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 2022-01-01, 15:04 #1 Alberico Lepore     May 2017 ITALY 2×32×29 Posts 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?
2022-01-03, 14:02   #2
Alberico Lepore

May 2017
ITALY

2·32·29 Posts

Quote:
 Originally Posted by Viliam Furik 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.

 2022-01-03, 14:21 #3 Alberico Lepore     May 2017 ITALY 2·32·29 Posts 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
2022-01-05, 13:20   #4
Alberico Lepore

May 2017
ITALY

2·32·29 Posts

Quote:
 Originally Posted by Alberico Lepore 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

2022-01-06, 15:18   #5
Alberico Lepore

May 2017
ITALY

2×32×29 Posts

Quote:
 Originally Posted by Alberico Lepore mi è venuta un IDEA

PROOF of factorizazion in polynomial time [8 digit]

reference

PART I
&

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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