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Old 2020-02-20, 18:20   #12
Alberico Lepore
 
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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
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Old 2020-02-20, 19:19   #13
retina
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So go ahead and factor one of the as-yet-unfactored RSA numbers.

Less pontificating, more factoring.
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Old 2020-02-20, 21:31   #14
Alberico Lepore
 
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Quote:
Originally Posted by retina View Post
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
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Old 2020-02-21, 14:09   #15
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UNCORRECT POST

Last fiddled with by Alberico Lepore on 2020-02-21 at 23:05 Reason: UNCORRECT POST
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Old 2020-02-22, 20:07   #16
Alberico Lepore
 
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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.
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Old 2020-02-22, 22:37   #17
Alberico Lepore
 
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Quote:
Originally Posted by Alberico Lepore View Post
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
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Old 2020-02-23, 00:04   #18
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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))]]]
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Old 2020-02-23, 01:13   #19
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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.
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Old 2020-02-24, 09:37   #20
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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
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Old 2020-02-24, 14:21   #21
Alberico Lepore
 
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Quote:
Originally Posted by mathwiz View Post
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
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Old 2020-02-24, 18:42   #22
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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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