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
1






active

oldest

votes


















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









$endgroup$













    Your Answer





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

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

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

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


    }
    });














    draft saved

    draft discarded


















    StackExchange.ready(
    function () {
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3033910%2felliptic-curve-scalar-multiplication-on-mathbbr%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    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

        95.7k5132258




        95.7k5132258






























            draft saved

            draft discarded




















































            Thanks for contributing an answer to Mathematics Stack Exchange!


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

            But avoid …



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

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


            Use MathJax to format equations. MathJax reference.


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




            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3033910%2felliptic-curve-scalar-multiplication-on-mathbbr%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown





















































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown

































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown







            -

            Popular posts from this blog

            Berounka

            Image
            Berounka.mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Caractéristiques Longueur 139  km Bassin 8 861  km 2 Bassin collecteur Elbe Débit moyen ? Cours Source Confluence de la Mže et de la Radbuza · Localisation Pilsen · Altitude 305  m · Coordonnées 49° 45′ 13″ N, 13° 23′ 24″ E Confluence Vltava · Localisation Prague · Altitude 187  m · Coordonnées 49° 59′ 43″ N, 14° 24′ 04″ E Géographie Pays traversés   République tchèque modifier   La Berounka est une rivière de la République tchèque. Elle porte le nom de Mže depuis sa source en Allemagne près de la frontière jusqu'à sa confluence avec la Radbuza à Plzeň. Elle continue ensuite sous le nom de Berounka jusqu'à sa confluence avec la Vltava, près de Prague. La rivière est une destination prisée pour les amateurs de canoë, qui appréc...
            Read more

            Sphinx de Gizeh

            Image
            Pour les articles homonymes, voir Sphinx. Sphinx de Gizeh.mw-parser-output .entete.egypte-antique{background-image:url("//upload.wikimedia.org/wikipedia/commons/c/cc/Picto_infobox_ancient_Egypt.png")} Site d'Égypte antique Le visage du sphinx de Gizeh. Noms en arabe أبو الهول , Abou al-Hôl Localisation Région Basse-Égypte Coordonnées 29° 58′ 31″ nord, 31° 08′ 16″ est Géolocalisation sur la carte : Égypte Sphinx de Gizeh modifier   Le sphinx de Gizeh est la statue thérianthrope qui se dresse devant les grandes pyramides du plateau de Gizeh, en Basse-Égypte. Sculpture monumentale monolithique la plus grande du monde [ 1 ] avec 73,5 mètres de longueur, 14 mètres de largeur et 20,22 mètres de hauteur, elle représente un sphinx couchant. Réalisée vers -2500, elle serait attribuée à Djédefrê, l'un des pharaons de la IV e  dynastie, représentant son père le pharaon Khéops. Sommaire 1 Étymologie 2 Desc...
            Read more

            Different font size/position of beamer's navigation symbols template's content depending on regular/plain...

            Image
            3 Let's suppose I'd like to divert the navigation symbols template (e.g. in order replace them by the frame numbers, see this bug report ;) I would be facing troubles regarding the font size/position, depending on regular/plain frames, as shown by the following MCE. How could I get rid of these troubles? documentclass{beamer} usetheme{PaloAlto} setbeamertemplate{navigation symbols}{insertframenumber/inserttotalframenumber} begin{document} begin{frame} frametitle{A regular frame} end{frame} begin{frame}[plain] frametitle{A plain frame} end{frame} end{document} beamer header-footer share | improve this question asked Dec 10 '18 a...
            Read more