Let $\mathbb{K} = \mathbb{F}_p$ bei a Field, where $p$ is a prime number $p > 3$ or $\text{char}(\mathbb{K}) = 2$.
Let $E(a, b,\mathbb{K})$ be a corresponding elliptic curve. Show that, $E(a, b, \mathbb{K})$ has at most three points of order two and no more than eight points of order three.
May attemp:
An element $P$ of any group is seid to have order $m$ if
$$mP = \underbrace{P + P + P + \cdots + P}_{m} = \mathcal{O}$$
Let $(x,y) = P \in E(a, b,\mathbb{K})$, with of order two, then $2P = \mathcal{O}\implies P = -P$.
Sience $-P = -(x,\,y)$ is just $(x,\, -y)$, these are the points with $y = 0$:
$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad$
Therefore, only $P_1$, $P_2$ and $P_3$ have order $2$. Is this correct?
Now, how can I prove, that in $E(a,\,b,\,\mathbb{K})$ there are maximum $8$ points of order $3$?!
Some tipps? Thanks.
In the case $p>3$ show that $(x,y)$ has order $2$ iff $y=0$.
Also in this case, $(x,y)$ is a point of order $3$ iff it's an inflection point of the cubic curve, and the inflection points are where the cubic curve (degree $3$) meets its Hessian (also degree $3$). In how many points can two curves of degree $3$ meet?
In the $p=2$ case there's a bit more fiddling around to do.