Elliptic Curve scalar multiplication on $\mathbb{R}$

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.

• The article is correct. Your $x$-coordinate gives a non-rational $y$. So just a miscalculation with the formula. – Dietrich Burde Dec 10 '18 at 13:30
• formulas I have taken are not from this article. Is it right formulas for this curve? – aid78 Dec 10 '18 at 14:00
• No, it is not correct, because it assumes the short Weierstrass form $y^2=x^3+Ax+b$ and not the one with term $x^2$. – Dietrich Burde Dec 10 '18 at 15:49
• @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. – aid78 Dec 10 '18 at 16:39

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 $$\begin{cases} x_2 &= x_3 &= s^2 - 2s_1 - a\\ y_2 &= -y_3 &= -y_1 + s(x_1 - x_2) \end{cases} \quad\text{ where }\quad s = \frac{3x_1^2 + 2ax_1 + b}{2y_1}$$

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

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