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Old 2019-12-11, 15:35   #1
Alberico Lepore
 
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Default RSA numerology in O(log n)

"free math in free world"


RSA factoring in O(log n)

if N=p*q && (q-p) mod 8 == 0

W=(N-1)/8 ed n=q-p


W+[-1+(n-8)/8*(n-16)/8/2-n/8*(p-3)/2]=M , W-[(p-1)/2+n/4]+[-1+(n-8)/8*(n-16)/8/2-n/8*((p-2)-3)/2]=B , p^2+n*p=8*46+1,M-B=J

you value n and observe J

Example

N=369 -> M=46


46+[-1+(n-8)/8*(n-16)/8/2-n/8*(p-3)/2]=M , 46-[(p-1)/2+n/4]+[-1+(n-8)/8*(n-16)/8/2-n/8*((p-2)-3)/2]=B , p^2+n*p=8*46+1,M-B=J , n=8
-> J1=8,3107

46+[-1+(n-8)/8*(n-16)/8/2-n/8*(p-3)/2]=M , 46-[(p-1)/2+n/4]+[-1+(n-8)/8*(n-16)/8/2-n/8*((p-2)-3)/2]=B , p^2+n*p=8*46+1,M-B=J , n=16
-> J2=7,9043

46+[-1+(n-8)/8*(n-16)/8/2-n/8*(p-3)/2]=M , 46-[(p-1)/2+n/4]+[-1+(n-8)/8*(n-16)/8/2-n/8*((p-2)-3)/2]=B , p^2+n*p=8*46+1,M-B=J , n=24
-> J3=7,8247

46+[-1+(n-8)/8*(n-16)/8/2-n/8*(p-3)/2]=M , 46-[(p-1)/2+n/4]+[-1+(n-8)/8*(n-16)/8/2-n/8*((p-2)-3)/2]=B , p^2+n*p=8*46+1,M-B=J , n=32
-> J4=8 && p=9

46+[-1+(n-8)/8*(n-16)/8/2-n/8*(p-3)/2]=M , 46-[(p-1)/2+n/4]+[-1+(n-8)/8*(n-16)/8/2-n/8*((p-2)-3)/2]=B , p^2+n*p=8*46+1,M-B=J , n=40
-> J5=8,3654

46+[-1+(n-8)/8*(n-16)/8/2-n/8*(p-3)/2]=M , 46-[(p-1)/2+n/4]+[-1+(n-8)/8*(n-16)/8/2-n/8*((p-2)-3)/2]=B , p^2+n*p=8*46+1,M-B=J , n=48
-> J6=8,8704

46+[-1+(n-8)/8*(n-16)/8/2-n/8*(p-3)/2]=M , 46-[(p-1)/2+n/4]+[-1+(n-8)/8*(n-16)/8/2-n/8*((p-2)-3)/2]=B , p^2+n*p=8*46+1,M-B=J , n=56
-> J7=9,4779

J1>J2>J3

J5<J6<J7

what do you think?
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Old 2019-12-11, 15:43   #2
mathwiz
 
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Please factor RSA-1024 using your technique. Thanks!
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Old 2019-12-11, 15:53   #3
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Quote:
Originally Posted by mathwiz View Post
Please factor RSA-1024 using your technique. Thanks!
I'm not a programmer and not even a mathematician to tell the truth.
I'm trying to learn
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Old 2019-12-11, 15:53   #4
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Huh. "Factoring" N by doing math with its factors. And a totally fictional big-O. Title changed appropriately.
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Old 2019-12-11, 15:56   #5
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Quote:
Originally Posted by bsquared View Post
Huh. "Factoring" N by doing math with its factors. And a totally fictional big-O. Title changed appropriately.
Do you mean that it can turn into O (1) or that it doesn't work?
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Old 2019-12-11, 16:06   #6
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Quote:
Originally Posted by Alberico Lepore View Post
Do you mean that it can turn into O (1) or that it doesn't work?
Sure. You must already know the factors to run through the steps of the method, so, Bob's your uncle, you're done!
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Old 2019-12-12, 08:41   #7
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What features should N have for this method to work?
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Old 2019-12-12, 09:37   #8
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Quote:
Originally Posted by Alberico Lepore View Post
What features should N have for this method to work?
What is your method actually trying to accomplish?

"Factoring" N means you ONLY know N, and you are trying to determine p and q.
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Old 2019-12-12, 09:39   #9
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Quote:
Originally Posted by mathwiz View Post
What is your method actually trying to accomplish?

"Factoring" N means you ONLY know N, and you are trying to determine p and q.
Yes
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Old 2019-12-12, 16:01   #10
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Quote:
Originally Posted by Alberico Lepore View Post
Yes
But your method uses p and/or q in the definition of n, M (via n), B, and J.
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Old 2019-12-12, 16:27   #11
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Quote:
Originally Posted by CRGreathouse View Post
But your method uses p and/or q in the definition of n, M (via n), B, and J.
In the method described in O (log n)
we assign to n values 8 * i and 8 * (i + 1) and we confornt the J
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