Cooperazione per la fattorizzazione

[ET_]

Originally Posted by CRGreathouse

For a > 46, both of the floored expressions are 0, so you have 0 = 187/a - 1 which has a solution at a = 187. It suffices to check smaller cases individually:

check(a)=46\a*a==(46*a-187)\(a^2-4*a)*(a-4)+187\a-1;
for(a=1,3, if(check(a), print1(a, ", ")))
for(a=5,46, if(check(a), print1(a, ", ")))

I get a total of 11 solutions. Note that the expression is undefined at a = 4.

You revealed the solution too soon
That's what I meant when I said "tabulations".

[LaurV]

Originally Posted by ET_

You revealed the solution too soon
That's what I meant when I said "tabulations".

[rds] +1. Let the guy do his own shkit.. otherwise he will never learn and will continue indefinitely to pollute the forum with banalities, especially with the attention he got in the last time... [/rds]

[CRGreathouse]

If Alberico downloads and installs gp and runs the program I will consider my time well-spent as a learning exercise.

[ET_]

Originally Posted by CRGreathouse

If Alberico downloads and installs gp and runs the program I will consider my time well-spent as a learning exercise.

[Alberico Lepore]

Originally Posted by CRGreathouse

If Alberico downloads and installs gp and runs the program I will consider my time well-spent as a learning exercise.

Yes, I will do it in the future, but now I have my fight.




If N = p * q = 4 * G + 3 where p = 4 * h + 3 and q = 4 * k + 1

I have shown that

x^2 - 2*y^2 = - (N + 1) / 4 [Generalized Pell equation]

where y = (p + 1) / 2

[Alberico Lepore]

Originally Posted by Alberico Lepore

Yes, I will do it in the future, but now I have my fight.




If N = p * q = 4 * G + 3 where p = 4 * h + 3 and q = 4 * k + 1

I have shown that

x^2 - 2*y^2 = - (N + 1) / 4 [Generalized Pell equation]

where y = (p + 1) / 2

2*y^2-x^2=(N+1)/4
,
y=(p+1)/2
,
x=[p-[(m-6)/8+(p+1)/2]]
,
p^2+m*p=N


#RSA #GAMEOVER

[Alberico Lepore]

Originally Posted by Alberico Lepore

2*y^2-x^2=(N+1)/4
,
y=(p+1)/2
,
x=[p-[(m-6)/8+(p+1)/2]]
,
p^2+m*p=N


#RSA #GAMEOVER

FALSE

[Alberico Lepore]

Rossella - factorization algorithm in O(log_2)

this is a copy of

https://www.academia.edu/37471570/Ro...hm_in_O_log_2_

for mersenneforum friends

[Alberico Lepore]

correct is here

[Alberico Lepore]

Post mute

Factoring RSA in O(log N)

N=45

solve X+a*(a-1)-3*a^2=45-a^2 ,a^2+n*a=45, n=2
a=-7.7823 , X=97.7823
a=5.7823 , X=84.2177
[45-(5.7823)^2]/5.7823<[97.7823-(7.7823)^2]/7.7823

solve X+a*(a-1)-3*a^2=45-a^2 ,a^2+n*a=45, n=4
a=-9 , X=117
a=5 , X=75
[45-5^2]/5=[117-9^2]/9

solve X+a*(a-1)-3*a^2=45-a^2 ,a^2+n*a=45, n=6
a=-10.348 , X=141.74
a=4.348 , X=68.257
[45-(4.348)^2]/4.348>[141.74-(10.348)^2]/10.348

[Alberico Lepore]

I found a method that factored 8.33333333% of the factorizable numbers
but I can't establish the percentage of numbers factored in acceptable times
could you help me by doing a data analysis?

This type of factorization factorizes the numbers N into the form N = 4 * G + 1 therefore 16.6666666% of this type of N

********************************************************
to factorise N make x vary in the underlying system from 0 onwards, {0,1,2,3,4,5,6, etc.}

solve
(Q-1)^2/4-sqrt(Q^2-N)/2-sqrt[(Q-1)^2/4]*[sqrt[(Q-1)^2/4]-1]=(p-1)/2
,
K=3*b
,
((3+b)/2-1)^2/4-(m/4)=(p-1)/2
,
((3+b)/2)^2-(m/2)^2=K
,
m/4=x
********************************************************