 mersenneforum.org HOW TO BYPASS RSA DIFFICULTY... not!
 Register FAQ Search Today's Posts Mark Forums Read 2020-04-28, 09:01 #1 Alberico Lepore   May 2017 ITALY 5·97 Posts HOW TO BYPASS RSA DIFFICULTY... not! HOW TO BYPASS RSA DIFFICULTY N=p*q and p+q-4 = 0 (mod 8) and x > 2 (3*N-1)/8=3*x*(x+1)/2-3*y*(y-1)/2+(3*x+1)*(3*x+2)/2 where p=[2*(3*x+1-(x-y+1))+1-(4*y-2)] and q=[2*(3*x+1-(x-y+1))+1] BYPASS in this way y=1/2*(sqrt(4*m^2-8*((3*N-1)/8-1)/3-131)+1)=(Z-1)/2 solve Z^2-4*m^2=-8*((3*N-1)/8-1)/3-131 4*m^2-Z^2=8*((3*N-1)/8-1)/3+131 EXAMPLE N=187 4*m^2-Z^2=315 factorize 315=3^2*5*7=15*21 2*m=(15+21)/2 Z=(21-15)/2 -> m=9 Z=3 y=(Z-1)/2=1 x=3 p=11 q=17  2020-04-28, 13:42 #2 Alberico Lepore   May 2017 ITALY 5×97 Posts in general RSA is of four types 2 types N = 4 * G + 1 2 types N = 4 * H + 3 of the latter 2 types we are interested in p + q-4 = 0 (mod 8) now to go from 4 * G + 1 to 4 * H + 3 just multiply by 3 to pass from p + q = 0 (mod 8) to p + q-4 = 0 (mod 8) just multiply by 5 so we will have two 2 ^ (to something) now you can try to factor the new generated number 8 * ((3 * N-1) / 8-1) / 3 + 131 or try to factor in for example the first P 1000 primes to the remaining number if of form 4 * G + 1 multiply also by 3 therefore we will have 1000 * [2 ^ (log in base r)] where r is a sort of coefficient that goes according to the number N and the choice of P  2020-04-28, 14:39   #3
VBCurtis

"Curtis"
Feb 2005
Riverside, CA

2×2,393 Posts Quote:
 Originally Posted by Alberico Lepore EXAMPLE N=187
How is your method faster than trial division? I can factor 187 by dividing by 2,3,5,7,11 and presto! I'm finished. Or I could start at the sqrt of 187 rounded down: divide by 13, then 11. Presto! Two steps.

Hint It's not. But if you're really clever, you might stumble onto a quadratic sieve. You might try reading about that one, rather than wandering in the dark like you do.  2020-04-29, 16:28   #4
Alberico Lepore

May 2017
ITALY

5×97 Posts Quote:
 Originally Posted by VBCurtis How is your method faster than trial division? I can factor 187 by dividing by 2,3,5,7,11 and presto! I'm finished. Or I could start at the sqrt of 187 rounded down: divide by 13, then 11. Presto! Two steps. Hint It's not. But if you're really clever, you might stumble onto a quadratic sieve. You might try reading about that one, rather than wandering in the dark like you do.
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