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Date: Thu, 26 Apr 2007 10:53:56 +0400 From: Eugene Chukhlomin <chukh29ru@...oline.su> To: full-disclosure@...ts.grok.org.uk Subject: Rapid integer factorization = end of RSA? Hi list! I discovered a new method of integer factorization for any precision numbers, probable it should be an end of RSA era. Details: Let N - the ring and N = p*q Then, (-p) in terms of ring(N) is equal (N-p) Lemma: p*(-q)=p*q*(-p) and respective: (-p)*q=p*q*(-q) Proof: p*(-q)=p*(N-q) - by the data, then p*(-q)=p*(p*q-q)=p*pq-p*q=p*q*p-p*q=(p-1)*(p*q) (-p)*q=q*(N-p) - by the data, then (-p)*q=(p*q-p)*q=p*q*q-p*q=p*q*q-p*q=(q-1)*(p*q) Q. E. D. 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)=p*q*(p-1) or A1*(-B1)+A2*(-B2)...+An*(-Bn) = (-p)*q=p*q*(q-1) If found such (A1...An) and (B1...Bn), we can find p or q by dividing p*(q-1) on p*q: p*(q-1)=p*q*(p-1) => (p*(q-1))/(p*q)=(p-1) => (p-1)+1 = p or (p-1)*q=p*q*(q-1)=>((-p)*q)/(p*q)=(q-1) => (q-1)+1 = q Sample: 21 = 3*7 Let's view a binary representation of this number: 10101 => 2^4 + 2^2 + 1 => 4*4+2*2+1*1 Then, we can try to find 7*(-3) in terms of ring(21): 4*(-4) + 2(-2) + 1*(-1) => 4*(21-4)+2*(21-2)+1*(21-1)=>4*17+2*19+1*20 = 68+38+20=> 68+38+20 = 126 = 6*21 6+1=7 This implementation of my gypothesis has very hard complexity (about a log2(N)! comparations), but exists a short way with fixed complexity for implementation of hypothesis ("plan B") - but, by ethical reason, I'll not post it here. Regards, Eugene Chukhlomin _______________________________________________ Full-Disclosure - We believe in it. Charter: http://lists.grok.org.uk/full-disclosure-charter.html Hosted and sponsored by Secunia - http://secunia.com/
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