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I'm reading Elliptic Tales.

I'm trying to understand why addition of $(0, 4)$ with itself on the elliptic curve $y^2 = x^3 + 1$ over the field $F_5$ results in $(0, 1)$.

Here's my attempt at solving the exercise:

Because we are adding 2 points that are the same, I need $\lambda$ to be the slope of the tangent line.

The slope of the tangent is:

$$ \frac{{3x_1}^2 + A}{2y_1} $$

if the equation is in the form $y^2 = x^3 + Ax + B$

We know $x_1$ is $0$ and $y_1$ is 4.

So the slope of the tangent is $\frac{1}{8} = \frac{1}{3} = 1 \cdot 3^{-1} = 1 \cdot 2 = 2$

We then use the formula: $x_3 = \lambda^2 - x_1 - x_2$, which gives us a x-coordinate of 4.

However, the book says the answer is $(0, 1)$, so the x-coordinate cannot be 4.

Does anyone know where I'm slipping up?

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    $\begingroup$ You switched $A$ and $B$. $A=0$ so the numerator for $\lambda$ is $3*0^2+0 = 0$ not $1$. $\endgroup$
    – AHusain
    Commented Jan 15, 2022 at 0:07

1 Answer 1

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Your lambda is not correct; $$y^2 = x^3 + A \cdot x + B$$ then $A=0$

$\lambda = \frac{{3x_1}^2 + A}{2y_1} = \frac{0 + 0}{1} = 0$

With the doubling formulas;

  • $x_3 = \lambda^2 - 2 x_1$
  • $y_3 = \lambda (x_1 -x_3) -y_1$

then

  • $x_3 = 0^2 - 2 \cdot 0 = 0$
  • $y_3 = 0 (0 - 0) - 4 = 1$

Therefore the result is $(0,1)$

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