^2=X^ 3+aX+b$, so that $kappa$ does not involve $a_1,a_3$, then checked the OP in the given case.






                        share|cite|improve this answer









                        $endgroup$
















                          0












                          0








                          0





                          $begingroup$

                          In essence, all solutions will hit the same point. Here is one more solution with the danger to become "lost in notations "... we first introduce some notations with a reference, then conclude immediately. The "general" elliptic curve (over some good ring, here $Bbb Z$, considered but in the same time by abuse also over the fraction field, here $Bbb Q$) is of the shape:
                          $$
                          y^2+a_1xy+a_3y=x^3+a_2x^2+a_4x+a_6 .
                          $$

                          It is usual to denote by $kappa$ the $y$-derivative of the corresponding polynomial extracted from the equation (by moving everything on the LHS), and then, using usual notations...
                          $$
                          begin{aligned}
                          kappa &= 2y +a_1 +a_3 ,\
                          kappa^2 &= 4x^3 + b_2x^2 + 2b_4x^2 + b_6 ,\
                          &qquadtext{ and one defines the following division polynomials:}\
                          psi_0 &= 0 ,\
                          psi_1 &= 1 ,\
                          psi_2 &= kappa ,\
                          psi_3 &= 3x^4+b_2x^3+3b_4x^2+3b_6x+b_8 ,\
                          &qquadtext{ and so on, and one defines recursively polynomials $psi_m$,}\[2mm]
                          psi_{2m+1} &= psi_{m+2}psi_m^3 - psi_{m-1}psi_{m+1}^3 ,\
                          psi_{2m} &= (psi_{m+2}psi_{m-1}^2 - psi_{m-2}psi_{m+1}^2)psi_m/kappa ,\
                          &qquadtext{ and then we introduce the "numerator" $phi_m$ by}\
                          phi_m &= xpsi_m^2-psi_{m-1}psi_{m+1}\[2mm]
                          &qquadtext{ such that one has in case of $mPne0$ the relation}\
                          x(nP)
                          &=
                          frac{phi_m(P)}{psi_m(P)^2} ,\
                          Delta
                          &=
                          {color{blue}{(sigma_2-x(2(x,y))tau_2)}}; kappa^2 ,\
                          &qquadtext{ where}\[2mm]
                          sigma_2 &=12x^3-b_2x^2-10b_4x+b_2b_4-27b_6\
                          tau_2 &=48x^2+8b_2x+32b_4-b_2^2 .
                          end{aligned}
                          $$

                          See for instance section 1.7.2, Propositions 1.7.8, 1.7.12 in the excellent book



                          Ian Connel, Elliptic curve Handbook



                          (this is around page 150).





                          Proof of the implicit claim in the OP:




                          If $P(x,y)$ is a torsion point with $2Pne O$ on an elliptic curve of equation $Y^2 = X^3 +a_2X^2+a_4X+a_6$ and discriminant $Delta$, consider $kappa=2y$, then $kappa^2$ divides $Delta$.




                          More general:




                          If $P,2Pne O$ have integer coordinates, then with the same notations $kappa^2$ divides $Delta$.




                          The above relation $(*)$, since the blue term above ${color{blue}{(sigma_2-x(2(x,y))tau_2)}}$ is an integer. This is for the first statement the case because twice the given point $P$ is also a torsion point, not $O$, thus integral.



                          $square$





                          Usually it is good in such cases to have an explicit example, here we have one using sagemath:



                          sage: import random
                          sage: alfa = random.choice([10^4..10^5])
                          sage: alfa
                          40882
                          sage: c = alfa^2-alfa
                          sage: b = alfa*c
                          sage: E = EllipticCurve( QQ, [1-c, -b, -b, 0, 0] )
                          sage: E
                          Elliptic Curve defined by y^2 - 1671297041*x*y - 68325965671044*y = x^3 - 68325965671044*x^2 over Rational Field
                          sage: EE = EllipticCurve( QQ, [ -27*E.c4(), -54*E.c6() ] )
                          sage: EE
                          Elliptic Curve defined by y^2 = x^3 - 210616964878176082678138017988707694683*x + 1176489725128296463454256557863234189321847831792178736374 over Rational Field
                          sage: EE.discriminant().factor()
                          2^19 * 3^19 * 13627^7 * 20441^7 * 68314266509987
                          sage: EE.torsion_points()
                          [(0 : 1 : 0),
                          (8378881486178014515 : -7379204292472752 : 1),
                          (8378881486178014515 : 7379204292472752 : 1),
                          (8378881546344708027 : -301661871656786182296 : 1),
                          (8378881546344708027 : 301661871656786182296 : 1),
                          (8381341220942172099 : -12332842306323413283199584 : 1),
                          (8381341220942172099 : 12332842306323413283199584 : 1)]
                          sage: P = _[-1]
                          sage: P
                          (8381341220942172099 : 12332842306323413283199584 : 1)
                          sage: P.xy()
                          (8381341220942172099, 12332842306323413283199584)
                          sage: y = P.xy()[1]
                          sage: y
                          12332842306323413283199584
                          sage: ((2*y)^2).factor()
                          2^12 * 3^10 * 13627^4 * 20441^6
                          sage: ((2*y)^2).divides( EE.discriminant() )
                          True


                          In words, we have constructed an elliptic curve with $Bbb Z/7$-torsion, then looked for a version of it of the shape $Y^2=X^ 3+aX+b$, so that $kappa$ does not involve $a_1,a_3$, then checked the OP in the given case.






                          share|cite|improve this answer









                          $endgroup$



                          In essence, all solutions will hit the same point. Here is one more solution with the danger to become "lost in notations "... we first introduce some notations with a reference, then conclude immediately. The "general" elliptic curve (over some good ring, here $Bbb Z$, considered but in the same time by abuse also over the fraction field, here $Bbb Q$) is of the shape:
                          $$
                          y^2+a_1xy+a_3y=x^3+a_2x^2+a_4x+a_6 .
                          $$

                          It is usual to denote by $kappa$ the $y$-derivative of the corresponding polynomial extracted from the equation (by moving everything on the LHS), and then, using usual notations...
                          $$
                          begin{aligned}
                          kappa &= 2y +a_1 +a_3 ,\
                          kappa^2 &= 4x^3 + b_2x^2 + 2b_4x^2 + b_6 ,\
                          &qquadtext{ and one defines the following division polynomials:}\
                          psi_0 &= 0 ,\
                          psi_1 &= 1 ,\
                          psi_2 &= kappa ,\
                          psi_3 &= 3x^4+b_2x^3+3b_4x^2+3b_6x+b_8 ,\
                          &qquadtext{ and so on, and one defines recursively polynomials $psi_m$,}\[2mm]
                          psi_{2m+1} &= psi_{m+2}psi_m^3 - psi_{m-1}psi_{m+1}^3 ,\
                          psi_{2m} &= (psi_{m+2}psi_{m-1}^2 - psi_{m-2}psi_{m+1}^2)psi_m/kappa ,\
                          &qquadtext{ and then we introduce the "numerator" $phi_m$ by}\
                          phi_m &= xpsi_m^2-psi_{m-1}psi_{m+1}\[2mm]
                          &qquadtext{ such that one has in case of $mPne0$ the relation}\
                          x(nP)
                          &=
                          frac{phi_m(P)}{psi_m(P)^2} ,\
                          Delta
                          &=
                          {color{blue}{(sigma_2-x(2(x,y))tau_2)}}; kappa^2 ,\
                          &qquadtext{ where}\[2mm]
                          sigma_2 &=12x^3-b_2x^2-10b_4x+b_2b_4-27b_6\
                          tau_2 &=48x^2+8b_2x+32b_4-b_2^2 .
                          end{aligned}
                          $$

                          See for instance section 1.7.2, Propositions 1.7.8, 1.7.12 in the excellent book



                          Ian Connel, Elliptic curve Handbook



                          (this is around page 150).





                          Proof of the implicit claim in the OP:




                          If $P(x,y)$ is a torsion point with $2Pne O$ on an elliptic curve of equation $Y^2 = X^3 +a_2X^2+a_4X+a_6$ and discriminant $Delta$, consider $kappa=2y$, then $kappa^2$ divides $Delta$.




                          More general:




                          If $P,2Pne O$ have integer coordinates, then with the same notations $kappa^2$ divides $Delta$.




                          The above relation $(*)$, since the blue term above ${color{blue}{(sigma_2-x(2(x,y))tau_2)}}$ is an integer. This is for the first statement the case because twice the given point $P$ is also a torsion point, not $O$, thus integral.



                          $square$





                          Usually it is good in such cases to have an explicit example, here we have one using sagemath:



                          sage: import random
                          sage: alfa = random.choice([10^4..10^5])
                          sage: alfa
                          40882
                          sage: c = alfa^2-alfa
                          sage: b = alfa*c
                          sage: E = EllipticCurve( QQ, [1-c, -b, -b, 0, 0] )
                          sage: E
                          Elliptic Curve defined by y^2 - 1671297041*x*y - 68325965671044*y = x^3 - 68325965671044*x^2 over Rational Field
                          sage: EE = EllipticCurve( QQ, [ -27*E.c4(), -54*E.c6() ] )
                          sage: EE
                          Elliptic Curve defined by y^2 = x^3 - 210616964878176082678138017988707694683*x + 1176489725128296463454256557863234189321847831792178736374 over Rational Field
                          sage: EE.discriminant().factor()
                          2^19 * 3^19 * 13627^7 * 20441^7 * 68314266509987
                          sage: EE.torsion_points()
                          [(0 : 1 : 0),
                          (8378881486178014515 : -7379204292472752 : 1),
                          (8378881486178014515 : 7379204292472752 : 1),
                          (8378881546344708027 : -301661871656786182296 : 1),
                          (8378881546344708027 : 301661871656786182296 : 1),
                          (8381341220942172099 : -12332842306323413283199584 : 1),
                          (8381341220942172099 : 12332842306323413283199584 : 1)]
                          sage: P = _[-1]
                          sage: P
                          (8381341220942172099 : 12332842306323413283199584 : 1)
                          sage: P.xy()
                          (8381341220942172099, 12332842306323413283199584)
                          sage: y = P.xy()[1]
                          sage: y
                          12332842306323413283199584
                          sage: ((2*y)^2).factor()
                          2^12 * 3^10 * 13627^4 * 20441^6
                          sage: ((2*y)^2).divides( EE.discriminant() )
                          True


                          In words, we have constructed an elliptic curve with $Bbb Z/7$-torsion, then looked for a version of it of the shape $Y^2=X^ 3+aX+b$, so that $kappa$ does not involve $a_1,a_3$, then checked the OP in the given case.







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Jan 15 at 13:01









                          dan_fuleadan_fulea

                          6,4281312




                          6,4281312






























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