# Point addition of $(0, 4)$ with itself on elliptic curve over finite field $F_5$

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?

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

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 $$(0,1)$$