Elliptic Curve scalar multiplication on $mathbb{R}$












0












$begingroup$


I have an elliptic curve $ y^2=x^3+109x^2+224x$ and a point $P(-100;260)$ on it. And I need to find point $2P$. I took a formulas $$x_2=left(frac{ax_1-b}{y_1}right)^2 -a+x_1$$ and $$y_2=-y_1+frac{ax_1-b}{y_1}(x_1-x_2)$$ put into this formulas $a=109$, $b=224$, $x_1=-100$, $y_1=260$ and have $2P=(frac{6850936}{4225}; frac{37736919137}{514164}) $ but this point is not lie on curve. In the article the result is $2P=(frac{8836}{25}; -frac{950716}{125})$ but I have no idea how can I find them.






elliptic-curves







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    The article is correct. Your $x$-coordinate gives a non-rational $y$. So just a miscalculation with the formula.
    $endgroup$
    – Dietrich Burde
    Dec 10 '18 at 13:30










  • $begingroup$
    formulas I have taken are not from this article. Is it right formulas for this curve?
    $endgroup$
    – aid78
    Dec 10 '18 at 14:00










  • $begingroup$
    No, it is not correct, because it assumes the short Weierstrass form $y^2=x^3+Ax+b$ and not the one with term $x^2$.
    $endgroup$
    – Dietrich Burde
    Dec 10 '18 at 15:49












  • $begingroup$
    @DietrichBurde Can you show me the right formulas. I tried to make substitution $x:=x-frac{109}{3}$ got equation $ y^2=x^3-frac{11209}{3}x+frac{2370314}{27}$ and tried to get point P in Mathcad but have an error point.
    $endgroup$
    – aid78
    Dec 10 '18 at 16:39


















0












$begingroup$


I have an elliptic curve $ y^2=x^3+109x^2+224x$ and a point $P(-100;260)$ on it. And I need to find point $2P$. I took a formulas $$x_2=left(frac{ax_1-b}{y_1}right)^2 -a+x_1$$ and $$y_2=-y_1+frac{ax_1-b}{y_1}(x_1-x_2)$$ put into this formulas $a=109$, $b=224$, $x_1=-100$, $y_1=260$ and have $2P=(frac{6850936}{4225}; frac{37736919137}{514164}) $ but this point is not lie on curve. In the article the result is $2P=(frac{8836}{25}; -frac{950716}{125})$ but I have no idea how can I find them.






elliptic-curves







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    The article is correct. Your $x$-coordinate gives a non-rational $y$. So just a miscalculation with the formula.
    $endgroup$
    – Dietrich Burde
    Dec 10 '18 at 13:30










  • $begingroup$
    formulas I have taken are not from this article. Is it right formulas for this curve?
    $endgroup$
    – aid78
    Dec 10 '18 at 14:00










  • $begingroup$
    No, it is not correct, because it assumes the short Weierstrass form $y^2=x^3+Ax+b$ and not the one with term $x^2$.
    $endgroup$
    – Dietrich Burde
    Dec 10 '18 at 15:49












  • $begingroup$
    @DietrichBurde Can you show me the right formulas. I tried to make substitution $x:=x-frac{109}{3}$ got equation $ y^2=x^3-frac{11209}{3}x+frac{2370314}{27}$ and tried to get point P in Mathcad but have an error point.
    $endgroup$
    – aid78
    Dec 10 '18 at 16:39
















0












0








0





$begingroup$


I have an elliptic curve $ y^2=x^3+109x^2+224x$ and a point $P(-100;260)$ on it. And I need to find point $2P$. I took a formulas $$x_2=left(frac{ax_1-b}{y_1}right)^2 -a+x_1$$ and $$y_2=-y_1+frac{ax_1-b}{y_1}(x_1-x_2)$$ put into this formulas $a=109$, $b=224$, $x_1=-100$, $y_1=260$ and have $2P=(frac{6850936}{4225}; frac{37736919137}{514164}) $ but this point is not lie on curve. In the article the result is $2P=(frac{8836}{25}; -frac{950716}{125})$ but I have no idea how can I find them.






elliptic-curves







share|cite|improve this question









$endgroup$




I have an elliptic curve $ y^2=x^3+109x^2+224x$ and a point $P(-100;260)$ on it. And I need to find point $2P$. I took a formulas $$x_2=left(frac{ax_1-b}{y_1}right)^2 -a+x_1$$ and $$y_2=-y_1+frac{ax_1-b}{y_1}(x_1-x_2)$$ put into this formulas $a=109$, $b=224$, $x_1=-100$, $y_1=260$ and have $2P=(frac{6850936}{4225}; frac{37736919137}{514164}) $ but this point is not lie on curve. In the article the result is $2P=(frac{8836}{25}; -frac{950716}{125})$ but I have no idea how can I find them.






elliptic-curves




elliptic-curves






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 10 '18 at 13:25









aid78aid78

1,2911315




1,2911315








  • 1




    $begingroup$
    The article is correct. Your $x$-coordinate gives a non-rational $y$. So just a miscalculation with the formula.
    $endgroup$
    – Dietrich Burde
    Dec 10 '18 at 13:30










  • $begingroup$
    formulas I have taken are not from this article. Is it right formulas for this curve?
    $endgroup$
    – aid78
    Dec 10 '18 at 14:00










  • $begingroup$
    No, it is not correct, because it assumes the short Weierstrass form $y^2=x^3+Ax+b$ and not the one with term $x^2$.
    $endgroup$
    – Dietrich Burde
    Dec 10 '18 at 15:49












  • $begingroup$
    @DietrichBurde Can you show me the right formulas. I tried to make substitution $x:=x-frac{109}{3}$ got equation $ y^2=x^3-frac{11209}{3}x+frac{2370314}{27}$ and tried to get point P in Mathcad but have an error point.
    $endgroup$
    – aid78
    Dec 10 '18 at 16:39
















  • 1




    $begingroup$
    The article is correct. Your $x$-coordinate gives a non-rational $y$. So just a miscalculation with the formula.
    $endgroup$
    – Dietrich Burde
    Dec 10 '18 at 13:30










  • $begingroup$
    formulas I have taken are not from this article. Is it right formulas for this curve?
    $endgroup$
    – aid78
    Dec 10 '18 at 14:00










  • $begingroup$
    No, it is not correct, because it assumes the short Weierstrass form $y^2=x^3+Ax+b$ and not the one with term $x^2$.
    $endgroup$
    – Dietrich Burde
    Dec 10 '18 at 15:49












  • $begingroup$
    @DietrichBurde Can you show me the right formulas. I tried to make substitution $x:=x-frac{109}{3}$ got equation $ y^2=x^3-frac{11209}{3}x+frac{2370314}{27}$ and tried to get point P in Mathcad but have an error point.
    $endgroup$
    – aid78
    Dec 10 '18 at 16:39










1




1




$begingroup$
The article is correct. Your $x$-coordinate gives a non-rational $y$. So just a miscalculation with the formula.
$endgroup$
– Dietrich Burde
Dec 10 '18 at 13:30




$begingroup$
The article is correct. Your $x$-coordinate gives a non-rational $y$. So just a miscalculation with the formula.
$endgroup$
– Dietrich Burde
Dec 10 '18 at 13:30












$begingroup$
formulas I have taken are not from this article. Is it right formulas for this curve?
$endgroup$
– aid78
Dec 10 '18 at 14:00




$begingroup$
formulas I have taken are not from this article. Is it right formulas for this curve?
$endgroup$
– aid78
Dec 10 '18 at 14:00












$begingroup$
No, it is not correct, because it assumes the short Weierstrass form $y^2=x^3+Ax+b$ and not the one with term $x^2$.
$endgroup$
– Dietrich Burde
Dec 10 '18 at 15:49






$begingroup$
No, it is not correct, because it assumes the short Weierstrass form $y^2=x^3+Ax+b$ and not the one with term $x^2$.
$endgroup$
– Dietrich Burde
Dec 10 '18 at 15:49














$begingroup$
@DietrichBurde Can you show me the right formulas. I tried to make substitution $x:=x-frac{109}{3}$ got equation $ y^2=x^3-frac{11209}{3}x+frac{2370314}{27}$ and tried to get point P in Mathcad but have an error point.
$endgroup$
– aid78
Dec 10 '18 at 16:39






$begingroup$
@DietrichBurde Can you show me the right formulas. I tried to make substitution $x:=x-frac{109}{3}$ got equation $ y^2=x^3-frac{11209}{3}x+frac{2370314}{27}$ and tried to get point P in Mathcad but have an error point.
$endgroup$
– aid78
Dec 10 '18 at 16:39












1 Answer
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$begingroup$

For any elliptic curve $E: y^2 = f(x)$ where $f(x) = x^3 + ax^2 + bx +c$.



If $P = (x_1,y_1)$ is a point on it, then the tangent line of $E$ through $P$ has the form
$$y = y_1 + s(x - x_1)$$



where $$2y_1 s = f'(x_1) iff s = frac{f'(x_1)}{2y_1} = frac{3x_1^2 + 2a x_1 + b}{2y_1}$$



If $(x_3,y_3)$ is the other intersection of $P$ with $E$, $x_3$ will be a root
of the cubic equation



$$big(y_1 + s(x- x_1)big)^2 = f(x) = x^3 +ax^2 + bx + c$$



Notice $x_1$ is a double root for same cubic equation. Apply Vieta's formula to
the coefficient of $x^2$, we obtain



$$2 x_1 + x_3 = s^2 - a implies x_3 = s^2 - 2x_1 - a$$



The point $2P = (x_2,y_2)$ is the image of $(x_3,y_3)$ under reflection of $x$-axis.
This means
$$begin{cases}
x_2 &= x_3 &= s^2 - 2s_1 - a\
y_2 &= -y_3 &= -y_1 + s(x_1 - x_2)
end{cases}
quadtext{ where }quad
s = frac{3x_1^2 + 2ax_1 + b}{2y_1}$$



For $(a,b,c) = (109,224,0)$ and $(x_1,y_1) = (-100,260)$, we get
$s = frac{81}{5}$ and



$$
begin{cases}
x_2 &= left(frac{81}{5}right)^2 - 2(-100) - 109 = frac{8836}{25}\
y_2 &= -260 + frac{81}{5}left(-100 - frac{8836}{25}right) = -frac{950716}{125}
end{cases}$$






share|cite|improve this answer









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    1












    $begingroup$

    For any elliptic curve $E: y^2 = f(x)$ where $f(x) = x^3 + ax^2 + bx +c$.



    If $P = (x_1,y_1)$ is a point on it, then the tangent line of $E$ through $P$ has the form
    $$y = y_1 + s(x - x_1)$$



    where $$2y_1 s = f'(x_1) iff s = frac{f'(x_1)}{2y_1} = frac{3x_1^2 + 2a x_1 + b}{2y_1}$$



    If $(x_3,y_3)$ is the other intersection of $P$ with $E$, $x_3$ will be a root
    of the cubic equation



    $$big(y_1 + s(x- x_1)big)^2 = f(x) = x^3 +ax^2 + bx + c$$



    Notice $x_1$ is a double root for same cubic equation. Apply Vieta's formula to
    the coefficient of $x^2$, we obtain



    $$2 x_1 + x_3 = s^2 - a implies x_3 = s^2 - 2x_1 - a$$



    The point $2P = (x_2,y_2)$ is the image of $(x_3,y_3)$ under reflection of $x$-axis.
    This means
    $$begin{cases}
    x_2 &= x_3 &= s^2 - 2s_1 - a\
    y_2 &= -y_3 &= -y_1 + s(x_1 - x_2)
    end{cases}
    quadtext{ where }quad
    s = frac{3x_1^2 + 2ax_1 + b}{2y_1}$$



    For $(a,b,c) = (109,224,0)$ and $(x_1,y_1) = (-100,260)$, we get
    $s = frac{81}{5}$ and



    $$
    begin{cases}
    x_2 &= left(frac{81}{5}right)^2 - 2(-100) - 109 = frac{8836}{25}\
    y_2 &= -260 + frac{81}{5}left(-100 - frac{8836}{25}right) = -frac{950716}{125}
    end{cases}$$






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      For any elliptic curve $E: y^2 = f(x)$ where $f(x) = x^3 + ax^2 + bx +c$.



      If $P = (x_1,y_1)$ is a point on it, then the tangent line of $E$ through $P$ has the form
      $$y = y_1 + s(x - x_1)$$



      where $$2y_1 s = f'(x_1) iff s = frac{f'(x_1)}{2y_1} = frac{3x_1^2 + 2a x_1 + b}{2y_1}$$



      If $(x_3,y_3)$ is the other intersection of $P$ with $E$, $x_3$ will be a root
      of the cubic equation



      $$big(y_1 + s(x- x_1)big)^2 = f(x) = x^3 +ax^2 + bx + c$$



      Notice $x_1$ is a double root for same cubic equation. Apply Vieta's formula to
      the coefficient of $x^2$, we obtain



      $$2 x_1 + x_3 = s^2 - a implies x_3 = s^2 - 2x_1 - a$$



      The point $2P = (x_2,y_2)$ is the image of $(x_3,y_3)$ under reflection of $x$-axis.
      This means
      $$begin{cases}
      x_2 &= x_3 &= s^2 - 2s_1 - a\
      y_2 &= -y_3 &= -y_1 + s(x_1 - x_2)
      end{cases}
      quadtext{ where }quad
      s = frac{3x_1^2 + 2ax_1 + b}{2y_1}$$



      For $(a,b,c) = (109,224,0)$ and $(x_1,y_1) = (-100,260)$, we get
      $s = frac{81}{5}$ and



      $$
      begin{cases}
      x_2 &= left(frac{81}{5}right)^2 - 2(-100) - 109 = frac{8836}{25}\
      y_2 &= -260 + frac{81}{5}left(-100 - frac{8836}{25}right) = -frac{950716}{125}
      end{cases}$$






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        For any elliptic curve $E: y^2 = f(x)$ where $f(x) = x^3 + ax^2 + bx +c$.



        If $P = (x_1,y_1)$ is a point on it, then the tangent line of $E$ through $P$ has the form
        $$y = y_1 + s(x - x_1)$$



        where $$2y_1 s = f'(x_1) iff s = frac{f'(x_1)}{2y_1} = frac{3x_1^2 + 2a x_1 + b}{2y_1}$$



        If $(x_3,y_3)$ is the other intersection of $P$ with $E$, $x_3$ will be a root
        of the cubic equation



        $$big(y_1 + s(x- x_1)big)^2 = f(x) = x^3 +ax^2 + bx + c$$



        Notice $x_1$ is a double root for same cubic equation. Apply Vieta's formula to
        the coefficient of $x^2$, we obtain



        $$2 x_1 + x_3 = s^2 - a implies x_3 = s^2 - 2x_1 - a$$



        The point $2P = (x_2,y_2)$ is the image of $(x_3,y_3)$ under reflection of $x$-axis.
        This means
        $$begin{cases}
        x_2 &= x_3 &= s^2 - 2s_1 - a\
        y_2 &= -y_3 &= -y_1 + s(x_1 - x_2)
        end{cases}
        quadtext{ where }quad
        s = frac{3x_1^2 + 2ax_1 + b}{2y_1}$$



        For $(a,b,c) = (109,224,0)$ and $(x_1,y_1) = (-100,260)$, we get
        $s = frac{81}{5}$ and



        $$
        begin{cases}
        x_2 &= left(frac{81}{5}right)^2 - 2(-100) - 109 = frac{8836}{25}\
        y_2 &= -260 + frac{81}{5}left(-100 - frac{8836}{25}right) = -frac{950716}{125}
        end{cases}$$






        share|cite|improve this answer









        $endgroup$



        For any elliptic curve $E: y^2 = f(x)$ where $f(x) = x^3 + ax^2 + bx +c$.



        If $P = (x_1,y_1)$ is a point on it, then the tangent line of $E$ through $P$ has the form
        $$y = y_1 + s(x - x_1)$$



        where $$2y_1 s = f'(x_1) iff s = frac{f'(x_1)}{2y_1} = frac{3x_1^2 + 2a x_1 + b}{2y_1}$$



        If $(x_3,y_3)$ is the other intersection of $P$ with $E$, $x_3$ will be a root
        of the cubic equation



        $$big(y_1 + s(x- x_1)big)^2 = f(x) = x^3 +ax^2 + bx + c$$



        Notice $x_1$ is a double root for same cubic equation. Apply Vieta's formula to
        the coefficient of $x^2$, we obtain



        $$2 x_1 + x_3 = s^2 - a implies x_3 = s^2 - 2x_1 - a$$



        The point $2P = (x_2,y_2)$ is the image of $(x_3,y_3)$ under reflection of $x$-axis.
        This means
        $$begin{cases}
        x_2 &= x_3 &= s^2 - 2s_1 - a\
        y_2 &= -y_3 &= -y_1 + s(x_1 - x_2)
        end{cases}
        quadtext{ where }quad
        s = frac{3x_1^2 + 2ax_1 + b}{2y_1}$$



        For $(a,b,c) = (109,224,0)$ and $(x_1,y_1) = (-100,260)$, we get
        $s = frac{81}{5}$ and



        $$
        begin{cases}
        x_2 &= left(frac{81}{5}right)^2 - 2(-100) - 109 = frac{8836}{25}\
        y_2 &= -260 + frac{81}{5}left(-100 - frac{8836}{25}right) = -frac{950716}{125}
        end{cases}$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 11 '18 at 3:19









        achille huiachille hui

        96k5132258




        96k5132258






























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            Musa

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            Para el género de plantas, véase Musa (planta). Apolo y la Nueve Musas En la mitología griega, las musas (en griego antiguo μοῦσαι « mousai» ) son, según los escritores más antiguos, las divinidades inspiradoras de las Artes: cada una de ellas está relacionada con ramas artísticas y del conocimiento. Son hijas de Zeus y de Mnemósine, compañeras del séquito de Apolo, dios olímpico de la música y patrón de las bellas artes, quien tuvo romances con cada una de ellas, dejando descendientes. Bajaban a la tierra a susurrar ideas e inspirar a aquellos mortales que las invocaran. En la época más arcaica eran las ninfas inspiradoras de las fuentes, en las cuales eran adoradas. Finalmente, alrededor de los siglos VIII-VII a. C. [ 1 ] ​ prevaleció en todo el territorio de la Hélade la adoración de las nueve Musas, que son Calíope, Clío, Erato, Euterpe, Melpómene, Polimnia, Talía, Terpsícore y Urania. El culto a las musas era originalmente de Tracia y Beocia, y fueron de vital impo...
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