20200428, 09:01  #1 
May 2017
ITALY
5·97 Posts 
HOW TO BYPASS RSA DIFFICULTY... not!
HOW TO BYPASS RSA DIFFICULTY
N=p*q and p+q4 = 0 (mod 8) and x > 2 (3*N1)/8=3*x*(x+1)/23*y*(y1)/2+(3*x+1)*(3*x+2)/2 where p=[2*(3*x+1(xy+1))+1(4*y2)] and q=[2*(3*x+1(xy+1))+1] BYPASS in this way y=1/2*(sqrt(4*m^28*((3*N1)/81)/3131)+1)=(Z1)/2 solve Z^24*m^2=8*((3*N1)/81)/3131 4*m^2Z^2=8*((3*N1)/81)/3+131 EXAMPLE N=187 4*m^2Z^2=315 factorize 315=3^2*5*7=15*21 2*m=(15+21)/2 Z=(2115)/2 > m=9 Z=3 y=(Z1)/2=1 x=3 p=11 q=17 
20200428, 13:42  #2 
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 + q4 = 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 + q4 = 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 * N1) / 81) / 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 
20200428, 14:39  #3 
"Curtis"
Feb 2005
Riverside, CA
2×2,393 Posts 
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. 
20200429, 16:28  #4  
May 2017
ITALY
5×97 Posts 
Quote:


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