S₃ Discriminants and Their Role in secp256k1

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#1Mar 31, 2025, 04:39 PM
Hey everyone I found something interesting while messing with some polynomials and automorphisms. So, there’s this S₃ Perfect-Square Discriminant. If you do x2 = β * x1 (where β³ = 1 when j = 0), the discriminant for the S₃ polynomial in z turns out to be Δ = 16 * (x₁³ + b)² (mod p). Does anyone know if this is linked to algorithms like Pollard Rho? I just started learning cryptography and I could use some help.
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#2Apr 1, 2025, 02:55 PM
Interesting point. You can cube x and get a value that represents P, P*lamba, P*lambda^2. This basically triples the effective table size. But the speedup only matters if a point’s equally likely to be in one of those groups. For secp256k1, DLP solving usually occurs in a limited range, right? The whole group is huge, so chances are slim unless x is within a small range. It's a real thing in some cases, but also tons of pointless challenges.
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degen23Member
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#3Apr 1, 2025, 07:31 PM
Oh hey I asked some AI out of curiosity about this acceleration topic, and it mentioned that it’s really more about ECDLP issues.
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sigma2016Newbie
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#4Apr 1, 2025, 11:34 PM
Yikes. AI can’t even count: https://www.facebook.com/reel/1138880084809988 And it struggles with ECDLP in small examples too: https://here
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falcon2019Full Member
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#5Apr 1, 2025, 11:44 PM
That's a fortunate case. I asked it to count 1 to 100 and it skipped 6. It also repeated lots of numbers and skipped random gaps after them, so that must be a clue to the secrets of the multiverse. So what I get, is that 6 does not exist, and arithmetic as we know it is wrong, ok? Though I don't see how this would be an issue for most people that ask it to break crypto.
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