20060810, 15:09  #1 
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) 
20060810, 16:38  #2 
Jul 2005
602_{8} Posts 
*plonk* (of ignore equivalent of)

20060812, 09:17  #3 
Sep 2005
177_{8} 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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