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What happens if the the next point generated in and ECC system is not an integer by integer coordinate? I'm new to ECC, so this is probably a dumb question, but any explanation would be appreciated.

William Grannis
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  • A curve as used in cryptography is over a finite field, so all points on the curve are $(x,y) \in F^2$ where $F$ is a finite field, most commonly a field of the form $F_p$ where $p$ is prime (not all such pairs, just the one obeying the EC equation). This has a group law $+$ that computes the sum of two points on the curve and always gives a point on the curve. There is no definedness problem. It's not like pictures we can draw in the plane, more like a "cloud" of points in a huge finite grid.. – Henno Brandsma Mar 27 '18 at 21:10

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Elliptic-curve cryptography is over finite fields, not over integers. Because $E(\mathbb{F}_q)$ is a group, for an elliptic curve $E$, we need not worry about this.

Dietrich Burde
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  • So are you saying that if the generating points are integer pairs, the output will be too? – William Grannis Mar 27 '18 at 14:28
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    They are not integer pairs, they are pairs $(x,y)$ in $\mathbb{F}_q^2$. It is customary for $\mathbb{F}_p$ to write the elements as $0,1,2,\ldots ,p-1$, but certainly $\mathbb{F}_p\not\subset \mathbb{Z}$. – Dietrich Burde Mar 27 '18 at 15:07
  • If they are not integer pairs, are they rationals? If they're neither, then they are irrational, so how do the computers store them without rounding errors after the 256 bit number of times a is dotted? – William Grannis Mar 27 '18 at 19:23
  • Also, if you know the starting point, can't you just go through and try each one until you get the ending point and discover the key? Isn't recursively calculating the next one how Alice figures out what to send anyway? – William Grannis Mar 27 '18 at 19:26