Cooperazione per la fattorizzazione - Page 12

Alberico Lepore · 2019-05-17, 18:35

the solution in O (log) is too simple

N=377

solve

(377/p+p)/2=Q , sqrt[(Q-1)^2/4-(Q-1)+1]=A ,sqrt[(A-1)^2/4-(A-1)+1]=B , (B-1)^2/4=1

explanation

Q=21

(Q-1)^2/4=100

sqrt[(Q-1)^2/4-(Q-1)+1]=9

Q=9

(Q-1)^2/4=16

sqrt[(Q-1)^2/4-(Q-1)+1]=3

Q=3

(Q-1)^2/4=1

I wanted to find in O (1)

so it will be for next time

sorry

CRGreathouse · 2019-05-18, 00:34

Quote:

Originally Posted by Alberico Lepore

the solution in O (log) is too simple

[...]

so it will be for next time

sorry

I'm not sure what you're saying, but let me know if you address my question

Quote:

Originally Posted by CRGreathouse

Are you saying this method can always find a factor of the form 4n + 1 (when one exists), or just in easy cases?

or take up my challenge

Quote:

Originally Posted by CRGreathouse

How about this: I generate 100 numbers, then you use your method to find the factorization of one of those numbers. I promise that each one will have at least one factor that is 1 mod 4. We can do a practice run at 100 digits (which you could quickly factor with other methods, but at least it lets us set up the problem) and then a real run at 200 digits (which should let you show your method's ability, since this is hard with general methods). Pass those and I'd be happy to help with analysis.

Alberico Lepore · 2019-05-19, 14:53

if N=4*G+1

Q^2-(n/2)^2=N
,
(n-6)-n/2=A
,
(Q^2-N)-(A/2)^2=X
,
(A-6)-(A/2-1)=B
,
[((n/2)^2)-X]-(B/2)^2=Y
,
(B-6)-B/2=C
,
[((A/2)^2)-Y]-(C/2)^2=Z
,
(C-6)-(C/2-1)=D
,
.........
.........
.........

if N=4*G+3

Q^2-(n/2)^2=N
,
(n-6)-(n/2-1)=A
,
(Q^2-N)-(A/2)^2=X
,
(A-6)-A/2=B
,
[((n/2)^2)-X]-(B/2)^2=Y
,
(B-6)-(B/2-1)=C
,
[((A/2)^2)-Y]-(C/2)^2=Z
,
(C-6)-C/2=D
........
........
........

hey
last two questions:
to whom should I compare A, B, C, D, .......?
what is the computational complexity?

Alberico Lepore · 2019-05-19, 20:34

I found the first solution on 23 September 2017

https://www.academia.edu/34655158/Tr...ne_in_un_gioco

x=(y+1)*(y+1-2)-(y+1-2)^2

2 =(2+1)*(2+1-2)-(2+1-2)^2
4 =(3+1)*(3+1-2)-(3+1-2)^2
6 =(4+1)*(4+1-2)-(4+1-2)^2
8 =(5+1)*(5+1-2)-(5+1-2)^2
10=(6+1)*(6+1-2)-(6+1-2)^2
12=(7+1)*(7+1-2)-(7+1-2)^2
.....
.....
.....
.....

23 September 2017

LaurV · 2019-05-20, 03:03

Your main problem is that you think this is "un gioco"...

Try factoring some 100 digits number with your method. In fact, we would be happy with some 50-70 digits too..

Alberico Lepore · 2019-05-20, 22:18

I show you one of the solutions in O(256 * (log_32 () ))

x=(y+1)*(y+1-2)-(y+1-2)^2 A=(y+1)+(y+1-2)+(y+1-2)

2 =(2+1)*(3-2)-(3-2)^2
4 =(3+1)*(4-2)-(4-2)^2
6 =(4+1)*(5-2)-(5-2)^2
8 =(5+1)*(6-2)-(6-2)^2
10=(6+1)*(7-2)-(7-2)^2
12=(7+1)*(8-2)-(8-2)^2
14=(8+1)*(9-2)-(9-2)^2
16=(9+1)*(10-2)-(10-2)^2
18=(10+1)*(11-2)-(11-2)^2
20=(11+1)*(12-2)-(12-2)^2
22=(12+1)*(13-2)-(13-2)^2
24=(13+1)*(14-2)-(14-2)^2
.....
.....

search p=256*G+3 and z mod 32 = 0

p^2+n*p=N , (p-3)/2=(x+1)*(x+1-2)-(x+1-2)^2 , (p-11)/2=(y+1)*(y+1-2)-(y+1-2)^2 ,(x+y)/4=z

example

p^2+n*p=66051^2 , (p-3)/2=(x+1)*(x+1-2)-(x+1-2)^2 , (p-11)/2=(y+1)*(y+1-2)-(y+1-2)^2 ,(x+y)/4=z , p=66051

to move from one p to the other p-2*k you do so N-(2*k)^2-2*k*n

(p-2*k)^2+(n+4*k)*(p-2*k)=N-(2*k)^2-(2*k*n) , (p-2*k-3)/2=(x+1)*(x+1-2)-(x+1-2)^2 , (p-2*k-11)/2=(y+1)*(y+1-2)-(y+1-2)^2 ,(x+y)/4=z

example

(p-2*1)^2+(n+4*1)*(p-2*1)=66053^2-(2*1)^2-(2*1*n) , (p-2*1-3)/2=(x+1)*(x+1-2)-(x+1-2)^2 , (p-2*1-11)/2=(y+1)*(y+1-2)-(y+1-2)^2 ,(x+y)/4=z , p=66053

what do you think?

VBCurtis · 2019-05-20, 22:36

What number did this factor? How much more work would it take if you doubled the number of digits?

LaurV · 2019-05-21, 03:39

Quote:

Originally Posted by Alberico Lepore

what do you think?

We think you need treatment, at least against some trolling syndrome (we can't imagine that it is something else, any other sane person in the world, over a certain age, like 10 years old or so, would have understood up to now, after so much talk in these threads, and after so many hints from people who know what they are talking about).

This is even not misc math, and we will propose on the mod's board to limit your posting to the blog area in the future.

Alberico Lepore · 2019-07-31, 15:35

a*b=N=4*Z+1
,
2*sqrt[a*b-(b-a)*a)]=A
,
2*((A-v)*v)+2*v^2+((A-v-v)/2)^2=B^2
,
2*((B-w)*w)+2*w^2+((B-w-w)/2)^2=C^2
,
2*((C-z)*z)+2*z^2+((C-z-z)/2)^2=D^2
,
........
.........

putting in place of v, w, z, x, y, ....... once 3 once 5
and testing A, B, C, D, E, F, ...... = 10

RSA is GAME OVER

2019-05-17, 18:35	#122
Alberico Lepore May 2017 ITALY 2·11·13 Posts	the solution in O (log) is too simple N=377 solve (377/p+p)/2=Q , sqrt[(Q-1)^2/4-(Q-1)+1]=A ,sqrt[(A-1)^2/4-(A-1)+1]=B , (B-1)^2/4=1 explanation Q=21 (Q-1)^2/4=100 sqrt[(Q-1)^2/4-(Q-1)+1]=9 Q=9 (Q-1)^2/4=16 sqrt[(Q-1)^2/4-(Q-1)+1]=3 Q=3 (Q-1)^2/4=1 I wanted to find in O (1) so it will be for next time sorry Last fiddled with by Alberico Lepore on 2019-05-17 at 18:59

2019-05-19, 14:53	#124
Alberico Lepore May 2017 ITALY 2·11·13 Posts	if N=4G+1 Q^2-(n/2)^2=N , (n-6)-n/2=A , (Q^2-N)-(A/2)^2=X , (A-6)-(A/2-1)=B , [((n/2)^2)-X]-(B/2)^2=Y , (B-6)-B/2=C , [((A/2)^2)-Y]-(C/2)^2=Z , (C-6)-(C/2-1)=D , ......... ......... ......... if N=4G+3 Q^2-(n/2)^2=N , (n-6)-(n/2-1)=A , (Q^2-N)-(A/2)^2=X , (A-6)-A/2=B , [((n/2)^2)-X]-(B/2)^2=Y , (B-6)-(B/2-1)=C , [((A/2)^2)-Y]-(C/2)^2=Z , (C-6)-C/2=D ........ ........ ........ hey last two questions: to whom should I compare A, B, C, D, .......? what is the computational complexity? Last fiddled with by Alberico Lepore on 2019-05-19 at 15:19 Reason: correction

2019-05-19, 20:34	#125
Alberico Lepore May 2017 ITALY 2·11·13 Posts	I found the first solution on 23 September 2017 https://www.academia.edu/34655158/Tr...ne_in_un_gioco x=(y+1)(y+1-2)-(y+1-2)^2 2 =(2+1)(2+1-2)-(2+1-2)^2 4 =(3+1)(3+1-2)-(3+1-2)^2 6 =(4+1)(4+1-2)-(4+1-2)^2 8 =(5+1)(5+1-2)-(5+1-2)^2 10=(6+1)(6+1-2)-(6+1-2)^2 12=(7+1)*(7+1-2)-(7+1-2)^2 ..... ..... ..... ..... 23 September 2017

2019-05-20, 03:03	#126
LaurV Romulan Interpreter Jun 2011 Thailand 7³×23 Posts	Your main problem is that you think this is "un gioco"... Try factoring some 100 digits number with your method. In fact, we would be happy with some 50-70 digits too..

2019-05-20, 22:18	#127
Alberico Lepore May 2017 ITALY 436₈ Posts	I show you one of the solutions in O(256 * (log_32 () )) x=(y+1)(y+1-2)-(y+1-2)^2 A=(y+1)+(y+1-2)+(y+1-2) 2 =(2+1)(3-2)-(3-2)^2 4 =(3+1)(4-2)-(4-2)^2 6 =(4+1)(5-2)-(5-2)^2 8 =(5+1)(6-2)-(6-2)^2 10=(6+1)(7-2)-(7-2)^2 12=(7+1)(8-2)-(8-2)^2 14=(8+1)(9-2)-(9-2)^2 16=(9+1)(10-2)-(10-2)^2 18=(10+1)(11-2)-(11-2)^2 20=(11+1)(12-2)-(12-2)^2 22=(12+1)(13-2)-(13-2)^2 24=(13+1)(14-2)-(14-2)^2 ..... ..... search p=256G+3 and z mod 32 = 0 p^2+np=N , (p-3)/2=(x+1)(x+1-2)-(x+1-2)^2 , (p-11)/2=(y+1)(y+1-2)-(y+1-2)^2 ,(x+y)/4=z example p^2+np=66051^2 , (p-3)/2=(x+1)(x+1-2)-(x+1-2)^2 , (p-11)/2=(y+1)(y+1-2)-(y+1-2)^2 ,(x+y)/4=z , p=66051 to move from one p to the other p-2k you do so N-(2k)^2-2kn (p-2k)^2+(n+4k)(p-2k)=N-(2k)^2-(2kn) , (p-2k-3)/2=(x+1)(x+1-2)-(x+1-2)^2 , (p-2k-11)/2=(y+1)(y+1-2)-(y+1-2)^2 ,(x+y)/4=z example (p-21)^2+(n+41)(p-21)=66053^2-(21)^2-(21n) , (p-21-3)/2=(x+1)(x+1-2)-(x+1-2)^2 , (p-21-11)/2=(y+1)(y+1-2)-(y+1-2)^2 ,(x+y)/4=z , p=66053 what do you think?

2019-05-20, 22:36	#128
VBCurtis "Curtis" Feb 2005 Riverside, CA 3·29·41 Posts	What number did this factor? How much more work would it take if you doubled the number of digits?

2019-07-31, 15:35	#130
Alberico Lepore May 2017 ITALY 2×11×13 Posts	ab=N=4Z+1 , 2sqrt[ab-(b-a)a)]=A , 2((A-v)v)+2v^2+((A-v-v)/2)^2=B^2 , 2((B-w)w)+2w^2+((B-w-w)/2)^2=C^2 , 2((C-z)z)+2z^2+((C-z-z)/2)^2=D^2 , ........ ......... putting in place of v, w, z, x, y, ....... once 3 once 5 and testing A, B, C, D, E, F, ...... = 10 RSA is GAME OVER