Ninjastic
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Scraped on 27/08/2025, 17:35:33 UTC
Quote from: PROFIGROK on August 27, 2025, 10:15:15 AM
Points A and B have a common Y coordinate. For each point P on the curve, there is a pair of points A and B such that A-B=P. Does this give us an advantage when computing the private key?

Yes. But just when the only information you have is that the private key is anywhere between 1 and n - 1.

In other words: good luck finding the specific pair of points that add up to P (or one of its equivalent X endos).
Original archived Re: P = A[X1, Y] - B[X2,Y]
Scraped on 27/08/2025, 17:30:08 UTC
Quote from: PROFIGROK on August 27, 2025, 10:15:15 AM
Points A and B have a common Y coordinate. For each point P on the curve, there is a pair of points A and B such that A-B=P. Does this give us an advantage when computing the private key?

Yes. But just when the only information you have is that the private key is anywhere between 1 and n - 1.

In other words: good luck finding the specific pair of points that add up to P (or one of its equivalent X endos).