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Let an elliptic curve $ E(a,b )$ $$E(a,b ) = \{(x,y)\,|\,y^2=x^3+ax + b\}$$ Where the points of the line $xm + n$: $$P =(x_1, y_1),\, Q =(x_2, y_2),\, R =(x_3, y_3) \in E(a, b)$$
How can you calculate the product between them? $$Q\cdot P =\, ?$$

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J. M. ain't a mathematician
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Darío A. Gutiérrez
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  • more or less by definition, the product is $(x_3, -y_3)$ (if the origin is the point at infinity). We have $x_1 + x_2 +x_3$ equals the coefficient of $x^2$ in $(xm +n)^2 = x^3 + ax + b$, i.e. $m^2$. so $x_3=m^2-x_1 - x_2$ – usr0192 Jun 12 '17 at 20:52
  • if I understood that right $(x_3, -y_3) = P + Q$ and not $ P \cdot Q$ – Darío A. Gutiérrez Jun 12 '17 at 20:56
  • Oh. What is the definition of $P\cdot Q$? – usr0192 Jun 12 '17 at 20:57
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    Group operation is typically written additively, so there usually is no $P\cdot Q$. Multiplication by an integer is ok though, so $nP$ – sharding4 Jun 12 '17 at 21:01

1 Answers1

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Say we have the line $(x_0+ct,y_0+dt)$ which intersects the curve in 3 points : at $t=-1,t=1$ and $t= ?$

Finding $?$ means solving the 3rd root of the cubic polynomial $$P(t) = (y_0+td)^2- (x_0+tc)^3-a(x_0+tc)-b$$ And obtain the factorization $$P(t) = -c^3 (t+1) (t-1)(t-?)$$ Therefore $$? = \frac{P(0)}{c^3}=\frac{y_0^2-x_0^3-ax_0-b}{c^3}$$ From which we obtain the group law $$(x_0-c,y_0-d)+(x_0+c,y_0+d) = (x_0+\frac{y_0^2-x_0^3-ax_0-b}{c^2}, -y_0-d\frac{y_0^2-x_0^3-ax_0-b}{c^3})$$

reuns
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