 mersenneforum.org RSA factorization... NOT
 Register FAQ Search Today's Posts Mark Forums Read  2020-02-10, 09:05 #1 Alberico Lepore   May 2017 ITALY 192 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

Jun 2011
Thailand

26×131 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

1011010012 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

5518 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

122708 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

1011010012 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 192 Posts The RSA secret is revealed by this image Attached Thumbnails   2020-02-11, 18:08 #8 bsquared   "Ben" Feb 2007 5×647 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   Jun 2011 Thailand 26×131 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 192 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 1111100111012 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).   Thread Tools Show Printable Version Email this Page Similar Threads Thread Thread Starter Forum Replies Last Post Robert Holmes Factoring 19 2010-11-08 18:46 kurtulmehtap Math 25 2010-09-12 14:13 dleclair NFSNET Discussion 1 2006-03-21 05:11 Wacky NFSNET Discussion 1 2006-03-20 23:43 Jeff Gilchrist NFSNET Discussion 7 2005-02-23 19:46

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