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 2020-02-20, 18:20 #12 Alberico Lepore     May 2017 ITALY 192 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)/2-3*y*(y-1)/2+(3*x+1)*(3*x+2)/2-1)/3-2-k=2*(3*x*(x+1)/2)-2-[2*x-(x-1)]-2*(x-y+1)^2 , 2*(3*x*(x+1)/2)-2-[2*x-(x-1)]=A and we will find integer values for k , x1 , y1 [log n] here [2*((3*N-1)/8-1)/3+k]=[3*(2*x+1)*[3*(2*x+1)+2+8*(x-y+1)]-3]/12 , 3*x*(x+1)/2-3*y*(y-1)/2+(3*x+1)*(3*x+2)/2=(3*N-1)/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
 2020-02-20, 19:19 #13 retina Undefined     "The unspeakable one" Jun 2006 My evil lair 529910 Posts So go ahead and factor one of the as-yet-unfactored RSA numbers. Less pontificating, more factoring.
2020-02-20, 21:31   #14
Alberico Lepore

May 2017
ITALY

192 Posts

Quote:
 Originally Posted by retina So go ahead and factor one of the as-yet-unfactored RSA numbers. Less pontificating, more factoring.
i think i will try to implement this

I will assign a an integer and compare it with x

k=(a+1)*(2*a+1)+(a-1)*a

[2*(((3*N-1)/8-1)/3+y*(y-1)/2)+k]=[3*(2*x+1)*[3*(2*x+1)+2+8*(x-1+1)]-3]/12
,
3*x*(x+1)/2-3*y*(y-1)/2+(3*x+1)*(3*x+2)/2=(3*N-1)/8

now I start to implement it

 2020-02-21, 14:09 #15 Alberico Lepore     May 2017 ITALY 192 Posts UNCORRECT POST Last fiddled with by Alberico Lepore on 2020-02-21 at 23:05 Reason: UNCORRECT POST
 2020-02-22, 20:07 #16 Alberico Lepore     May 2017 ITALY 192 Posts would you kindly tell me: How do you calculate the interval of x? solve h=[x^2-[4*x-1+4*x-1-3*(y-2)]*(y-1)/2] , k=(x+1)*(2*(x-y+1)+1)+(x-y)*(x-y+1) , (3*N-1)/8+h=2*((3*N-1)/8-1)/3+k , k-2*h=(x+1)^2+2*[x^2-(x-y+1)^2] , k>0 , y>0 , x>0 , y Example solve h=[x^2-[4*x-1+4*x-1-3*(y-2)]*(y-1)/2] , k=(x+1)*(2*(x-y+1)+1)+(x-y)*(x-y+1) , 70+h=2*23+k , k-2*h=(x+1)^2+2*[x^2-(x-y+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.
2020-02-22, 22:37   #17
Alberico Lepore

May 2017
ITALY

192 Posts

Quote:
 Originally Posted by Alberico Lepore would you kindly tell me: How do you calculate the interval of x? solve h=[x^2-[4*x-1+4*x-1-3*(y-2)]*(y-1)/2] , k=(x+1)*(2*(x-y+1)+1)+(x-y)*(x-y+1) , (3*N-1)/8+h=2*((3*N-1)/8-1)/3+k , k-2*h=(x+1)^2+2*[x^2-(x-y+1)^2] , k>0 , y>0 , x>0 , y Example solve h=[x^2-[4*x-1+4*x-1-3*(y-2)]*(y-1)/2] , k=(x+1)*(2*(x-y+1)+1)+(x-y)*(x-y+1) , 70+h=2*23+k , k-2*h=(x+1)^2+2*[x^2-(x-y+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.
in particular I would need k> h

UPDATE

k>x

Last fiddled with by Alberico Lepore on 2020-02-22 at 23:04 Reason: UPDATE

 2020-02-23, 00:04 #18 Alberico Lepore     May 2017 ITALY 192 Posts in this range look at this 3*x*(x+1)/2-3*y*(y-1)/2+(3*x+1)*(3*x+2)/2=(3*N-1)/8 and by varying the x you will see that if x > x(0) [[[h(n)-h(n-1)]-[h(n-1)-(h(n-2))]]-[[h(n-1)-h(n-2)]-[h(n-2)-(h(n-3))]]] > [[[h(n-1)-h(n-2)]-[h(n-2)-(h(n-3))]]-[[h(n-2)-h(n-3)]-[h(n-3)-(h(n-4))]]] if x < x(0) [[[h(n)-h(n-1)]-[h(n-1)-(h(n-2))]]-[[h(n-1)-h(n-2)]-[h(n-2)-(h(n-3))]]] < [[[h(n-1)-h(n-2)]-[h(n-2)-(h(n-3))]]-[[h(n-2)-h(n-3)]-[h(n-3)-(h(n-4))]]]
 2020-02-23, 01:13 #19 mathwiz   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.
 2020-02-24, 09:37 #20 Alberico Lepore     May 2017 ITALY 192 Posts In summary Given an N = p * q in the form (p + q-4) mod 8 = 0 with p and q not necessarily prime but odd numbers then given the system solve h=[x^2-[4*x-1+4*x-1-3*(y-2)]*(y-1)/2] , k=(x+1)*(2*(x-y+1)+1)+(x-y)*(x-y+1) , (3*N-1)/8+h=2*((3*N-1)/8-1)/3+k , k-2*h=(x+1)^2+2*[x^2-(x-y+1)^2] , k>0 , y>0 , x>0 , y you will have that q=2*(3*x+1-(x-y+1))+1 e p=2*(3*x+1-(x-y+1))+1-(4*y-2) 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(n-1)]-[h(n-1)-(h(n-2))]]-[[h(n-1)-h(n-2)]-[h(n-2)-(h(n-3))]]] > [[[h(n-1)-h(n-2)]-[h(n-2)-(h(n-3))]]-[[h(n-2)-h(n-3)]-[h(n-3)-(h(n-4))]]] if x < x(0) [[[h(n)-h(n-1)]-[h(n-1)-(h(n-2))]]-[[h(n-1)-h(n-2)]-[h(n-2)-(h(n-3))]]] < [[[h(n-1)-h(n-2)]-[h(n-2)-(h(n-3))]]-[[h(n-2)-h(n-3)]-[h(n-3)-(h(n-4))]]] 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
2020-02-24, 14:21   #21
Alberico Lepore

May 2017
ITALY

192 Posts

Quote:
 Originally Posted by mathwiz 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.
I try to give you a demonstration

The rank of this system system

solve
h=[x^2-[4*x-1+4*x-1-3*(y-2)]*(y-1)/2]
,
k=(x+1)*(2*(x-y+1)+1)+(x-y)*(x-y+1)
,
(3*N-1)/8+h=2*((3*N-1)/8-1)/3+k
,
k-2*h=(x+1)^2+2*[x^2-(x-y+1)^2]

does not increase by adding this 3*x*(x+1)/2-3*y*(y-1)/2+(3*x+1)*(3*x+2)/2=(3*N-1)/8

and the rank of

3*x*(x+1)/2-3*y*(y-1)/2+(3*x+1)*(3*x+2)/2=(3*N-1)/8

does not increase by adding this [2*(3*x+1-(x-y+1))+1]*[2*(3*x+1-(x-y+1))+1-(4*y-2)]=N

therefore [2*(3*x+1-(x-y+1))+1] and [2*(3*x+1-(x-y+1))+1-(4*y-2)] 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 2020-02-24 at 14:24

 2020-02-24, 18:42 #22 Alberico Lepore     May 2017 ITALY 192 Posts for x<7 5+(8+8+4*(x-2))*(x-1)/2>(3*N-1)/8-2*((3*N-1)/8-1)/3>4+(6+6+3*(x-2))*(x-1)/2 Last fiddled with by Alberico Lepore on 2020-02-24 at 18:48

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