20171220, 17:27  #12 
Aug 2006
5979_{10} Posts 
If it's not a factorization algorithm that would explain why you seem to start with a factorization (which of course you can't do when you don't yet know it).

20171221, 02:56  #13 
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
10011001011110_{2} Posts 

20171221, 14:41  #14 
May 2017
ITALY
2·3^{2}·29 Posts 
CRGreathouse if i post the solution in O (log (N)) you promise me that if you find it interesting,you implement it?

20171221, 14:53  #15 
May 2017
ITALY
2·3^{2}·29 Posts 
you implememt it?

20171221, 16:59  #16 
Banned
"Luigi"
Aug 2002
Team Italia
2^{2}·7·173 Posts 
I still can't understand (my fault...) what you are trying to conjecture.
a) Factorisation? b) A hard way to write N=187? c) A generalization in writing a subclass of polynomials having the same behavio(u)r? (and in this case, what is such behavio(u)r? In other, simpler words, what should your conjecture demonstrate? Feel free to answer in Italian if you feel more apt to do it. Last fiddled with by ET_ on 20171221 at 17:01 
20171221, 17:12  #17  
May 2017
ITALY
20A_{16} Posts 
Quote:
Effettivamente è tutto dimostrabile solo che non sono pratico con la complessità computazionale che a me sembra log_2() Ah c'è un piccolo errore bisogna testare anche il 19 

20171221, 17:28  #18  
Aug 2006
3·1,993 Posts 
Quote:
Letting K = 2m+1 the first equation becomes sqrt[3(2m+1)+2*(3(2m+1)/9)^2+3(2m+1)+((3(2m+1)/99)/2)^2] = sqrt[Nn*a] which can be simplified to sqrt((m + 5)^2) = sqrt[Nn*a] and so presumably (m + 5)^2 = N  n*a where N is the number we're trying to factor, a is some factor of N, and n = N/a  a. Now let's use that last equation to simplify the right side: (m + 5)^2 = N  (N/a  a)*a (m + 5)^2 = N  (N  a^2) (m + 5)^2 = a^2 m + 5 = ±a So your method is just checking numbers m + 5 until they divide N. This is a primitive version of trial division and its running time (with schoolbook division) is O(sqrt(n)*log(n)^2), exponentially slower than you suggest. 

20171221, 17:33  #19  
May 2017
ITALY
2×3^{2}×29 Posts 
Quote:
I get better 

20171221, 18:23  #20 
May 2017
ITALY
2×3^{2}×29 Posts 
now it's okay
16° Primality test and factorization of Lepore 
20171221, 18:39  #21 
May 2017
ITALY
2·3^{2}·29 Posts 
we miss a passage that I'm studying
in a while I post it 
20171221, 18:52  #22 
May 2017
ITALY
522_{10} Posts 
now is correct

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