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Old 2006-08-10, 15:09   #1
bearnol
 
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Sep 2005

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Default 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)
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Old 2006-08-10, 16:38   #2
Greenbank
 
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*plonk* (of ignore equivalent of)
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Old 2006-08-12, 09:17   #3
bearnol
 
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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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