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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

5518 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 36110 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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