Elliptic Curve scalar multiplication on $mathbb{R}$
$begingroup$
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
asked Dec 10 '18 at 13:25
aid78aid78
1,2911315
$endgroup$
add a comment |
$begingroup$
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
asked Dec 10 '18 at 13:25
aid78aid78
1,2911315
$endgroup$
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
add a comment |
$begingroup$
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
asked Dec 10 '18 at 13:25
aid78aid78
1,2911315
$endgroup$
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
asked Dec 10 '18 at 13:25
aid78aid78
1,2911315
asked Dec 10 '18 at 13:25
aid78aid78
1,2911315
asked Dec 10 '18 at 13:25
aid78aid78
1,2911315
asked Dec 10 '18 at 13:25
aid78aid78
1,2911315
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
add a comment |
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
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
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
$$
begin{cases}
x_2 &= left(frac{81}{5}right)^2 - 2(-100) - 109 = frac{8836}{25}\
y_2 &= -260 + frac{81}{5}left(-100 - frac{8836}{25}right) = -frac{950716}{125}
end{cases}$$
answered Dec 11 '18 at 3:19
achille huiachille hui
95.7k5132258
$endgroup$
add a comment |
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1 Answer
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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
$$
begin{cases}
x_2 &= left(frac{81}{5}right)^2 - 2(-100) - 109 = frac{8836}{25}\
y_2 &= -260 + frac{81}{5}left(-100 - frac{8836}{25}right) = -frac{950716}{125}
end{cases}$$
answered Dec 11 '18 at 3:19
achille huiachille hui
95.7k5132258
$endgroup$
add a comment |
$begingroup$
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
$$
begin{cases}
x_2 &= left(frac{81}{5}right)^2 - 2(-100) - 109 = frac{8836}{25}\
y_2 &= -260 + frac{81}{5}left(-100 - frac{8836}{25}right) = -frac{950716}{125}
end{cases}$$
answered Dec 11 '18 at 3:19
achille huiachille hui
95.7k5132258
$endgroup$
add a comment |
$begingroup$
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
$$
begin{cases}
x_2 &= left(frac{81}{5}right)^2 - 2(-100) - 109 = frac{8836}{25}\
y_2 &= -260 + frac{81}{5}left(-100 - frac{8836}{25}right) = -frac{950716}{125}
end{cases}$$
answered Dec 11 '18 at 3:19
achille huiachille hui
95.7k5132258
$endgroup$
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
$$
begin{cases}
x_2 &= left(frac{81}{5}right)^2 - 2(-100) - 109 = frac{8836}{25}\
y_2 &= -260 + frac{81}{5}left(-100 - frac{8836}{25}right) = -frac{950716}{125}
end{cases}$$
answered Dec 11 '18 at 3:19
achille huiachille hui
95.7k5132258
answered Dec 11 '18 at 3:19
achille huiachille hui
95.7k5132258
answered Dec 11 '18 at 3:19
achille huiachille hui
95.7k5132258
answered Dec 11 '18 at 3:19
achille huiachille hui
95.7k5132258
95.7k5132258
add a comment |
add a comment |
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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