20200220, 18:20  #12 
May 2017
ITALY
19^{2} Posts 
I THINK I found a solution in O[K*(log n)^2] where K is the bruteforce to locate any k that is not to be underestimated
[BRUTE FORCE k ] X [log n] here 2*(3*x*(x+1)/23*y*(y1)/2+(3*x+1)*(3*x+2)/21)/32k=2*(3*x*(x+1)/2)2[2*x(x1)]2*(xy+1)^2 , 2*(3*x*(x+1)/2)2[2*x(x1)]=A and we will find integer values for k , x1 , y1 [log n] here [2*((3*N1)/81)/3+k]=[3*(2*x+1)*[3*(2*x+1)+2+8*(xy+1)]3]/12 , 3*x*(x+1)/23*y*(y1)/2+(3*x+1)*(3*x+2)/2=(3*N1)/8 recapping bruteforce on k in our brain and at each assignment of a k a log n to see if k integer gives x and y once identified , log n by observing x1 and y1 with x2 and y2 I will be more specific in the near future 
20200220, 19:19  #13 
Undefined
"The unspeakable one"
Jun 2006
My evil lair
5299_{10} Posts 
So go ahead and factor one of the asyetunfactored RSA numbers.
Less pontificating, more factoring. 
20200220, 21:31  #14  
May 2017
ITALY
19^{2} Posts 
Quote:
I will assign a an integer and compare it with x k=(a+1)*(2*a+1)+(a1)*a [2*(((3*N1)/81)/3+y*(y1)/2)+k]=[3*(2*x+1)*[3*(2*x+1)+2+8*(x1+1)]3]/12 , 3*x*(x+1)/23*y*(y1)/2+(3*x+1)*(3*x+2)/2=(3*N1)/8 now I start to implement it 

20200221, 14:09  #15 
May 2017
ITALY
19^{2} Posts 
UNCORRECT POST
Last fiddled with by Alberico Lepore on 20200221 at 23:05 Reason: UNCORRECT POST 
20200222, 20:07  #16 
May 2017
ITALY
19^{2} Posts 
would you kindly tell me:
How do you calculate the interval of x? solve h=[x^2[4*x1+4*x13*(y2)]*(y1)/2] , k=(x+1)*(2*(xy+1)+1)+(xy)*(xy+1) , (3*N1)/8+h=2*((3*N1)/81)/3+k , k2*h=(x+1)^2+2*[x^2(xy+1)^2] , k>0 , y>0 , x>0 , y Example solve h=[x^2[4*x1+4*x13*(y2)]*(y1)/2] , k=(x+1)*(2*(xy+1)+1)+(xy)*(xy+1) , 70+h=2*23+k , k2*h=(x+1)^2+2*[x^2(xy+1)^2] , k>0 , y>0 , x>0 , y https://www.wolframalpha.com/input/?...0%2Cx%3E0+%2Cy In the example, the only possible integer is 3. 
20200222, 22:37  #17  
May 2017
ITALY
19^{2} Posts 
Quote:
UPDATE k>x Last fiddled with by Alberico Lepore on 20200222 at 23:04 Reason: UPDATE 

20200223, 00:04  #18 
May 2017
ITALY
19^{2} Posts 
in this range look at this
3*x*(x+1)/23*y*(y1)/2+(3*x+1)*(3*x+2)/2=(3*N1)/8 and by varying the x you will see that if x > x(0) [[[h(n)h(n1)][h(n1)(h(n2))]][[h(n1)h(n2)][h(n2)(h(n3))]]] > [[[h(n1)h(n2)][h(n2)(h(n3))]][[h(n2)h(n3)][h(n3)(h(n4))]]] if x < x(0) [[[h(n)h(n1)][h(n1)(h(n2))]][[h(n1)h(n2)][h(n2)(h(n3))]]] < [[[h(n1)h(n2)][h(n2)(h(n3))]][[h(n2)h(n3)][h(n3)(h(n4))]]] 
20200223, 01:13  #19 
Mar 2019
3×19 Posts 
Just so you know, no one is paying attention to you any more.
You need to test your own equations, and demonstrate them on larger numbers. 
20200224, 09:37  #20 
May 2017
ITALY
19^{2} Posts 
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 h=[x^2[4*x1+4*x13*(y2)]*(y1)/2] , k=(x+1)*(2*(xy+1)+1)+(xy)*(xy+1) , (3*N1)/8+h=2*((3*N1)/81)/3+k , k2*h=(x+1)^2+2*[x^2(xy+1)^2] , k>0 , y>0 , x>0 , y you will have that q=2*(3*x+1(xy+1))+1 e p=2*(3*x+1(xy+1))+1(4*y2) I put forward a hypothesis that in the range sought as x changes, h behaves in this way called x (0) the x sought then if x > x(0) [[[h(n)h(n1)][h(n1)(h(n2))]][[h(n1)h(n2)][h(n2)(h(n3))]]] > [[[h(n1)h(n2)][h(n2)(h(n3))]][[h(n2)h(n3)][h(n3)(h(n4))]]] if x < x(0) [[[h(n)h(n1)][h(n1)(h(n2))]][[h(n1)h(n2)][h(n2)(h(n3))]]] < [[[h(n1)h(n2)][h(n2)(h(n3))]][[h(n2)h(n3)][h(n3)(h(n4))]]] so if the hypothesis were right we should have another interval (maybe) and at least the factorization of N should be solved in at least logarithmic times Someone could kindly deny or confirm this hypothesis 
20200224, 14:21  #21  
May 2017
ITALY
19^{2} Posts 
Quote:
The rank of this system system solve h=[x^2[4*x1+4*x13*(y2)]*(y1)/2] , k=(x+1)*(2*(xy+1)+1)+(xy)*(xy+1) , (3*N1)/8+h=2*((3*N1)/81)/3+k , k2*h=(x+1)^2+2*[x^2(xy+1)^2] does not increase by adding this 3*x*(x+1)/23*y*(y1)/2+(3*x+1)*(3*x+2)/2=(3*N1)/8 and the rank of 3*x*(x+1)/23*y*(y1)/2+(3*x+1)*(3*x+2)/2=(3*N1)/8 does not increase by adding this [2*(3*x+1(xy+1))+1]*[2*(3*x+1(xy+1))+1(4*y2)]=N therefore [2*(3*x+1(xy+1))+1] and [2*(3*x+1(xy+1))+1(4*y2)] are factors of N If they are integer N factors then x and y are integers so it is also h and therefore also k k>0 if p>= (p+q)/4 It remains to prove my hypothesis Last fiddled with by Alberico Lepore on 20200224 at 14:24 

20200224, 18:42  #22 
May 2017
ITALY
19^{2} Posts 
for x<7
5+(8+8+4*(x2))*(x1)/2>(3*N1)/82*((3*N1)/81)/3>4+(6+6+3*(x2))*(x1)/2 Last fiddled with by Alberico Lepore on 20200224 at 18:48 
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