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Old 2020-02-10, 09:05   #1
Alberico Lepore
 
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Default 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
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Old 2020-02-10, 10:04   #2
LaurV
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Quote:
Originally Posted by Alberico Lepore View Post
which has 16 solutions to establish the infinite solutions
This sounds so garbage... (even without reading the rest of the text)
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Old 2020-02-10, 10:10   #3
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Quote:
Originally Posted by LaurV View Post
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
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Old 2020-02-10, 10:14   #4
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The only error is

Quote:
Originally Posted by Alberico Lepore View Post

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)
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Old 2020-02-10, 10:31   #5
retina
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Quote:
Originally Posted by Alberico Lepore View Post
Example N=91
Still using those tiny numbers. Boring.
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Old 2020-02-11, 08:26   #6
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Quote:
Originally Posted by retina View Post
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
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Old 2020-02-11, 16:29   #7
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The RSA secret is revealed by this image
Attached Thumbnails
Click image for larger version

Name:	RSA_SOLUTION.jpg
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Old 2020-02-11, 18:08   #8
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Outstanding. Now just make the x and y axes of that chart 150 orders of magnitude larger.
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Old 2020-02-13, 06:43   #9
LaurV
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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
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Old 2020-02-14, 19:02   #10
Alberico Lepore
 
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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
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Old 2020-02-14, 20:12   #11
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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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