I've been troubled by this, given any random number as the number of times we'll multiply a point by itself, will we always be guaranteed to land on the curve again? Is there a case where we'll have a miss?

Also, is the question to above not always true unless we are in finite fields defined by specific curves? Is this part of the reason why we need to choose specific curves?

  • $\begingroup$ I'm voting to close this question as off-topic because it is about general mathematics. $\endgroup$ – fkraiem Sep 12 '17 at 0:11

Adding two points on a curve following the addition law will always give another point on the curve. That's part of what it means for the curve to form a group.

The addition law is defined relative to a particular curve and a particular field. For example, consider the field $\mathbb F_3$, i.e. the integers $\{0, 1, 2\}$ with arithmetic—addition and multiplication—performed modulo 3, and consider the curve $$E/\mathbb F_3 : x^2 + y^2 = 1 + 2 x^2 y^2.$$ The point $(1, 0)$ (i.e., $x = 1, y = 0$) is on this curve, which you can check by a simple calculation. The point $(2, 0)$, however, is not. (Exercise: List, for every possible $x, y \in \mathbb F_3$, whether $(x, y)$ is on the curve. There are only nine of them!) The addition law $$(x_0, y_0) + (x_1, y_1) = \left(\frac{x_0 y_1 + x_1 y_0}{1 + d x_0 x_1 y_0 y_1}, \frac{y_0 y_1 - x_0 x_1}{1 - d x_0 x_1 y_0 y_1}\right)$$ always gives another point on the curve. For example, $(2, 1)$ is also on the curve, and $$(1, 0) + (2, 1) = \left(\frac{1 + 0}{1 + 0}, \frac{0 - 2}{1 - 0}\right) = \left(1, \frac{-2}{1}\right) = (1, 1),$$ which you can check by another simple calculation.

That addition law always works for the points $(x, y)$ on $E$ for $x, y \in \mathbb F_3$, but does not always work for every curve or every field. For example, it does not work for $$E/\mathbb F_7 : x^2 + y^2 = 1 + 4 x^2 y^2$$ because $4 = 2\cdot 2$ is a quadratic residue in $\mathbb F_7$. (Try it with $(1, 5)$ and $(4, 5)$—you'll get a denominator which is a multiple of 7, which is equivalent to zero in $\mathbb F_7$, which you can't divide by.) Not every curve can even be written in this form, which is called Edwards form; for example, curves of prime order, such as NIST P-256, have no Edwards form. Other forms of curves have different addition laws, which work on different subsets of points and in different cases.


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