full-disclosure - Re: Rapid integer factorization = end of RSA?

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Date: Thu, 26 Apr 2007 13:07:45 +0200
From: Stanislaw Klekot <dozzie@...amit.im.pwr.wroc.pl>
To: Eugene Chukhlomin <chukh29ru@...oline.su>
Cc: Full-Disclosure <full-disclosure@...ts.grok.org.uk>
Subject: Re: Rapid integer factorization = end of RSA?

On Thu, Apr 26, 2007 at 03:04:39PM +0400, Eugene Chukhlomin wrote:
> >#v+
> >gap> p;
> >163473364580925384844313388386509085984178367003309231218111085238933310010450\
> >8151212118167511579
> >gap> q;
> >190087128166482211312685157393541397547189678996851549366663853908802710380210\
> >4498957191261465571
> >gap> n := p * q;
> >310741824049004372135075003588856793003734602284272754572016194882320644051808\
> >150455634682967172328678243791627283803341547107310850191954852900733772482278\
> >3525742386454014691736602477652346609
> >gap> (p * (n - q)) mod n;
> >0
> >gap> 
> >#v-
> >
> >What is it supposed to proove?
> 
>   
> 
> My gypothesis: if exists subsets(A1...An) and (B1...Bn) which satisfies 
> equality: A1*B1 +...An*Bn =  N = p*q, then exists some of them, which 
> satisfies equality A1*(-B1)+...An*(-Bn)=p*q*(q-1)

But what does that proof have to do with your gypothesis? Except that
  p*q * (q-1) = p*q = p*q * (-1) = p*q * (N-1) = 0   (mod N)
what is obvious equality.

-- 
Stanislaw Klekot

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