20200224, 20:20  #23 
Mar 2019
111001_{2} Posts 

20200224, 21:01  #24  
"Curtis"
Feb 2005
Riverside, CA
2×3×5×7×19 Posts 
Quote:
At 20 digits, OP would notice how terrible his methods are, and that even if his methods could factor numbers, they're worse than trial division. 

20200226, 09:08  #25  
May 2017
ITALY
19^{2} Posts 
Quote:
Now I will test this k=(x+1)*(2*(xy+1)+1)+(xy)*(xy+1) , 3*x*(x+1)/23*y*(y1)/2+(3*x+1)*(3*x+2)/2=(3*N1)/8 I will assign k integers I, xk = I1 && yk = I I have to test if xk < x0 && k < k0 then x > xk && y > yk if xk > x0 && k < k0 then x < xk && y < yk if xk > x0 && k > k0 then x < xk && y < yk the problem is this: as you can see from the link they are huge equations https://www.wolframalpha.com/input/?...%2F8+%2C+x%2Cy Last fiddled with by Alberico Lepore on 20200226 at 09:38 

20200226, 20:39  #26  
May 2017
ITALY
551_{8} Posts 
Quote:
it would seem to work from the first hand tests 

20200227, 04:57  #27 
Aug 2006
16DE_{16} Posts 

20200227, 07:14  #28 
May 2017
ITALY
19^{2} Posts 
The following may be wrong but it may also be right
In summary Given an N = p * q in the form (p + q4) mod 8 = 0 with p and q not necessarily prime but odd numbers then given the system solve k=(x+1)*(2*(xy+1)+1)+(xy)*(xy+1) , 3*x*(x+1)/23*y*(y1)/2+(3*x+1)*(3*x+2)/2=(3*N1)/8 you will have that q=2*(3*x+1(xy+1))+1 e p=2*(3*x+1(xy+1))+1(4*y2) Now I will test this k=(x+1)*(2*(xy+1)+1)+(xy)*(xy+1) , 3*x*(x+1)/23*y*(y1)/2+(3*x+1)*(3*x+2)/2=(3*N1)/8 I will assign k integers I, xk = I1 && yk = I I have to test if xk < x0 && k < k0 then x > xk && y > yk if xk > x0 && k < k0 then x < xk && y < yk if xk > x0 && k > k0 then x < xk && y < yk it would seem to work from the first hand tests the problem is this: as you can see from the link they are huge equations https://www.wolframalpha.com/input/?...%2F8+%2C+x%2Cy I don't understand that much demonstrations, now I try solve (k)=(x+1)*(2*(xy+1)+1)+(xy)*(xy+1) > x=(1/3)*[sqrt(3*k+(y3)*y)+2*y3] if xk > x0 && k < k0 then x < xk && y < yk if xk > x0 && k > k0 then x < xk && y < yk solve x=(1/3)*[sqrt(3*k+(y3)*y)+2*y3],3*x*(x+1)/23*y*(y1)/2+(3*x+1)*(3*x+2)/2=(3*N1)/8 , k=x+2 , y>=k impossible if xk < x0 && k < k0 then x > xk && y > yk solve x=(1/3)*[sqrt(3*k+(y3)*y)+2*y3],3*x*(x+1)/23*y*(y1)/2+(3*x+1)*(3*x+2)/2=(3*N1)/8 , k=x , y<=k > x<=1.5 
20200227, 07:24  #29 
Aug 2006
2·2,927 Posts 
I was hoping for more along the lines of "here is a big number and how I factored it with my method". But here's a small number to get you started: 596085571408853. Could you show us how your method factors it? It takes less than a millisecond with Shanks' SQUFOF.

20200227, 07:32  #30  
May 2017
ITALY
19^{2} Posts 
Quote:
(p + q4) mod 8 = 0 k>0 if p>= (p+q)/4 Last fiddled with by Alberico Lepore on 20200227 at 07:35 

20200227, 09:53  #31 
May 2017
ITALY
19^{2} Posts 
I have tried with larger numbers and I have seen that it was another wrong hypothesis but I have found this
range 1/2 *[sqrt[2*[(3*N1)/81]/3+1]1]<= x <= 1/6*[sqrt[3*N]3] By assigning k integer numbers > 0 if xk = k1 < x0 then x > 1/6*[sqrt[3*N]3] I don't know if it's another wrong hypothesis, I'll test it ************************** UPDATE This hypothesis is also wrong Last fiddled with by Alberico Lepore on 20200227 at 10:25 
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