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| Subject: | Re: [Full-disclosure] Rapid integer factorization = end of RSA? |
|---|---|
| Date: | Fri, 27 Apr 2007 20:42:16 +0200 (MET DST) |
On Thu, 26 Apr 2007, e.chukhlomin wrote:
Gypothesis: Let N = p*q = A1*B1 + A2*B2... + An*Bn Then exists some subset(A1...An) and respective subset(B1...Bn), which satisfies for equality: A1*B1+A2*B2...+An*Bn = p*q and: A1*(-B1)+A2*(-B2)...+An*(-Bn) = p*(-q)=p*q*(p-1) or A1*(-B1)+A2*(-B2)...+An*(-Bn) = (-p)*q=p*q*(q-1)
Let n = 1, A1 = p, B1 = q. Then
1. A1B1 = pq = N.
2. A1(-B1) = p(-q) =
[let's pretend this careless mixing of equalities in Z an
congruences in Z_N makes any sense and assume -X stands for N-X]
= p(N-q) = p(pq-q) = p(p-1)q = pq(p-1).
QED.
Ok. Your "gypothesis" holds (sort of). We can factor N when we know its
factors. What a breakthrough. Perhaps Bill Gates will mention it in
"The Road Ahead II".
--Pavel Kankovsky aka Peak [ Boycott Microsoft--http://www.vcnet.com/bms ]
"Resistance is futile. Open your source code and prepare for assimilation."
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