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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 ?

MAS
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  • An interesting fact is that some elliptic curves have some special properties that make point doubling and addition "easier" (for example a montgomery curve). You can check here for some of them (https://en.wikipedia.org/wiki/Table_of_costs_of_operations_in_elliptic_curves) – Alex Them Oct 03 '22 at 09:55
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    Not sure, but it seems that $(33/4,-841/16)$ is not on the elliptic curve. You used $25/4$ instead of $5/2$… – Aphelli Oct 03 '22 at 09:57
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    My answer here might be of help: https://math.stackexchange.com/questions/4158446/how-to-find-2p-and-3p-of-an-elliptic-curve/4175577#4175577 – Douglas Molin Oct 03 '22 at 09:57
  • @Aphelli, thanks. I made mistake in calculation – MAS Oct 03 '22 at 10:14
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    If you are learning about elliptic curves, you may want to use PARI/GP for computations. For example, in your case, try E=ellinit(ellfromeqn(-y^2+x^3+2)); P=[-1,1]; for(n=1,4,print(n," ",ellmul(E,P,n))). This will find $2P=(17/4,-71/8)$ and $3P,4P$. – Somos Oct 03 '22 at 10:48
  • @Somos, thanks. Yes I have used PARI/GP in other purpose. But why my calculation by hand didn't work ? – MAS Oct 03 '22 at 10:59
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    Here's the LMFDB entry for this specific elliptic curve: https://www.lmfdb.org/EllipticCurve/Q/1728/n/4 – Daniel Hast Oct 03 '22 at 17:11
  • @DanielHast, thanks. LMFDB is very informative. Can you please comment on my above approach ? what went wrong here ? Because at first I have to learn by hand calculation. – MAS Oct 04 '22 at 05:30

1 Answers1

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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

A = 0
B = 2
E = EllipticCurve([A,B])

P = E(-1,1)
2*P

outputs

(17/4 : -71/8 : 1)
kelalaka
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