# Find all points order 3 on an elliptic curve

Let $$P=(x,y)\in E(\mathbb{F}_p)$$ by a Weierstrass equation $$y^2=x^3+ax+b$$. Show that $$3P=\mathcal{O}$$ iff $$3x^4+6ax^2+12bx-a^2=0$$.

I derived that every point in $$\{P\in E(\mathbb{F}_p)|3P=\mathcal{O}\}$$ is a root of the above equation, then no further clue. I also tried to from $$y^2=x^3+ax+b$$ to yield $$3x^4+6ax^2+12bx-a^2=0$$, but not sure is the right approach because I can't find a way to do it.

Any help or hints will be great, thanks.

• $3P=O$ iff $2P=-P$, so I suggest using the formula for $2P$. Apr 3, 2020 at 5:04

$$\textit{Proof:}$$
First, note that $$3P=\infty$$ iff $$2P=-P$$. Then let $$P=\mathbb{F}_p$$. If $$P$$ is the finite point $$(x,y)$$, then $$-P=(x,-y)$$; iff $$P=\infty$$, then $$2P=3P=\infty$$. If $$y=0$$, then $$2P=\infty\neq-P$$. Otherwise, we can calculate $$2P=(x', y')$$ by $$\lambda=\frac{3x^2+a}{2y}, x'=\lambda-2x, y'=\lambda(x-x')-y$$ So $$2P=-P$$ iff $$x'=x$$ and $$y'=y$$. Now let's consider: \begin{align*} 3x&=\lambda^2\\ 3x&=\frac{(3x^2+a)^2}{4y^2} \\ 3x&=\frac{3x^2+a)^2}{4(x^3+ax+b)}\\ 12x(x^3+ax+b)&=(3x^2+a)^2\\ 12x^4+12ax^2+12bx&=9x^4+6ax^2+a^2\\ 3x^4+6ax^2+12bx-a^2&=0 \end{align*} And vice versa for the other direction of the proof. So in conclusion, $$P=(x,y)\in E(\mathbb{F}_p)$$ satisfies $$3P=\infty$$ iff $$3x^4+6ax^2+12bx-a^2=0$$.