Why is it true that on an elliptic curve a point on it $P$ is of order 3 iff $P$ is an inflection point on the curve?
I'm taking a course on elliptic curves and this is something that I see in proofs everywhere, but I can't seem to find a proof of the fact itself. I've tried to work it out myself but I just can't seem to figure it out. Any help would be greatly appreciated.
Jyrki Lahtonen, provided the formulation;
$$[3]P = \mathcal{O} \iff [2]P=-P$$
Let see this first geometrically;
First, we need to find a curve with a 3-torsion element;
After 200K random searches, this provided
You may get a different curve ( feel free to add some more samples of such curves)
Now, more SageMath to execute the calculations
Now the final output;
From the figure, we can observe the tangent line doesn't intersect with the curve other than the point $P$. We need to verify this observation with Math.
Mathematically to see this we need substitution of the tangent line
$$y = -6(x − 12 ) + -153$$
with the curve equation
$$y^2 = x^3 + 1404 x + 4833$$ to get
$$(-6(x − 12 ) + -153)^2 = x^3 + 1404 x + 4833$$
We can continue to use SageMath for this to find the roots (i.e. the intersection points of the curve and the tangent), too
and this outputs a triple root at $12$
$$(x - 12)^3$$
and, this clearly explains that the tangent line only intersects at point $(12,-153)$ making the point $P$ as a torsion $3$ element and an inflection point.