Quote from: kTimesG on May 28, 2025, 02:11:37 PM

Quote from: fixedpaul on May 27, 2025, 11:53:19 PM

If I understood your question right, maybe one possible way to simplify the check is to first verify whether both points are n-torsion points (nP = O and nQ = O). If not, you can already rule out that they belong to the same n-order subgroup. If they both are, i think you have to build up the entire subgroup as you said, i don't think there is a direct way to check. But at least the n-torsion check can help "quickly" exclude cases that definitely don’t match, or at least use less computation. Could this make sense?

I used the wrong terminology. I meant quick-checking whether two points belong to the same orbit. Since n - 1 divides by 2, 3, 149, 631, .... So all points on the curve can be grouped in orbits of size 2, 3, 149, 631. The sum of all points (and scalars) of each distinct orbit is O (0).

In the end, any valid point has a unique 7-dimensional scalar. I think a trial and error search of common coefficients on lower dimensions (same orbit, or finding the distance between orbits) may simplify the ECDLP (with some sort of pollard rho). It might also be nothin'.

Can you elaborate more on this? Send me a DM. I am working on this idea.