20200210, 09:05  #1 
May 2017
ITALY
111111100_{2} Posts 
RSA factorization... NOT
if N = p * q such that (p+q4) mod 8 = 0
then (M^2+2*(M3)*M1)/83*y*(y1)/2=(3*N1)/8 and (3*N1)/8=3*x*(x+1)/23*y*(y1)/2+(3*x+1)*(3*x+2)/2 where M=4*x+3 and qp =4*y2 Example N=91 (M^2+2*(M3)*M1)/83*y*(y1)/2=34 (M^2+2*(M3)*M1)/8y*(y1)/2=34+y*(y1) therefore solve integer (M^2+2*(M3)*M1)/8y*(y1)/2=H which has 16 solutions to establish the infinite solutions so let's say we try them all and get here H=4*(6*c^2+3*c8*d3*d) , M=8*c+3 ,y=8*d+2 H=34+(8*d+2)*(8*d+21)=4*(6*c^2+3*c8*d3*d) solve 34=3*x*(x+1)/23*y*(y1)/2+(3*x+1)*(3*x+2)/2 , y=8*d+2 , 8*c+3=4*x+3 , 34+(8*d+2)*(8*d+21)=4*(6*c^2+3*c8*d3*d) and obtain c=1 ,d=0 , x=2, y=2 
20200210, 10:04  #2 
Romulan Interpreter
"name field"
Jun 2011
Thailand
2·4,903 Posts 

20200210, 10:10  #3  
May 2017
ITALY
1FC_{16} Posts 
Quote:
M is definitely in the form 8 * c + 3 or 8 * c + 7 then giving y values 8 * d or 8 * d + 1 or .... or 8 * d + 7 there are 16 solutions 

20200210, 10:14  #4 
May 2017
ITALY
774_{8} Posts 

20200210, 10:31  #5 
Undefined
"The unspeakable one"
Jun 2006
My evil lair
14234_{8} Posts 

20200211, 08:26  #6 
May 2017
ITALY
2^{2}·127 Posts 
N=91
34=3*x*(x+1)/23*y*(y1)/2+(3*x+1)*(3*x+2)/2 , (3*(H1)1)/8+3*(2*(3*(H1)1)/82)/3*(2*(3*(H1)1)/82+3)/3/23*y*(y1)/2=34 , 2*[2*[3*x+1(xy+1)]+1](4*y2)=H then q=2*[3*x+1(xy+1)]+1 p=2*[3*x+1(xy+1)]+1(4*y2) Last fiddled with by Alberico Lepore on 20200211 at 08:36 
20200211, 16:29  #7 
May 2017
ITALY
508_{10} Posts 
The RSA secret is revealed by this image

20200211, 18:08  #8 
"Ben"
Feb 2007
3·1,193 Posts 
Outstanding. Now just make the x and y axes of that chart 150 orders of magnitude larger.

20200213, 06:43  #9 
Romulan Interpreter
"name field"
Jun 2011
Thailand
2·4,903 Posts 
RSA secret? ... hm.. not interested.
Please post an image with the NSA secret. Last fiddled with by LaurV on 20200213 at 06:43 
20200214, 19:02  #10 
May 2017
ITALY
2^{2}×127 Posts 
if N = p * q such that (p+q4) mod 8 = 0
if this system admits solutions [((3*N1)/81)/3+y*(y1)/21[4*x+4]*x/2]+4*x*(4*x+1)/2c=(3*N1)/8+3*y*(y1)/23 , [((3*N1)/81)/3+y*(y1)/214*x[4*(x1)+4]*(x1)/2]+4*(x1)*(4*(x1)+1)/2[c+2sqrt(8*c+8)]=(3*N1)/8+3*y*(y1)/2312*x , (3*N1)/8=3*x*(x+1)/23*y*(y1)/2+(3*x+1)*(3*x+2)/2 then q=2*[3*x+1(xy+1)]+1 please someone help me solve the system 
20200214, 20:12  #11 
"Curtis"
Feb 2005
Riverside, CA
11711_{8} Posts 
Why can't you solve your own systems? If you can't, you should conclude that your methods are far too complicated to be useful to anyone (even yourself, since you can't solve them).

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