mersenneforum.org RSA factorization... NOT
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 2020-02-10, 09:05 #1 Alberico Lepore     May 2017 ITALY 1111111002 Posts RSA factorization... NOT if N = p * q such that (p+q-4) mod 8 = 0 then (M^2+2*(M-3)*M-1)/8-3*y*(y-1)/2=(3*N-1)/8 and (3*N-1)/8=3*x*(x+1)/2-3*y*(y-1)/2+(3*x+1)*(3*x+2)/2 where M=4*x+3 and q-p =4*y-2 Example N=91 (M^2+2*(M-3)*M-1)/8-3*y*(y-1)/2=34 (M^2+2*(M-3)*M-1)/8-y*(y-1)/2=34+y*(y-1) therefore solve integer (M^2+2*(M-3)*M-1)/8-y*(y-1)/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*c-8*d-3*d) , M=8*c+3 ,y=8*d+2 H=34+(8*d+2)*(8*d+2-1)=4*(6*c^2+3*c-8*d-3*d) solve 34=3*x*(x+1)/2-3*y*(y-1)/2+(3*x+1)*(3*x+2)/2 , y=8*d+2 , 8*c+3=4*x+3 , 34+(8*d+2)*(8*d+2-1)=4*(6*c^2+3*c-8*d-3*d) and obtain c=1 ,d=0 , x=2, y=2
2020-02-10, 10:04   #2
LaurV
Romulan Interpreter

"name field"
Jun 2011
Thailand

2·4,903 Posts

Quote:
 Originally Posted by Alberico Lepore which has 16 solutions to establish the infinite solutions
This sounds so garbage... (even without reading the rest of the text)

2020-02-10, 10:10   #3
Alberico Lepore

May 2017
ITALY

1FC16 Posts

Quote:
 Originally Posted by LaurV This sounds so garbage... (even without reading the rest of the text)
https://www.wolframalpha.com/input/?...y-1%29%2F2%3DH

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

2020-02-10, 10:14   #4
Alberico Lepore

May 2017
ITALY

7748 Posts

The only error is

Quote:
 Originally Posted by Alberico Lepore H=4*(6*c^2+3*c-8*d-3*d)
correct is

H=4*(6*c^2+3*c-8*d^2-3*d)

2020-02-10, 10:31   #5
retina
Undefined

"The unspeakable one"
Jun 2006
My evil lair

142348 Posts

Quote:
 Originally Posted by Alberico Lepore Example N=91
Still using those tiny numbers. Boring.

2020-02-11, 08:26   #6
Alberico Lepore

May 2017
ITALY

22·127 Posts

Quote:
 Originally Posted by retina Still using those tiny numbers. Boring.
N=91

34=3*x*(x+1)/2-3*y*(y-1)/2+(3*x+1)*(3*x+2)/2
,
(3*(H-1)-1)/8+3*(2*(3*(H-1)-1)/8-2)/3*(2*(3*(H-1)-1)/8-2+3)/3/2-3*y*(y-1)/2=34
,
2*[2*[3*x+1-(x-y+1)]+1]-(4*y-2)=H

then

q=2*[3*x+1-(x-y+1)]+1
p=2*[3*x+1-(x-y+1)]+1-(4*y-2)

Last fiddled with by Alberico Lepore on 2020-02-11 at 08:36

 2020-02-11, 16:29 #7 Alberico Lepore     May 2017 ITALY 50810 Posts The RSA secret is revealed by this image Attached Thumbnails
 2020-02-11, 18:08 #8 bsquared     "Ben" Feb 2007 3·1,193 Posts Outstanding. Now just make the x and y axes of that chart 150 orders of magnitude larger.
 2020-02-13, 06:43 #9 LaurV 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 2020-02-13 at 06:43
 2020-02-14, 19:02 #10 Alberico Lepore     May 2017 ITALY 22×127 Posts if N = p * q such that (p+q-4) mod 8 = 0 if this system admits solutions [((3*N-1)/8-1)/3+y*(y-1)/2-1-[4*x+4]*x/2]+4*x*(4*x+1)/2-c=(3*N-1)/8+3*y*(y-1)/2-3 , [((3*N-1)/8-1)/3+y*(y-1)/2-1-4*x-[4*(x-1)+4]*(x-1)/2]+4*(x-1)*(4*(x-1)+1)/2-[c+2-sqrt(8*c+8)]=(3*N-1)/8+3*y*(y-1)/2-3-12*x , (3*N-1)/8=3*x*(x+1)/2-3*y*(y-1)/2+(3*x+1)*(3*x+2)/2 then q=2*[3*x+1-(x-y+1)]+1 please someone help me solve the system
 2020-02-14, 20:12 #11 VBCurtis     "Curtis" Feb 2005 Riverside, CA 117118 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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