20191211, 15:35  #1 
May 2017
ITALY
2×3^{2}×29 Posts 
RSA numerology in O(log n)
"free math in free world"
RSA factoring in O(log n) if N=p*q && (qp) mod 8 == 0 W=(N1)/8 ed n=qp W+[1+(n8)/8*(n16)/8/2n/8*(p3)/2]=M , W[(p1)/2+n/4]+[1+(n8)/8*(n16)/8/2n/8*((p2)3)/2]=B , p^2+n*p=8*46+1,MB=J you value n and observe J Example N=369 > M=46 46+[1+(n8)/8*(n16)/8/2n/8*(p3)/2]=M , 46[(p1)/2+n/4]+[1+(n8)/8*(n16)/8/2n/8*((p2)3)/2]=B , p^2+n*p=8*46+1,MB=J , n=8 > J1=8,3107 46+[1+(n8)/8*(n16)/8/2n/8*(p3)/2]=M , 46[(p1)/2+n/4]+[1+(n8)/8*(n16)/8/2n/8*((p2)3)/2]=B , p^2+n*p=8*46+1,MB=J , n=16 > J2=7,9043 46+[1+(n8)/8*(n16)/8/2n/8*(p3)/2]=M , 46[(p1)/2+n/4]+[1+(n8)/8*(n16)/8/2n/8*((p2)3)/2]=B , p^2+n*p=8*46+1,MB=J , n=24 > J3=7,8247 46+[1+(n8)/8*(n16)/8/2n/8*(p3)/2]=M , 46[(p1)/2+n/4]+[1+(n8)/8*(n16)/8/2n/8*((p2)3)/2]=B , p^2+n*p=8*46+1,MB=J , n=32 > J4=8 && p=9 46+[1+(n8)/8*(n16)/8/2n/8*(p3)/2]=M , 46[(p1)/2+n/4]+[1+(n8)/8*(n16)/8/2n/8*((p2)3)/2]=B , p^2+n*p=8*46+1,MB=J , n=40 > J5=8,3654 46+[1+(n8)/8*(n16)/8/2n/8*(p3)/2]=M , 46[(p1)/2+n/4]+[1+(n8)/8*(n16)/8/2n/8*((p2)3)/2]=B , p^2+n*p=8*46+1,MB=J , n=48 > J6=8,8704 46+[1+(n8)/8*(n16)/8/2n/8*(p3)/2]=M , 46[(p1)/2+n/4]+[1+(n8)/8*(n16)/8/2n/8*((p2)3)/2]=B , p^2+n*p=8*46+1,MB=J , n=56 > J7=9,4779 J1>J2>J3 J5<J6<J7 what do you think? 
20191211, 15:43  #2 
Mar 2019
3×101 Posts 
Please factor RSA1024 using your technique. Thanks!

20191211, 15:53  #3 
May 2017
ITALY
2·3^{2}·29 Posts 

20191211, 15:53  #4 
"Ben"
Feb 2007
E60_{16} Posts 
Huh. "Factoring" N by doing math with its factors. And a totally fictional bigO. Title changed appropriately.

20191211, 15:56  #5 
May 2017
ITALY
2×3^{2}×29 Posts 

20191211, 16:06  #6 
"Ben"
Feb 2007
3680_{10} Posts 

20191212, 08:41  #7 
May 2017
ITALY
20A_{16} Posts 
What features should N have for this method to work?

20191212, 09:37  #8 
Mar 2019
3·101 Posts 

20191212, 09:39  #9 
May 2017
ITALY
1000001010_{2} Posts 

20191212, 16:01  #10 
Aug 2006
5,987 Posts 

20191212, 16:27  #11 
May 2017
ITALY
2·3^{2}·29 Posts 

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