20201103, 11:51  #1 
May 2017
ITALY
2^{2}·11^{2} Posts 
WTF factorization. Do you think it is a good method?
case:
N=p*q & qp=n & n mod 8 = 0 & p+n/2=3*(2*x+1) > 288*Xn^232*((N1)/81)=0 288*X(n8)^232*((N1)/8+2*(n/41)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^232*(471)=0 , 288*(1152 k² + 32 k + 6)(n8)^232*(47+2*(n/41)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 ellipticcurve factorization]) Do you think it is a good method? 
20201103, 12:07  #2 
"Viliam Furík"
Jul 2018
Martin, Slovakia
111001000_{2} Posts 
If you show an example of it working, maybe it is good...

20201103, 12:11  #3 
Undefined
"The unspeakable one"
Jun 2006
My evil lair
6140_{10} Posts 
Alberico Lepore: Using N=377 is silly, stupid and pointless.
Solve the 18 digit challenge first. https://mersenneforum.org/showthread.php?t=25929 
20201103, 15:07  #4 
Mar 2019
3×53 Posts 

20201103, 15:21  #5  
Feb 2017
Nowhere
2^{3}×3^{4}×7 Posts 
If N is an RSA number you may assume that N = p*q.
Quote:
If N == 3, 5, or 7 (mod 8), you're done. Quote:
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:


20201103, 18:52  #6  
May 2017
ITALY
2^{2}×11^{2} Posts 
Quote:
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*(x1)1]+1=Z^2 , h,x 

20201103, 19:07  #7 
Aug 2006
3^{2}×5×7×19 Posts 

20201103, 19:26  #8 
6809 > 6502
"""""""""""""""""""
Aug 2003
101×103 Posts
5×1,913 Posts 
Last fiddled with by Uncwilly on 20201103 at 19:26 
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