0
$\begingroup$

Suppose there is an elliptic curve

$E:y^2=x^3+(p-1)\cdot x\bmod{p}$

with $p>3$.

The question is: What are the points that have an order of 2?

$\endgroup$
1
$\begingroup$

Let $E$ an elliptic curve defined by the equation $y^2 = f(x)$ over a prime field $\mathbf F_p$ with $p>3$, with $f$ a degree $3$ polynomial.

A point $P=(x_0,y_0)$ of order $2$ is a point that satisfies $2P = \mathcal O$ where $\mathcal O$ is the neutral element of the group law also known as the infinity point. Then we have $$ 2P = \mathcal O \quad \Longleftrightarrow \quad P + P = \mathcal O \quad \Longleftrightarrow \quad P = -P, $$ but $-P = (x_0,-y_0)$ so it means that $y_0 = - y_0$ so $y_0 = 0$. There we have it, a point of order $2$ must have its $y$-coordinate set to 0, and plugging it into the curve equation, then $x_0$ satisfies the equation $$ f(x_0) = 0. $$ To find all the $2$-torsion points you must find the roots of this polynomial.

Example: the curve $y^2 = x^3 + 10x + 6$ on the field $\mathbf F_{17}$ has three points of order $2$: $(1,0)$, $(2,0)$ and $(14,0)$, since we have $x^3 + 10x + 6 = (x-1)(x-2)(x+3)$.

| improve this answer | |
$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.