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 2020-08-18, 08:06 #23 Alberico Lepore   May 2017 ITALY 52×19 Posts see photo 8,14,2 Let's see who succeeds? I say that RSA is factored into O (log) Attached Thumbnails
2020-08-18, 08:10   #24
retina
Undefined

"The unspeakable one"
Jun 2006
My evil lair

587910 Posts

Quote:
 Originally Posted by Alberico Lepore see photo 8,14,2 Let's see who succeeds? I say that RSA is factored into O (log)
Well go on then. Show us how you factor a larger RSA number.

2020-08-18, 08:19   #25
Alberico Lepore

May 2017
ITALY

52·19 Posts

Quote:
 Originally Posted by retina Well go on then. Show us how you factor a larger RSA number.

in seven days.
upon expiry of this survey

2020-08-18, 10:35   #26
Alberico Lepore

May 2017
ITALY

52×19 Posts

Quote:
 Originally Posted by Alberico Lepore see photo 8,14,2 Let's see who succeeds? I say that RSA is factored into O (log)
studying studying I found this

I show you
1 of 9 cases

H=1/3*(8*a+1) && (N-1)/6 mod 7 =1

[[[2*((N-1)/6)^2-(N-1)/6] mod N]+5]/8 - (H+5)/8=X <-> (6*a+1) divide X

Example

N=259

15-(1/3*(8*a+1)+5)/8=X

43-a=3*X
Generalized Euclidean algorithm
->

a=3*n+1 ; X=14-n

replacement

(43-a)=b*(6*a+1)

(42-3*n)=b*(6*(3*n+1)+1)

let's take advantage of this (N-1)/6 mod 7 =1

N=7*(6-b)/(3*(6*b+1))

-> (6-b)/(6*b+1) -> (36-6*b)/(6*b+1) -> (6*b+1) divide 37

GCD(259,37)=37=q

2020-08-18, 10:38   #27
retina
Undefined

"The unspeakable one"
Jun 2006
My evil lair

5,879 Posts

Quote:
 Originally Posted by Alberico Lepore N=259

Something bigger. A lot bigger.

2020-08-18, 12:00   #28
mathwiz

Mar 2019

127 Posts

Quote:
 Originally Posted by Alberico Lepore N=259
Would you mind factoring for me:

Code:
135066410865995223349603216278805969938881475605667027524485143851526510604859533833940287150571909441798207282164471551373680419703964191743046496589274256239341020864383202110372958725762358509643110564073501508187510676594629205563685529475213500852879416377328533906109750544334999811150056977236890927563
Thanks!

Last fiddled with by bsquared on 2020-08-18 at 13:29 Reason: code tags

2020-08-18, 14:10   #29
Dr Sardonicus

Feb 2017
Nowhere

3,797 Posts

Quote:
 Originally Posted by mathwiz Would you mind factoring for me: Code: 135066410865995223349603216278805969938881475605667027524485143851526510604859533833940287150571909441798207282164471551373680419703964191743046496589274256239341020864383202110372958725762358509643110564073501508187510676594629205563685529475213500852879416377328533906109750544334999811150056977236890927563 Thanks!
Interesting, a 1024-bit number. I verified that it is composite.

I expect Alberico to have it factored within the hour.

2020-08-18, 14:20   #30
Alberico Lepore

May 2017
ITALY

52×19 Posts

Quote:
 Originally Posted by mathwiz Would you mind factoring for me: Code: 135066410865995223349603216278805969938881475605667027524485143851526510604859533833940287150571909441798207282164471551373680419703964191743046496589274256239341020864383202110372958725762358509643110564073501508187510676594629205563685529475213500852879416377328533906109750544334999811150056977236890927563 Thanks!
Quote:
 Originally Posted by Dr Sardonicus Interesting, a 1024-bit number. I verified that it is composite. I expect Alberico to have it factored within the hour.
have patience with me

It is more complex than I expected

In particular to find the general solution to case 1 of 9

you have to study this

[[[2*((N-1)/6)^2-(N-1)/6] mod N]+5]/8-(1/3*(8*a+1)+5)/8+c*(6*a+1)=X , a=3*n+1 -> X=(c*18-1)*n+(7*c+M-1) where M=[[[2*((N-1)/6)^2-(N-1)/6] mod N]+5]/8

and

(c*18-1) and (7*c+M-1) they must have one factor in common

so i thought of a little bruteforce [but ....]

(c*18-1)=s*z
,
(7*c+M-1)=t*z

Example

N=25*55

77-(1/3*(8*a+1)+5)/8+2*(6*a+1)=X , a=3*n+1

X=(35*n+90)-> 7*n+18

18*(7*n+18)-7*(6*(3*n+1)+1)=275

GCD(275,1375)=25

Last fiddled with by Alberico Lepore on 2020-08-18 at 14:21

2020-08-18, 15:04   #31
bsquared

"Ben"
Feb 2007

13×257 Posts

Quote:
 Originally Posted by Dr Sardonicus Interesting, a 1024-bit number. I verified that it is composite. I expect Alberico to have it factored within the hour.
Well, he has been trying to factor N=91 since at least May of 2017. Now, apparently working on 1375 (?). That's 3 years for a 2 digit increase in input size. So, even with the conjectured O(log N) complexity of the various "methods" we are looking at 3.25 years * log(2^1024) / log(1375) ~1664 more years of analysis.

2020-08-18, 15:05   #32
Alberico Lepore

May 2017
ITALY

47510 Posts

Quote:
 Originally Posted by Alberico Lepore have patience with me It is more complex than I expected In particular to find the general solution to case 1 of 9 you have to study this [[[2*((N-1)/6)^2-(N-1)/6] mod N]+5]/8-(1/3*(8*a+1)+5)/8+c*(6*a+1)=X , a=3*n+1 -> X=(c*18-1)*n+(7*c+M-1) where M=[[[2*((N-1)/6)^2-(N-1)/6] mod N]+5]/8 and (c*18-1) and (7*c+M-1) they must have one factor in common so i thought of a little bruteforce [but ....] (c*18-1)=s*z , (7*c+M-1)=t*z Example N=25*55 77-(1/3*(8*a+1)+5)/8+2*(6*a+1)=X , a=3*n+1 X=(35*n+90)-> 7*n+18 18*(7*n+18)-7*(6*(3*n+1)+1)=275 GCD(275,1375)=25

I think I'll just do a bruteforce on odd z starting from 3 and then Generalized Euclid's algorithm.
the problem is that I will have to learn the generalized Euclid algorithm so it will take at least a week.
then I'll have to see another 8 cases
and then I'll have to see this bruteforce what it brings me

In the meantime, you, more experienced than me, could test it.

What do you think?

Last fiddled with by Alberico Lepore on 2020-08-18 at 15:08 Reason: quote

2020-08-18, 15:32   #33
Dr Sardonicus

Feb 2017
Nowhere

3,797 Posts

Quote:
 Originally Posted by Alberico Lepore What do you think?
I think you're a bot.

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