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Let $P$ be a point on an elliptic curve over $\mathbb{R}$. Give a geometric condition that is equivalent to P being a point of order (a) $2$ , (b) $3 $ , (c) $ 4$ .

Could someone explain this to me in dummy terms? :) I have a test next week and the lecturer hinted a question similar to this may come up. Thanking you in advance!

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Given two points P and Q, how do I find their sum P+Q geometrically? Remember that "P is a point of order 2" means "P+P = 0", so if that's true, what geometric restriction does that place on P? – Billy Mar 21 '13 at 22:16
up vote 6 down vote accepted

This may help. Let $E/\mathbb{Q}$ be an elliptic curve with points $P$, $Q$, and $R$, such that $P$ is of order $2$, $Q$ is of order $3$, and $R$ is of order $4$. Now:

  • Plot $P$, and try to find $2P$ geometrically (i.e., show graphically that $2P=\mathcal{O}$).
  • Plot $Q$, find $2Q$ geometrically, plot it, and then try to find $3Q$ (and try to understand why $3Q=\mathcal{O}$).
  • Plot $R$, find $2R$, $3R$, and try to find $4R$ geometrically.

To get you started, here is such a curve $E: y^2 = x^3 - 157707x + 78888006$, with $$P=(-549,0),\quad Q=(27,8640),\quad \text{and} \quad R=(315,-7776).$$ If you use Sage, then you can plot $E$ like this:

E=EllipticCurve([- 157707, 78888006]);


Understanding the geometry behind points of order $2$ and $3$ may be easier in $E:y^2=x^3+1$, where $S=(2,3)$ has order $6$.

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