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Old 2022-01-01, 15:04   #1
Alberico Lepore
 
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Default WG FACTORIZAZION in polynomial time

Happy New Year !

WG FACTORIZAZION in polynomial time

Languages [ITA-MATH]
https://www.academia.edu/66788027/Fattorizzazione_wg

What do you think?
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Old 2022-01-03, 14:02   #2
Alberico Lepore
 
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Quote:
Originally Posted by Viliam Furik View Post
Maybe if you provided an English text with some example of usage, preferably not located at an external link, that would be nice.
the mathematical procedure is correct.
Only that it is computationally impossible to find a suitable B.
I am already studying another method which, as you suggested, I will write in English and show the example.
Sorry for the inconvenience.
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Old 2022-01-03, 14:21   #3
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more less I'm studying this:



to factor N = 27 * 65

you have to choose (65-p) mod 8 = 0

and

you have to choose (q-27) mod 8 = 0

suppose we choose 41 and 43

41 * 43 = 1763

the following W and w are in the form

W = 65 * n + (1763 - 27 * 65) / 8

w = 27 * m + (1763 - 27 * 65) / 8



(1763-3) / 8 = 220


220 - W- [4- (65-7) * (65-5) / 8] = 65 * X

W = - (65 * n + 1) = q * (65-p) / 8, p * q = 1763

q = 27-8 * n


220 - w- [4- (27-7) * (27-5) / 8] = 27 * X

w = (27 * m + 1) = p * (q-27) / 8, p * q = 1763

p = 65-8 * m


Later I test if binary search can work

Last fiddled with by Alberico Lepore on 2022-01-03 at 15:08 Reason: update
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Old 2022-01-05, 13:20   #4
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Quote:
Originally Posted by Alberico Lepore View Post
Happy New Year !

WG FACTORIZAZION in polynomial time

Languages [ITA-MATH]
https://www.academia.edu/66788027/Fattorizzazione_wg

What do you think?
mi è venuta un IDEA

N*25^F=(a*5^F)*(b*5^F)

scegliere B != 5*J


quando arriveremo alla forma

(t^2+u*t+v) mod (B^2) = 0


t=n*(a*5^F)

Z=n*a

25^F*Z^2 +5^F*Z+v mod (B^2)

se v=5^J

riformuliamo il tutto

(t^2+u*t+v) mod (5^J*B^2) = 0

e vediamo se rientra in coppermisth method
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Old 2022-01-06, 15:18   #5
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Quote:
Originally Posted by Alberico Lepore View Post
mi è venuta un IDEA

PROOF of factorizazion in polynomial time [8 digit]

reference

PART I
https://www.academia.edu/48848013/Le...rization_nr_88
&
https://www.academia.edu/66788027/Fattorizzazione_wg


Supponiamo di voler fattorizzare N=9967*6781=67586227

A=sqrt(N) B=2*sqrt(N/2)

log_5(2*sqrt(N/2))=log_5(11631)=6

67586227*25^6=16500543701171875

A=sqrt(N) B=2*sqrt(N/2)*25^6

C=A D=(B+8)*25^6


Scegliamo
A=8821 B=11631*25^6
C=8821 B=11639*25^6


solve
N=16500543701171875
,
M=(8821)*11631*25^6
,
H=(8821)*11639*25^6
,
(M-3)/8-(a*n+(M-N)/8)-[4-(a-7)*(a-5)/8]=a*(a+11631*25^6-12)/8
,
(H-3)/8-(a*m+(H-N)/8)-[4-(a-7)*(a-5)/8]=a*(a+11639*25^6-12)/8

->m=n-244140625


solve
N=16500543701171875
,
M=(8821)*11631*25^6
,
H=(8821)*11639*25^6
,
(M-3)/8-(a*n+(M-N)/8)-[4-(a-7)*(a-5)/8]=a*(a+11631*25^6-12)/8
,
(a*n+(M-N)/8)*(a*(n-244140625)+(H-N)/8)=X
,
a*n=t
,
a,X

->

11631^2*25^12*X=t^2+2136891113281250*t+1141575907505095005035400390625

Use Coppersmith method

PROOF

t=6781*(9967*25^6-11631*25^6)/8

-t=344347656250000<11631^2*25^12
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