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-   -   14° Primality test and factorization of Lepore ( conjecture ) (https://www.mersenneforum.org/showthread.php?t=22806)

 Alberico Lepore 2017-12-20 12:44

14° Primality test and factorization of Lepore ( conjecture )

1 Attachment(s)
Hey friends, could you take a look?

What do you think?

 Alberico Lepore 2017-12-20 14:37

Testing m, r, s, t, v, z,

n-m=1
m-r=1
r-s=1
s-t=1
t-v=1
v-z=1
etc.etc.

 CRGreathouse 2017-12-20 15:11

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 = q-1 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. :smile:

 Alberico Lepore 2017-12-20 15:14

[QUOTE=CRGreathouse;474436]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 = q-1 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. :smile:[/QUOTE]

There's something wrong, yet.
If I find the problem I will start the implementation.

 Alberico Lepore 2017-12-20 16:16

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^2-m*a=1829
A=a^2+r*a=1829-[3*a^2-1829] -> r=n-m
B=a^2+s*a=-[3*a^2-1829]+[1829-[3*a^2-1829]] ->s=r-m
a^2+t*a=-A-B -> s+r=t ERROR=WIN
s+r=t+1

I do not know exactly if I have to do X + Y or X-Y or -X + Y or -X-Y
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^2-m*a=1829
A=a^2+r*a=1829-[3*a^2-1829]
B=a^2+s*a=-[3*a^2-1829]+[1829-[3*a^2-1829]]
a^2+t*a=-A-B
s+r=t+1

test 1 and -1

 CRGreathouse 2017-12-20 16:33

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?

 Alberico Lepore 2017-12-20 16:41

[QUOTE=CRGreathouse;474444]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?[/QUOTE]

I discovered this magic number 3 * a ^ 2 today, it is not a random number.
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?

 CRGreathouse 2017-12-20 16:45

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.

 Alberico Lepore 2017-12-20 17:08

[QUOTE=CRGreathouse;474436]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 = q-1 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. :smile:[/QUOTE]

you could give me the factoring of this number, in order to test the procedure

 ET_ 2017-12-20 17:23

[QUOTE=Alberico Lepore;474449]you could give me the factoring of this number, in order to test the procedure[/QUOTE]

You could give us the factors coming out from your procedure, and we would tell you whether the result is correct...

 CRGreathouse 2017-12-20 17:25

[QUOTE=Alberico Lepore;474449]you could give me the factoring of this number, in order to test the procedure[/QUOTE]

I don't understand, I thought your procedure was supposed to generate a factorization? :confused: Perhaps I misunderstood the purpose of your algorithm. What is it supposed to do?

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