I've been trying to manually compute the doubling of the point (3, 6) on the elliptic curve $y^2 = x^3 + 2x + 3 \mod 97$. I have found the x coordinate, with the steps shown below:
- Differentiating implicitly
$2yy' = 3x^2 + 2 \mod 97$
- Substituting for x and y (the point (3, 6))
$12y' = 29 \mod 97$
Finding the multiplicative inverse using Euler's totient theorem, and using this to find the value of y' which I found to be 59
Substituting the value of y' (59) into the equation $x_3$ = $\lambda^2 - 2x_1 \mod 97$, where $\lambda$ is the slope, $x_1$ is the original point on the equation and $x_3$ is the new point on the equation.
Now that I found that $x_3$ is equal to 80 (which is correct, as seen in this simulation: https://cdn.rawgit.com/andreacorbellini/ecc/920b29a/interactive/modk-mul.html) I need to find the y value for the "doubled" point on the elliptic cruve.
First, did I make any mistakes in finding the x-coordinate of the "doubled" point? I don't think I did, as I got the correct answer according to the simulation I'm using to test.
Most importantly, how do I find the y value on the elliptic curve once I found the x value? To conclude, the curve is $y^2 = x^3 + 2x + 3 \mod 97$, the point I'm trying to double is (3, 6) and the $x$ coordinate of the new point is 80. Now, how do I find the y - coordinate of this point?
Thank you!
You can do all calculations as if they're in $\mathbb{Q}$.
For the elliptic curve $y^2 = x^3 + 2x + 3$ over $\mathbb{F}_{97}$, the derivative at $(3,6)$ is: $$2yy' = 3x^2+2 \implies 12y' = 3(3^2)+2 \implies y' = 59$$ Hence the tangent at $(3,6)$ is: $$y-6 = 59(x - 3) \implies y=59x + 23$$ Subsitute it into $$(59x+23)^2 = x^3+2x+3 \implies x^3 + \cdots + (3-23^2)=0$$ So product of three roots is $23^2-3 = 41$, two of them are $3$, thus the remaining one is $41/9 = 80$. Thus $y$-coordinate of third intersection is $59\times 80+23 = 87$.
Thus the double of $(3,6)$ is $(80,-87)=(80,10)$.