I am learning about elliptic curves.
Consider the elliptic curve $y^2=x^3+2$ over $\mathbb Q$. Then $P=(-1,1)$ is a point on it.
I want to find its multiples i.e., $2P,3P, \cdots$
The addition of two points $P=(x_1,y_1),~Q=(x_2,y_2)$ is obtained by intersecting the line $\bar{PQ}$ intersecting the elliptic curve, i.e., if $R=(x_3,y_3)$ be the addition of $P$ and $Q$, then it is given by
\begin{align} x_3&=\lambda^2-x_1-x_2 \\
y_3&=\lambda(x_1-x_3)-y_1, \\
\lambda&=\frac{y_2-y_1}{x_2-x_1} \end{align}
But when we double a point, the slope $\lambda$ will be as follows: $$\lambda=\frac{3x_1^2+2}{2y_1}.$$
So in our case $\lambda=\frac{5}{2}$. Hence:
$$x_3=25/4+1+1=33/4, ~y_3=5/2(-1-33/4)-1=-193/8.$$
So $2P=(\frac{33}{4},-\frac{193}{8})$.
Is it correct ?
Is it the only way ?
Edit: I already got the answer as mentioned in the folloing comment sections. But I couldn't understand why my approach didn't work ? Are the formulas that I used correct ?
Best Answer
Basically, your slope calculation was not correct.
Let $P=(-1,1)$
$$\lambda=\frac{3 x_1^2+2}{2y_1} = \frac{3\cdot 1^2+0}{2}= \color{red}{\frac{3}{2}}$$ The formula is
$$x_e = \lambda^2 - 2x_1$$
$$x_3=(9/4) - 2(-1) = 17/4$$
For $y_3$ formula is
$$y_3 = \lambda(x_1 − x_3 ) − y_1$$
\begin{align} y_3&=\left[(3/2)(-1-17/4)\right]-1\\ &=\left[(3/2)(-21/4)\right]-1\\ &=(-63/8)-1\\ &=(-63/8)-1\\ &=-71/8\\ \end{align}
Verifacation with SageMath on the SageMathCell
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