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The elliptic curve $ y^2 = x^3 -x + b $ has 2 points of inflection where $ y'' = 0 $. Visualized here.

It seems that $ P + P $ for at say the upper point would be $ - P $ since the tangent at $ P $ intersects "triply" at $ P $ and therefore $ P + P + P = 0 $.

If I am correct then it seems that $ 2P = - P $, $ 3P = 0 $, $ 4P = P $, $ 5P = -P $, $ 6P = 0 $, and so forth. This "feels unusual" since no other point except the left-most one does this. All other points seem to bounce around indefinitely when you compute $ nP $.

Is there any special term or treatment in elliptic curves for the point I have described for which $ nP \in \{ O, -P, P \} \forall n $? Or, is my math incorrect?

user2297550
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1 Answers1

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If one has a Weierstrass elliptic curve $$E:\quad y^2=x^3+ax+b$$ then a nonzero point $P$ satisfies $3P=O$ if and only if $P$ is an inflection point, that is the tangent meets $E$ with multiplicity $3$.

It's not true that every point "bounces around" when you compute $nP$. Over the complex numbers, an elliptic curve has $n^2$ points with $nP=O$. Over the real numbers, at least $n$ of them have real coordinates. These are the torsion points of $E$. However Barry Mazur proved that on a curve with rational cooefficients, any torsion point with rational coordinates has $nP=O$ with $n\le12$.

Angina Seng
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  • Thanks. That last part was slightly confusing. I guess it leaves open the fact that a curve with rational coefficients does have n points with real coordinates with nP = 0 for all n, but when n > 12 the points will not have rational coordinates? (+1) – user2297550 Jan 04 '18 at 10:47
  • An elliptic curve with real coefficients has either $n$ or $2n$ real solutions of $nP=O$. – Angina Seng Jan 04 '18 at 10:48