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 – 4x₁⁴ – 4ax₁² – 4bx₁ = 0

or that

3x₁⁴ + 6ax₁² + 12bx₁ – a² = 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 + 12bx – a² = 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:

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.