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Old 2020-05-02, 15:22   #1
Alberico Lepore
 
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Default yet another attempt to bring down RSA

to solve this factorization

N=p*q

just solve this

solve M=(3*(W+3)-1)/8 , W=N*x-3 , [W-(22+(24+24+8*(b-3))*(b-2)/2)]*W=16*[(M-1)/3]*[M] ,P=2*(3*b+1-(b-1+1))+1-2,Q=2*(3*b+1-(b-1+1))+1

GCD(N,P)= p || q

GCD(N,Q)= p || q

then

solve M=(3*(W+3)-1)/8 , W+4=Y^2,W

3*Y^2-8*M-4=0

solution

3 ⁢x² - 8 ⁢y - 4 = 0

x = 16 k
y = 96 k²

and also:
x = 16 k + 6
y = 96 k² + 72 k + 13

and also:
x = 16 k + 12
y = 96 k² + 144 k + 53

and also:
x = 16 k + 2
y = 96 k² + 24 k + 1

and also:
x = 16 k + 8
y = 96 k² + 96 k + 23

and also:
x = 16 k + 14
y = 96 k² + 168 k + 73

and also:
x = 16 k + 4
y = 96 k² + 48 k + 5

and also:
x = 16 k + 10
y = 96 k² + 120 k + 37


Example

N=65

M=(3*(W+3)-1)/8 , W=65*x-3 , [W-(22+(24+24+8*(b-3))*(b-2)/2)]*W=16*[(M-1)/3]*[M] ,P=2*(3*b+1-(b-1+1))+1-2 , Q=2*(3*b+1-(b-1+1))+1
,
M=73

-> Q=15

GCD(65,15)=5


we hope it is the right time
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Old 2020-05-03, 06:36   #2
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Quote:
Originally Posted by Alberico Lepore View Post
to solve this factorization

N=p*q

just solve this

solve M=(3*(W+3)-1)/8 , W=N*x-3 , [W-(22+(24+24+8*(b-3))*(b-2)/2)]*W=16*[(M-1)/3]*[M] ,P=2*(3*b+1-(b-1+1))+1-2,Q=2*(3*b+1-(b-1+1))+1

GCD(N,P)= p || q

GCD(N,Q)= p || q

then

solve M=(3*(W+3)-1)/8 , W+4=Y^2,W

3*Y^2-8*M-4=0

solution

3 ⁢x² - 8 ⁢y - 4 = 0

x = 16 k
y = 96 k²

and also:
x = 16 k + 6
y = 96 k² + 72 k + 13

and also:
x = 16 k + 12
y = 96 k² + 144 k + 53

and also:
x = 16 k + 2
y = 96 k² + 24 k + 1

and also:
x = 16 k + 8
y = 96 k² + 96 k + 23

and also:
x = 16 k + 14
y = 96 k² + 168 k + 73

and also:
x = 16 k + 4
y = 96 k² + 48 k + 5

and also:
x = 16 k + 10
y = 96 k² + 120 k + 37


Example

N=65

M=(3*(W+3)-1)/8 , W=65*x-3 , [W-(22+(24+24+8*(b-3))*(b-2)/2)]*W=16*[(M-1)/3]*[M] ,P=2*(3*b+1-(b-1+1))+1-2 , Q=2*(3*b+1-(b-1+1))+1
,
M=73

-> Q=15

GCD(65,15)=5


we hope it is the right time



[IMPORTANT UPDATE]


to solve this

65*X=(8*M+1)/3=[8*(96 k² + 168 k + 73)+1]/3

solve this
195*X=768 k² + 1344 k + 585



I took the following from the site
https://www.alpertron.com.ar/QUAD.HTM

768 ⁢x² + 1344 ⁢x - 195 ⁢y + 585 = 0
The discriminant is b² − 4⁢a⁢c = 0

Multiplying by 4⁢a

( 512 ⁢x )² + 458752 ⁢x - 66560 ⁢y + 199680 = 0

Let t = 512 ⁢x (1)

(t + d)² = ( t + 448 )² = 66560 ⁢y + 1024 (2)

where the linear coefficient is 2⁢(b⁢d − 2⁢ae) and the constant coefficient is d² − 4⁢af.

We have to solve T² = 1024 (mod 66560)

To solve this quadratic modular equation we have to factor the modulus and find the solution modulo the powers of the prime factors. Then we combine them by using the Chinese Remainder Theorem.

66560 = 2^10 × 5 × 13
Solutions modulo 2^10: 0, 32, 64, 96, 128, 160, 192, 224, 256, 288, 320, 352, 384, 416, 448, 480, 512, 544, 576, 608, 640, 672, 704, 736, 768, 800, 832, 864, 896, 928, 960 and 992

Solutions modulo 5: 2 and 3

Solutions modulo 13: 6 and 7

without factorizing 65 we see what happens



T = 448

t = T − d + 66560 ⁢k = 66560 k (7) (where k is any integer).

Replacing t in equation (2) we can get the value of y:

y = 66560 k² + 896 k + 3 (8)

From (1) and (7):

512 ⁢x = 66560 k

x = 130 k
y = 66560 k² + 896 k + 3


for x=3

M=(3*(W+3)-1)/8 , W=65*x-3 , [W-(22+(24+24+8*(b-3))*(b-2)/2)]*W=16*[(M-1)/3]*[M] ,P=2*(3*b+1-(b-1+1))+1-2 , Q=2*(3*b+1-(b-1+1))+1
,
x=3

-> Q=15

GCD(65,15)=5



what do you think?
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Old 2020-05-03, 06:42   #3
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Quote:
Originally Posted by Alberico Lepore View Post
what do you think?
I think that instead of wasting everyone's time with factoring tiny numbers like 65, you should be using larger numbers. Even 50 digits would be a start. But not 2 or 3 digits. That's just boring and useless.
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Old 2020-05-03, 13:15   #4
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I don't know if this is correct I have to do tests

Last fiddled with by Alberico Lepore on 2020-05-03 at 13:57
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Old 2020-05-04, 00:49   #5
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Quote:
Originally Posted by Alberico Lepore View Post
I don't know if this is correct I have to do tests
Then do your tests elsewhere (on your own machine) and save us from having to look at your playing around. Once you find something at works for a 20 digit number ask us for a 65 digit number and if it passes, then post here.
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Old 2020-05-04, 11:56   #6
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Quote:
Originally Posted by Uncwilly View Post
Then do your tests elsewhere (on your own machine) and save us from having to look at your playing around. Once you find something at works for a 20 digit number ask us for a 65 digit number and if it passes, then post here.
the first post is 100% correct.
it is the way to find an x that is difficult
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Old 2020-05-06, 20:08   #7
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x = 16 k
y = 96 k² -> NULL

and also:
x = 16 k + 6
y = 96 k² + 72 k + 13 -> F=(96 k² + 72 k + 13)*8+1=3*(16*k+5)*(16*k+7)

and also:
x = 16 k + 12
y = 96 k² + 144 k + 53 -> NULL

and also:
x = 16 k + 2
y = 96 k² + 24 k + 1 -> F=(96 k² + 24 k + 1)*8+1=3*(16*k+1)*(16*k+3)

and also:
x = 16 k + 8
y = 96 k² + 96 k + 23 -> NULL

and also:
x = 16 k + 14
y = 96 k² + 168 k + 73 -> F=(96 k² + 168 k + 73)*8+1=3*(16*k+13)*(16*k+15)

and also:
x = 16 k + 4
y = 96 k² + 48 k + 5 -> NULL

and also:
x = 16 k + 10
y = 96 k² + 120 k + 37 -> F=(96 k² + 120 k + 37)*8+1=3*(16*k+9)*(16*k+11)





if N=4*G+3

N*x*(x+4*h)=[8*((96 k² + 24 k + 1)-3*a*(a-1)/2)+1]/3

N*x*(x+4*h)=[8*((96 k² + 72 k + 13)-3*a*(a-1)/2)+1]/3

N*x*(x+4*h)=[8*((96 k² + 120 k + 37)-3*a*(a-1)/2)+1]/3

N*x*(x+4*h)=[8*((96 k² + 168 k + 73)-3*a*(a-1)/2)+1]/3


if N=4*G+1

N*x*(x+2+4*h)=[8*((96 k² + 24 k + 1)-3*a*(a-1)/2)+1]/3

N*x*(x+2+4*h)=[8*((96 k² + 72 k + 13)-3*a*(a-1)/2)+1]/3

N*x*(x+2+4*h)=[8*((96 k² + 120 k + 37)-3*a*(a-1)/2)+1]/3

N*x*(x+2+4*h)=[8*((96 k² + 168 k + 73)-3*a*(a-1)/2)+1]/3




Example

N=187=4*G+3

choose N*x*(x+4*h)=[8*((96 k² + 120 k + 37)-3*a*(a-1)/2)+1]/3

187*x*(x+4*h)=[8*((96 k² + 120 k + 37)-3*a*(a-1)/2)+1]/3

we vary h and choose [we have to decide whether to choose a or b or x to solve more easily]

h=2 , x=5 , a=42 ,k=8

sqrt[(8*(96 k² + 120 k + 37)+1)/3+1]-(2*a-1)=55

GCD(187,55)=11
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Old 2020-05-07, 07:40   #8
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One procedure could be this

To make things easier, take N ^ 2

Example N=187

chose h=1

solve 187^2*x*(x+2+4*1)=[8*((96 k² + 120 k + 37)-3*a*(a-1)/2)+1]/3 ,x

Last fiddled with by Alberico Lepore on 2020-05-07 at 11:22
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Old 2020-05-08, 13:52   #9
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Given N=p*q

Solving these four with odd x and N with h varying from 0 to 7 and with a> = 1 and X> = 3, assigning at a value to our choice such that

[1] [8*(96 k² + 24 k + 1)+1]/3 >= N^2*x*(x+2+4*h)

[2] [8*(96 k² + 72 k + 13)+1]/3 >= N^2*x*(x+2+4*h)

[3] [8*(96 k² + 120 k + 37)+1]/3 >= N^2*x*(x+2+4*h)

[4] [8*(96 k² + 168 k + 73)+1]/3 >= N^2*x*(x+2+4*h)

[1]

N^2*x*(x+2+4*h)=[8*((96 k² + 24 k + 1)-3*a*(a-1)/2)+1]/3

sqrt[(8*(96 k² + 24 k + 1)+1)/3+1]-(2*a-1)=P

GCD(N,P)= p || q || N


[2]

N^2*x*(x+2+4*h)=[8*((96 k² + 72 k + 13)-3*a*(a-1)/2)+1]/3

sqrt[(8*(96 k² + 72 k + 13)+1)/3+1]-(2*a-1)=P

GCD(N,P)= p || q || N

[3]

N^2*x*(x+2+4*h)=[8*((96 k² + 120 k + 37)-3*a*(a-1)/2)+1]/3

sqrt[(8*(96 k² + 120 k + 37)+1)/3+1]-(2*a-1)=P

GCD(N,P)= p || q || N

[4]

N^2*x*(x+2+4*h)=[8*((96 k² + 168 k + 73)-3*a*(a-1)/2)+1]/3

sqrt[(8*(96 k² + 168 k + 73)+1)/3+1]-(2*a-1)=P

GCD(N,P)= p || q || N


Example N=187


187^2*x*(x+2+4*1)=[8*((96 k² + 24 k + 1)-3*a*(a-1)/2)+1]/3

(187*x+561)^2=4*(78689-a^2+a+64*k^2+48*k)

(187*x+561)^2=(-4*a^2+4*a-1)+1+(256*k^2+64*k+4)-4+314724

-(256*k^2+64*k+4)+(187*x+561)^2=(-4*a^2+4*a-1)+314721

we must assign at a value that meets the conditions and that is easily factored the number (-4 * a ^ 2 + 4 * a-1) +314721

now I choose it at random just to make you understand the process

a=56

X^2-K^2=302400

follow

K^2=978^2=(256*k^2+64*k+4) -> k=61

sqrt[(8*(96 k² + 24 k + 1)+1)/3+1]-(2*a-1)=P , a=56 ,k=61

-> P=867

GCD(867,187)=17

How should I choose a to have a smaller computational complexity?

RSA GAME OVER ?
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Old 2020-05-09, 18:53   #10
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Quote:
Originally Posted by Alberico Lepore View Post
Given N=p*q

Solving these four with odd x and N with h varying from 0 to 7 and with a> = 1 and X> = 3, assigning at a value to our choice such that

[1] [8*(96 k² + 24 k + 1)+1]/3 >= N^2*x*(x+2+4*h)

[2] [8*(96 k² + 72 k + 13)+1]/3 >= N^2*x*(x+2+4*h)

[3] [8*(96 k² + 120 k + 37)+1]/3 >= N^2*x*(x+2+4*h)

[4] [8*(96 k² + 168 k + 73)+1]/3 >= N^2*x*(x+2+4*h)

[1]

N^2*x*(x+2+4*h)=[8*((96 k² + 24 k + 1)-3*a*(a-1)/2)+1]/3

sqrt[(8*(96 k² + 24 k + 1)+1)/3+1]-(2*a-1)=P

GCD(N,P)= p || q || N


[2]

N^2*x*(x+2+4*h)=[8*((96 k² + 72 k + 13)-3*a*(a-1)/2)+1]/3

sqrt[(8*(96 k² + 72 k + 13)+1)/3+1]-(2*a-1)=P

GCD(N,P)= p || q || N

[3]

N^2*x*(x+2+4*h)=[8*((96 k² + 120 k + 37)-3*a*(a-1)/2)+1]/3

sqrt[(8*(96 k² + 120 k + 37)+1)/3+1]-(2*a-1)=P

GCD(N,P)= p || q || N

[4]

N^2*x*(x+2+4*h)=[8*((96 k² + 168 k + 73)-3*a*(a-1)/2)+1]/3

sqrt[(8*(96 k² + 168 k + 73)+1)/3+1]-(2*a-1)=P

GCD(N,P)= p || q || N


Example N=187


187^2*x*(x+2+4*1)=[8*((96 k² + 24 k + 1)-3*a*(a-1)/2)+1]/3

(187*x+561)^2=4*(78689-a^2+a+64*k^2+48*k)

(187*x+561)^2=(-4*a^2+4*a-1)+1+(256*k^2+64*k+4)-4+314724

-(256*k^2+64*k+4)+(187*x+561)^2=(-4*a^2+4*a-1)+314721

we must assign at a value that meets the conditions and that is easily factored the number (-4 * a ^ 2 + 4 * a-1) +314721

now I choose it at random just to make you understand the process

a=56

X^2-K^2=302400

follow

K^2=978^2=(256*k^2+64*k+4) -> k=61

sqrt[(8*(96 k² + 24 k + 1)+1)/3+1]-(2*a-1)=P , a=56 ,k=61

-> P=867

GCD(867,187)=17

How should I choose a to have a smaller computational complexity?

RSA GAME OVER ?
If we assign x value [3,5,7,9] and h becomes a variable, then it should work for any value of a.

In the next few days I will study this version.

187^2*3*(3+2+4*h)=[8*((96 k² + 24 k + 1)-3*a*(a-1)/2)+1]/3 , a=1

187^2*3*(3+2+4*h)=[8*((96 k² + 24 k + 1)-3*a*(a-1)/2)+1]/3 , a=2

187^2*3*(3+2+4*h)=[8*((96 k² + 24 k + 1)-3*a*(a-1)/2)+1]/3 , a=3

etc,

etc
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Old 2020-05-15, 11:44   #11
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Quote:
Originally Posted by Alberico Lepore View Post
Given N=p*q

Solving these four with odd x and N with h varying from 0 to 7 and with a> = 1 and X> = 3, assigning at a value to our choice such that

[1] [8*(96 k² + 24 k + 1)+1]/3 >= N^2*x*(x+2+4*h)

[2] [8*(96 k² + 72 k + 13)+1]/3 >= N^2*x*(x+2+4*h)

[3] [8*(96 k² + 120 k + 37)+1]/3 >= N^2*x*(x+2+4*h)

[4] [8*(96 k² + 168 k + 73)+1]/3 >= N^2*x*(x+2+4*h)

[1]

N^2*x*(x+2+4*h)=[8*((96 k² + 24 k + 1)-3*a*(a-1)/2)+1]/3

sqrt[(8*(96 k² + 24 k + 1)+1)/3+1]-(2*a-1)=P

GCD(N,P)= p || q || N


[2]

N^2*x*(x+2+4*h)=[8*((96 k² + 72 k + 13)-3*a*(a-1)/2)+1]/3

sqrt[(8*(96 k² + 72 k + 13)+1)/3+1]-(2*a-1)=P

GCD(N,P)= p || q || N

[3]

N^2*x*(x+2+4*h)=[8*((96 k² + 120 k + 37)-3*a*(a-1)/2)+1]/3

sqrt[(8*(96 k² + 120 k + 37)+1)/3+1]-(2*a-1)=P

GCD(N,P)= p || q || N

[4]

N^2*x*(x+2+4*h)=[8*((96 k² + 168 k + 73)-3*a*(a-1)/2)+1]/3

sqrt[(8*(96 k² + 168 k + 73)+1)/3+1]-(2*a-1)=P

GCD(N,P)= p || q || N


Example N=187


187^2*x*(x+2+4*1)=[8*((96 k² + 24 k + 1)-3*a*(a-1)/2)+1]/3

(187*x+561)^2=4*(78689-a^2+a+64*k^2+48*k)

(187*x+561)^2=(-4*a^2+4*a-1)+1+(256*k^2+64*k+4)-4+314724

-(256*k^2+64*k+4)+(187*x+561)^2=(-4*a^2+4*a-1)+314721

we must assign at a value that meets the conditions and that is easily factored the number (-4 * a ^ 2 + 4 * a-1) +314721

now I choose it at random just to make you understand the process

a=56

X^2-K^2=302400

follow

K^2=978^2=(256*k^2+64*k+4) -> k=61

sqrt[(8*(96 k² + 24 k + 1)+1)/3+1]-(2*a-1)=P , a=56 ,k=61

-> P=867

GCD(867,187)=17

How should I choose a to have a smaller computational complexity?

RSA GAME OVER ?
so in our case to have any valid a we have to increase h
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