# Find Points of order 2 on elliptic curve [closed]

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?

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 $$x^3 + 10x + 6 = (x-1)(x-2)(x+3)$$.