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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.

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

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$\textit{Proof:}$

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 $3x^4+6ax^2+12bx-a^2=0$.

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