20171220, 12:44  #1 
May 2017
ITALY
2×3^{2}×29 Posts 
14° Primality test and factorization of Lepore ( conjecture )
Hey friends, could you take a look?
What do you think? 
20171220, 14:37  #2 
May 2017
ITALY
2×3^{2}×29 Posts 
Testing m, r, s, t, v, z,
nm=1 mr=1 rs=1 st=1 tv=1 vz=1 etc.etc. 
20171220, 15:11  #3 
Aug 2006
5988_{10} Posts 
Would you use this method to factor, say, 647978103069410806903326919883810380070087355654969148727717? It is of the form pq for primes p and q, so you can choose a = p and n = q1 to see that it is of the desired form a^2 + n*a.
This is a small number as far as factoring is concerned  yafu's SIQS cracks it in 2.8 seconds  but it should be enough to show us how your method works. Please don't use other factoring tools for this, we're interested in how well your method works not how well others work. 
20171220, 15:14  #4  
May 2017
ITALY
2×3^{2}×29 Posts 
Quote:
If I find the problem I will start the implementation. 

20171220, 16:16  #5 
May 2017
ITALY
2·3^{2}·29 Posts 
I will start the implementation (if I can) while I explain the algorithm
from a structural error I found a very quick solution a^2+n*a=1829 3*a^2m*a=1829 A=a^2+r*a=1829[3*a^21829] > r=nm B=a^2+s*a=[3*a^21829]+[1829[3*a^21829]] >s=rm a^2+t*a=AB > s+r=t ERROR=WIN s+r=t+1 I do not know exactly if I have to do X + Y or XY or X + Y or XY but I know there will be an error in the right path, using this error I find the solution a^2+n*a=1829 3*a^2m*a=1829 A=a^2+r*a=1829[3*a^21829] B=a^2+s*a=[3*a^21829]+[1829[3*a^21829]] a^2+t*a=AB s+r=t+1 test 1 and 1 
20171220, 16:33  #6 
Aug 2006
2^{2}×3×499 Posts 
So at each step there are four possibilities to explore, which suggests that with s steps the entire tree has 4^s possibilities with an average of 4^s / 2 before you find the desired one (if it is unique and you have no other information). One question that comes to mind: how many steps are there for a given number, and how does this grow with the size of the number?

20171220, 16:41  #7  
May 2017
ITALY
522_{10} Posts 
Quote:
I still do not have much information. I'm not a programmer so I could take weeks. Would anyone want to implement it? I would be happy to share it with someone else CRGreathouse implements it? 

20171220, 16:45  #8 
Aug 2006
2^{2}×3×499 Posts 
I don't understand what you're doing, and the pdf you posted doesn't appear to contain an algorithm. It's not at all clear to me where the numbers you post come from, and in any case they're too small for me to figure out what calculations they could result from. I asked for an example with a somewhat larger number so I could follow it more easily but you weren't able to give it.
Until you know what you're doing there is no hope of someone else implementing it for you. 
20171220, 17:08  #9  
May 2017
ITALY
2×3^{2}×29 Posts 
Quote:


20171220, 17:23  #10 
Banned
"Luigi"
Aug 2002
Team Italia
43·113 Posts 

20171220, 17:25  #11 
Aug 2006
2^{2}×3×499 Posts 

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