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Old 2020-08-18, 08:06   #23
Alberico Lepore
 
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8,14,2
Let's see who succeeds?
I say that RSA is factored into O (log)
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Old 2020-08-18, 08:10   #24
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Quote:
Originally Posted by Alberico Lepore View Post
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.
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Old 2020-08-18, 08:19   #25
Alberico Lepore
 
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Quote:
Originally Posted by retina View Post
Well go on then. Show us how you factor a larger RSA number.

in seven days.
upon expiry of this survey

https://twitter.com/albericolep/stat...35758427111426
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Old 2020-08-18, 10:35   #26
Alberico Lepore
 
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Quote:
Originally Posted by Alberico Lepore View Post
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
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Old 2020-08-18, 10:38   #27
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Quote:
Originally Posted by Alberico Lepore View Post
N=259


Something bigger. A lot bigger.
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Old 2020-08-18, 12:00   #28
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Quote:
Originally Posted by Alberico Lepore View Post
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
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Old 2020-08-18, 14:10   #29
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Quote:
Originally Posted by mathwiz View Post
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.
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Old 2020-08-18, 14:20   #30
Alberico Lepore
 
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Quote:
Originally Posted by mathwiz View Post
Would you mind factoring for me:

Code:
135066410865995223349603216278805969938881475605667027524485143851526510604859533833940287150571909441798207282164471551373680419703964191743046496589274256239341020864383202110372958725762358509643110564073501508187510676594629205563685529475213500852879416377328533906109750544334999811150056977236890927563
Thanks!
Quote:
Originally Posted by Dr Sardonicus View Post
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
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Old 2020-08-18, 15:04   #31
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Quote:
Originally Posted by Dr Sardonicus View Post
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.
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Old 2020-08-18, 15:05   #32
Alberico Lepore
 
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Quote:
Originally Posted by Alberico Lepore View Post
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
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Old 2020-08-18, 15:32   #33
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Quote:
Originally Posted by Alberico Lepore View Post
What do you think?
I think you're a bot.
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