Points of order three on elliptic curves

In an earlier post, we saw how it’s easy to tell which points on an elliptic curve y2 = x3 + ax + b have order 2. What about order 3? That’s not much harder. If we have 3P = O then 2P =  –P, and we can use what we know about points of order 2 to find out what happens for points of order 3.

Let’s write

P1= (x1,y1)

and

P2 = 2P1 = (x2,y2)

so that

 –P1 = (x1, –y1)

From the earlier post on point doubling we have that

x2 = (x14  – 2ax12  – 8bx1 + a2) / [4 (x13 + ax1 + b)]

If 2P1 =  –P1 then we have that

x2 = x1

or

(x14 – 2ax12 – 8bx1 + a2) / [4 (x13 + ax1 + b)] = x1

or

(x14 – 2ax12 – 8bx1 + a2) / [4 (x13 + ax1 + b)] – x1 = 0

so that

x14 – 2ax12 – 8bx1 + a2 – 4x144ax12 – 4bx1 = 0

or that

3x14 + 6ax12 + 12bx1a2 = 0

Example

For the elliptic curve y2 = x3 + 1 we have that the x-coordinates of the points of order 3 need to have the property that

3x4 + 6ax2 + 12bxa2 = 0

with a = 0 and b = 1 this means that we have

3 x4 + 12x  = 3x (x3 + 4) = 0

the only rational solution of which is x = 0. Thich corresponds to the points (0,1) and (0,-1) on the elliptic curve. Here’s what this looks like:

Image001

Another point of view

From the formula for doubling a point we get that

x2 = m2 – 2x1

where

m = y′ = (3x12 + a) / (2y1)

And because x2 = x1 we can write

x1 = m2 – 2x1

or that

m2 = 3x1

Now if we have that

y2 = x3 + ax + b

then we have that

2 y y′ = 3x2 + a

and that

2 y y′′  + 2 (y′)2 = 6x

so that

y′′ = (6x – 2 (y′)2) / (2y)

This means that we have y′′ = 0 when

6x – 2 (y′)2 = 0

or

(y′)2 = 3x

or

m2 = 3x

As we saw above, this happens at a point of order 3, so at a point of order 3 we have that y′′ = 0.

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