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Old 2020-05-15, 15:23   #12
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Quote:
Originally Posted by Alberico Lepore View Post
so in our case to have any valid a we have to increase h
Do you think it is better to enlarge h and choose a finite number of a and have a factoring of the transformed number easier or to try with a, keeping a finite number of h?
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Old 2020-05-15, 15:25   #13
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Quote:
Originally Posted by Alberico Lepore View Post
Do you think it is better to enlarge h and choose a finite number of a and have a factoring of the transformed number easier or to try with a, keeping a finite number of h?
No.
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Old 2020-05-15, 15:45   #14
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Quote:
Originally Posted by retina View Post
No.
could I know your opinion?
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Old 2020-05-15, 15:59   #15
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Quote:
Originally Posted by Alberico Lepore View Post
could I know your opinion?
I gave it. I don't think any of the options you suggest will be better.
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Old 2020-05-15, 16:08   #16
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Quote:
Originally Posted by retina View Post
I gave it. I don't think any of the options you suggest will be better.
And what is the best solution applied to these equations?
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Old 2020-05-15, 16:11   #17
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Quote:
Originally Posted by Alberico Lepore View Post
And what is the best solution applied to these equations?
There is no best solution. You're posting stuff that doesn't work on anything except your cherry picked values like 65 and 187.

Last fiddled with by retina on 2020-05-17 at 02:44
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Old 2020-05-15, 20:53   #18
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Quote:
Originally Posted by retina View Post
There is no best solution. Your posting stuff that doesn't work on anything except your cherry picked values like 65 and 187.
this is the algorithm related to [1]

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 ?


I would like to implement this algorithm
but I still can't implement it

N^2*x*(x+2+4*h)-[8*((96 k² + 24 k + 1)-3*a*(a-1)/2)+1]/3=0

[N^2*x^2+(4*h+2)*N^2*x]-[256*k^2+64*k]+[4*a^2-4*a]-3=0

[[N*x+(2*h+1)*N]^2-(2*h+1)^2*N^2]-[[16*k+2]^2-4]+[[2*a-1]^2-1]-3=0

[N*x+(2*h+1)*N]^2-[16*k+2]^2=(2*h+1)^2*N^2-[2*a-1]^2-4+3+1

now I put a semiprime in place of N to continue the explanation

solve [N*x+(2*h+1)*N]^2-[16*k+2]^2=(2*h+1)^2*N^2-[2*a-1]^2-4+3+1 , [N*x+(2*h+1)*N]-[16*k+2] =2*12*N,x=3,N=66390187 ,a

-> 8*k-66390187*h+663901871=0

apply the Generalized Euclidean algorithm

->

k= 24896320 + 66390187 t
h = 13 + 8 t

but h > 0

little bruteforce from -1 to .....
for t =-1

k=-41493867
h=5

->

a=165975468

->

k=-41493867 ,a=165975468 , sqrt[(8*(96 k² + 24 k + 1)+1)/3+1]-(2*a-1)=P

-> P=99915

GCD(66390187,99915)=6661


what do you think?
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Old 2020-05-16, 08:29   #19
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Generalized algorithm of [1]

V=1

While(GCD(N,P)!= p && GCD(N,P)!=q){


solve [N*x+(2*h+1)*N]^2-[16*k+2]^2=(2*h+1)^2*N^2-[2*a-1]^2-4+3+1 , [N*x+(2*h+1)*N]-[16*k+2] =2*12*V^2*N,x=3 ,a,k
->
a=1/2 (1 + sqrt(N^2 (1 + 4 h + 4 h^2 - 192 V^2 - 96 h V^2 + 576 V^4)))

8*k-N*h+(12*V^2-2)*N+1=0

apply the Generalized Euclidean algorithm 8*k-N*h+(12*V^2-2)*N+1=0 nel paramether t

vary t in h with h>=0 until a has complex roots

in the meantime, calculate

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

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

if GCD == p || q -> the algorithm ends

otherwise V=V+1

}


someone please could try it for pleasure

Last fiddled with by Alberico Lepore on 2020-05-16 at 09:01
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Old 2020-05-17, 00:08   #20
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[187*x+(2*h+1)*187]^2-[16*k+2]^2=(2*h+1)^2*187^2-[2*a-1]^2-4+3+1
,
x=3
,
h=1
,
sqrt[(8*(96 k² + 24 k + 1)+1)/3+1]+(2*a-1)=27*187*[2*(3*z+1-(z-y+1))+1-(4*y-2)]
,
sqrt[(8*(96 k² + 24 k + 1)+1)/3+1]-(2*a-1)=[2*(3*z+1-(z-y+1))+1]
,
70=3*z*(z+1)/2-3*y*(y-1)/2+(3*z+1)*(3*z+2)/2

Last fiddled with by Alberico Lepore on 2020-05-17 at 00:18
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Old 2020-05-17, 03:28   #21
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Quote:
Originally Posted by Alberico Lepore View Post
N=66390187
Good. We are getting somewhere.

Now apply this to a larger number, say 15 digits long. And report back how much "little bruteforce from -1 to ....." you needed to do.
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Old 2020-05-17, 10:33   #22
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Quote:
Originally Posted by retina View Post
Good. We are getting somewhere.

Now apply this to a larger number, say 15 digits long. And report back how much "little bruteforce from -1 to ....." you needed to do.
I understand now that GCD (N, P) is always equal to N

give values to a and apply Lenstra elliptic-curve factorization hoping that the factors of the transformed number are less than 40 digits



[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
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