20200818, 08:06  #23 
May 2017
ITALY
3×5×31 Posts 
see photo
8,14,2 Let's see who succeeds? I say that RSA is factored into O (log) 
20200818, 08:10  #24 
Undefined
"The unspeakable one"
Jun 2006
My evil lair
3^{2}×643 Posts 

20200818, 08:19  #25 
May 2017
ITALY
465_{10} Posts 

20200818, 10:35  #26  
May 2017
ITALY
465_{10} Posts 
Quote:
I show you 1 of 9 cases H=1/3*(8*a+1) && (N1)/6 mod 7 =1 [[[2*((N1)/6)^2(N1)/6] mod N]+5]/8  (H+5)/8=X <> (6*a+1) divide X Example N=259 15(1/3*(8*a+1)+5)/8=X 43a=3*X Generalized Euclidean algorithm > a=3*n+1 ; X=14n replacement (43a)=b*(6*a+1) (423*n)=b*(6*(3*n+1)+1) let's take advantage of this (N1)/6 mod 7 =1 N=7*(6b)/(3*(6*b+1)) > (6b)/(6*b+1) > (366*b)/(6*b+1) > (6*b+1) divide 37 GCD(259,37)=37=q 

20200818, 10:38  #27 
Undefined
"The unspeakable one"
Jun 2006
My evil lair
3^{2}×643 Posts 

20200818, 12:00  #28 
Mar 2019
2·7^{2} Posts 
Would you mind factoring for me:
Code:
135066410865995223349603216278805969938881475605667027524485143851526510604859533833940287150571909441798207282164471551373680419703964191743046496589274256239341020864383202110372958725762358509643110564073501508187510676594629205563685529475213500852879416377328533906109750544334999811150056977236890927563 Last fiddled with by bsquared on 20200818 at 13:29 Reason: code tags 
20200818, 14:10  #29  
Feb 2017
Nowhere
3×7×13^{2} Posts 
Quote:
I expect Alberico to have it factored within the hour. 

20200818, 14:20  #30  
May 2017
ITALY
3×5×31 Posts 
Quote:
Quote:
It is more complex than I expected In particular to find the general solution to case 1 of 9 you have to study this [[[2*((N1)/6)^2(N1)/6] mod N]+5]/8(1/3*(8*a+1)+5)/8+c*(6*a+1)=X , a=3*n+1 > X=(c*181)*n+(7*c+M1) where M=[[[2*((N1)/6)^2(N1)/6] mod N]+5]/8 and (c*181) and (7*c+M1) they must have one factor in common so i thought of a little bruteforce [but ....] (c*181)=s*z , (7*c+M1)=t*z Example N=25*55 77(1/3*(8*a+1)+5)/8+2*(6*a+1)=X , a=3*n+1 X=(35*n+90)> 7*n+18 18*(7*n+18)7*(6*(3*n+1)+1)=275 GCD(275,1375)=25 Last fiddled with by Alberico Lepore on 20200818 at 14:21 

20200818, 15:04  #31  
"Ben"
Feb 2007
3294_{10} Posts 
Quote:


20200818, 15:05  #32  
May 2017
ITALY
1D1_{16} Posts 
Quote:
I think I'll just do a bruteforce on odd z starting from 3 and then Generalized Euclid's algorithm. the problem is that I will have to learn the generalized Euclid algorithm so it will take at least a week. then I'll have to see another 8 cases and then I'll have to see this bruteforce what it brings me In the meantime, you, more experienced than me, could test it. What do you think? Last fiddled with by Alberico Lepore on 20200818 at 15:08 Reason: quote 

20200818, 15:32  #33 
Feb 2017
Nowhere
3·7·13^{2} Posts 

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