 mersenneforum.org WTF factorization. Do you think it is a good method?
 Register FAQ Search Today's Posts Mark Forums Read 2020-11-03, 11:51 #1 Alberico Lepore   May 2017 ITALY 22·112 Posts WTF factorization. Do you think it is a good method? case: N=p*q & q-p=n & n mod 8 = 0 & p+n/2=3*(2*x+1) -> 288*X-n^2-32*((N-1)/8-1)=0 288*X-(n-8)^2-32*((N-1)/8+2*(n/4-1)-1)=0 -> Example N=377 X=1152 h^2 + 64 h + 6 = 1152 k² + 32 k + 6 288*(1152 h^2 + 64 h + 6)-n^2-32*(47-1)=0 , 288*(1152 k² + 32 k + 6)-(n-8)^2-32*(47+2*(n/4-1)-1)=0 , (1152 h^2 + 64 h + 6)=(1152 k² + 32 k + 6) given n = 576 + 8 * w assigning w values from 0 to 71 we will have O (72 * 1152 ^ 2 * [Lenstra elliptic-curve factorization]) Do you think it is a good method?   2020-11-03, 12:07 #2 Viliam Furik   "Viliam Furík" Jul 2018 Martin, Slovakia 23×3×19 Posts If you show an example of it working, maybe it is good...   2020-11-03, 12:11 #3 retina Undefined   "The unspeakable one" Jun 2006 My evil lair 22·5·307 Posts Alberico Lepore: Using N=377 is silly, stupid and pointless. Solve the 18 digit challenge first. https://mersenneforum.org/showthread.php?t=25929   2020-11-03, 15:07   #4
mathwiz

Mar 2019

3×53 Posts Quote:
 Originally Posted by Alberico Lepore Do you think it is a good method?
No. But, by all means, keep posting random equations and asking people what they think.   2020-11-03, 15:21   #5
Dr Sardonicus

Feb 2017
Nowhere

7×647 Posts Quote:
 Originally Posted by Alberico Lepore case: N=p*q
If N is an RSA number you may assume that N = p*q.
Quote:
 & q-p=n & n mod 8 = 0
This implies that N == 1 (mod 8).

If N == 3, 5, or 7 (mod 8), you're done.
Quote:
 & p+n/2=3*(2*x+1)
This implies that p+q is divisible by 3, and therefore (assuming that N isn't 9) that

N == 2 (mod 3).

If N == 1 (mod 3), you're done.

So, you're saying that you're only interested in N == 17 (mod 24).
<snip>
Quote:
 Do you think it is a good method?
Method? I don't see any "method."   2020-11-03, 18:52   #6
Alberico Lepore

May 2017
ITALY

7448 Posts Quote:
 Originally Posted by Alberico Lepore case: N=p*q & q-p=n & n mod 8 = 0 & p+n/2=3*(2*x+1) -> 288*X-n^2-32*((N-1)/8-1)=0 288*X-(n-8)^2-32*((N-1)/8+2*(n/4-1)-1)=0 -> Example N=377 X=1152 h^2 + 64 h + 6 = 1152 k² + 32 k + 6 288*(1152 h^2 + 64 h + 6)-n^2-32*(47-1)=0 , 288*(1152 k² + 32 k + 6)-(n-8)^2-32*(47+2*(n/4-1)-1)=0 , (1152 h^2 + 64 h + 6)=(1152 k² + 32 k + 6) given n = 576 + 8 * w assigning w values from 0 to 71 we will have O (72 * 1152 ^ 2 * [Lenstra elliptic-curve factorization]) Do you think it is a good method?
Quote:
 Originally Posted by Viliam Furik If you show an example of it working, maybe it is good...
I lowered to O(72 *[Lenstra elliptic-curve factorization])

case: x odd

N=377

solve (1152 h^2 + 64 h + 6)=x*(x+1)/2
,
2/3*[9*(1152 h^2 + 64 h + 6)+1+sqrt[8*[9*(1152 h^2 + 64 h + 6)+1]+1]*3*(x-1)-1]+1=Z^2
,
h,x   2020-11-03, 19:07   #7
CRGreathouse

Aug 2006

32·5·7·19 Posts Quote:
 Originally Posted by Alberico Lepore I lowered to O(72 *[Lenstra elliptic-curve factorization])
You understand that this isn't logarithmic, though, right?    2020-11-03, 19:26 #8 Uncwilly 6809 > 6502   """"""""""""""""""" Aug 2003 101×103 Posts 955910 Posts  Last fiddled with by Uncwilly on 2020-11-03 at 19:26  Thread Tools Show Printable Version Email this Page Similar Threads Thread Thread Starter Forum Replies Last Post nesio Computer Science & Computational Number Theory 28 2019-12-23 22:21 datblubat Computer Science & Computational Number Theory 6 2018-12-25 17:29 bhelmes Computer Science & Computational Number Theory 7 2017-06-26 02:20 henryzz Miscellaneous Math 4 2017-04-13 12:41 10metreh Factoring 6 2010-04-08 11:51

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