
View Poll Results: Works or it does not work ?  
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20220101, 15:04  #1 
May 2017
ITALY
2^{3}·5·13 Posts 
WG FACTORIZAZION in polynomial time
Happy New Year !
WG FACTORIZAZION in polynomial time Languages [ITAMATH] https://www.academia.edu/66788027/Fattorizzazione_wg What do you think? 
20220103, 14:02  #2  
May 2017
ITALY
2^{3}×5×13 Posts 
Quote:
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. 

20220103, 14:21  #3 
May 2017
ITALY
2^{3}·5·13 Posts 
more less I'm studying this:
to factor N = 27 * 65 you have to choose (65p) mod 8 = 0 and you have to choose (q27) 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 (17633) / 8 = 220 220  W [4 (657) * (655) / 8] = 65 * X W =  (65 * n + 1) = q * (65p) / 8, p * q = 1763 q = 278 * n 220  w [4 (277) * (275) / 8] = 27 * X w = (27 * m + 1) = p * (q27) / 8, p * q = 1763 p = 658 * m Later I test if binary search can work Last fiddled with by Alberico Lepore on 20220103 at 15:08 Reason: update 
20220105, 13:20  #4  
May 2017
ITALY
2^{3}·5·13 Posts 
Quote:
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 

20220106, 15:18  #5 
May 2017
ITALY
208_{16} Posts 
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 , (M3)/8(a*n+(MN)/8)[4(a7)*(a5)/8]=a*(a+11631*25^612)/8 , (H3)/8(a*m+(HN)/8)[4(a7)*(a5)/8]=a*(a+11639*25^612)/8 >m=n244140625 solve N=16500543701171875 , M=(8821)*11631*25^6 , H=(8821)*11639*25^6 , (M3)/8(a*n+(MN)/8)[4(a7)*(a5)/8]=a*(a+11631*25^612)/8 , (a*n+(MN)/8)*(a*(n244140625)+(HN)/8)=X , a*n=t , a,X > 11631^2*25^12*X=t^2+2136891113281250*t+1141575907505095005035400390625 Use Coppersmith method PROOF t=6781*(9967*25^611631*25^6)/8 t=344347656250000<11631^2*25^12 
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