mersenneforum.org ECDLP cracked mathematically acc. Bearnol
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 2006-08-10, 15:09 #1 bearnol     Sep 2005 127 Posts ECDLP cracked mathematically acc. Bearnol Certicom was/is currently offering a challenge to crack some elliptic curves encrypted by discrete log... Perhaps you could tell me if the following is of merit? (I suppose I don't really mind if it gets moved to the Misc. Math section, like most of my posts seem to ( :-) ), so long as it does actually get evaluated by someone competent to judge at some point, and then the decision is marked as such - otherwise I feel like I'm posting into a black hole!) Also, if, as I hope, it now renders the Certicom challenges trivial (at least to elliptic curve pro) the cash would come in handy :) Anyway, Problem: solve B^x == y [mod p] Solution: B^x == y [mod p2^n] 2^n.B^x == 2^n.y [mod p2^n] x == log_B{(y.2^n)/2^n} [mod p2^n] x == log_B(y.2^n) - log_B(2^n) [mod p], where n is chosen s.t. y.2^n > p, thus converting the problem from one in Fp to one in reals. Note that this is all related to the rel. primality of 2^n, and p - see Fermat's Big Theorem (aka Wanless' Theorem) on my website, as I've cited before. Note also, that if dealing with the corresponding problem in F2^m, ie B^x == y [mod 2^m], the corresponding logic applies, where a q is chosen s.t. y.q > 2^m thanks, J (bearnol)
 2006-08-10, 16:38 #2 Greenbank     Jul 2005 2×193 Posts *plonk* (of ignore equivalent of)
 2006-08-12, 09:17 #3 bearnol     Sep 2005 127 Posts Hmmmm... I'm not sure if this my original post is quite right. Hopefully someone will see what I was _trying_ to do, and finish it off? thanks, J

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