1

I am trying to work out an example problem in which I have to find order of $P(1/2,1/2)$ on the elliptic curve $ y^2 = x^3 + x/4 $ .

So Far I have done the following:

1) Given equation of the curve, found out the equation of tangent at P (it is $y = x$). Then found the intersection with the curve at $(0,0)$. Verified this by using the formula for finding $2P$. So $2P$ is $(0,0)$

Now my problem is given $P$ and $2P$ how do I proceed further to find $3P$ and so on.

I have used the $P+Q$ result here to find $3P$ as $(1/2,0)$ but I cannot seem to obtain it using the geometrical interpretation.

Any explanation to proceed with this will be really helpful.

Raffaele
  • 26,371
Meseeks
  • 11
  • You should learn how to post using MathJax. But this is basically a well-posed question, showing some work on your part. – rogerl Nov 03 '17 at 18:16
  • Thanks..I am totally new to posting on this forum (i usually find my answers in other posts). But i'll work on MathJax to improve quality of the post. – Meseeks Nov 03 '17 at 18:22
  • The order of the point $P$ is $4$ $$3P=(1/2,-1/2);;4P=(0,1);;5P=(1/2,1/2)$$ I used SAGE, I advice you to get it, it's free and it can do a lot of things! http://www.sagemath.org/ – Raffaele Nov 03 '17 at 21:30
  • Whenever the equation of the curve is in the (short) Weierstrass form $$y^2=x^3+ax^2+bx+c,$$ the points of order two are exactly those with $y=0$ (implicit differentiation tells that those have a vertical tangent, so their doubles go to the point at infinity). Therefore your $2P$ has order two, and consequently $P$ has order four. – Jyrki Lahtonen Nov 03 '17 at 21:46
  • Thanks for the replies everyone. I am pretty sure I understand the concept better. – Meseeks Nov 04 '17 at 02:38

0 Answers0