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This might be a newbie question, In elliptic curve, I chosen a secret a and computed aG, where G is the base point. But here since I know G, I can compute inverse of G and multiply it with aG, to get a.

I knew I'm terribly missing something here, please explain me what am I missing here?

For example on curve25519, I'm doing the following:

Base Point: 9 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

Inverse of Base Point: 18 199 113 28 199 113 28 199 113 28 199 113 28 199 113 28 199 113 28 199 113 28 199 113 28 199 113 28 199 113 28 71

Base Point * It's Inverse: 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

Isn't it the inverse?

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    $\begingroup$ Multiplication of two elements does not exist in elliptuc curves, and therefore a multiplicative inverse of $G$ does not exist either. The elliptic curve is a group, not a ring (or even field). $\endgroup$ – tylo Nov 9 '17 at 15:22
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Group operations in elliptic curves such as $aG$ are point additions or point doubling. Addition of two points is done (geometrically) following the ''chord-and-tangent'' rule: Draw a line through both points - this line will intersect the curve in one other point. The sum of the two points is the reflection of this intersection point about the x-axis like this: (source: page 13 of Johnson et al. http://residentrf.ucoz.ru/_ld/0/34_Digital_Signatu.pdf)

In your terms, think of $aG$ as $G+G+G+G+...$ $a$ times in total, starting with $G+G$ where your line is the tangent on $G$ (through ''both points'').

The inverse you refer to is the elliptic curve discrete logarithm problem (''find $a$''), but as you see the term $aG$ is not a regular multiplication of two numbers, but a sequence of operations in a group of points on an elliptic curve over a finite field. Inverting the addition operation descibed above is extremely difficult for classical computers.

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  • $\begingroup$ thank you, I just edited my question with some inputs, I'm working on curve25519, in that some API is there that can compute the inverse of the point, with that I felt like I can compute the inverse of base point and use it nullify the base point to get the secret. $\endgroup$ – sg777 Nov 10 '17 at 4:47

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