 mersenneforum.org yet another attempt to bring down RSA
 Register FAQ Search Today's Posts Mark Forums Read  2020-05-15, 15:23   #12
Alberico Lepore

May 2017
ITALY

38810 Posts Quote:
 Originally Posted by Alberico Lepore 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?   2020-05-15, 15:25   #13
retina
Undefined

"The unspeakable one"
Jun 2006
My evil lair

25·173 Posts Quote:
 Originally Posted by Alberico Lepore 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.   2020-05-15, 15:45   #14
Alberico Lepore

May 2017
ITALY

22·97 Posts Quote:
 Originally Posted by retina No.   2020-05-15, 15:59   #15
retina
Undefined

"The unspeakable one"
Jun 2006
My evil lair

15A016 Posts Quote:
 Originally Posted by Alberico Lepore could I know your opinion?
I gave it. I don't think any of the options you suggest will be better.   2020-05-15, 16:08   #16
Alberico Lepore

May 2017
ITALY

22·97 Posts Quote:
 Originally Posted by retina 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?   2020-05-15, 16:11   #17
retina
Undefined

"The unspeakable one"
Jun 2006
My evil lair

10101101000002 Posts Quote:
 Originally Posted by Alberico Lepore 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   2020-05-15, 20:53   #18
Alberico Lepore

May 2017
ITALY

22×97 Posts Quote:
 Originally Posted by retina 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 

Quote:
 Originally Posted by Alberico Lepore 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  [8*(96 k² + 24 k + 1)+1]/3 >= N^2*x*(x+2+4*h)  [8*(96 k² + 72 k + 13)+1]/3 >= N^2*x*(x+2+4*h)  [8*(96 k² + 120 k + 37)+1]/3 >= N^2*x*(x+2+4*h)  [8*(96 k² + 168 k + 73)+1]/3 >= N^2*x*(x+2+4*h)  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  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  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  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?   2020-05-16, 08:29 #19 Alberico Lepore   May 2017 ITALY 22·97 Posts Generalized algorithm of  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   2020-05-17, 00:08 #20 Alberico Lepore   May 2017 ITALY 22·97 Posts [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   2020-05-17, 03:28   #21
retina
Undefined

"The unspeakable one"
Jun 2006
My evil lair

25·173 Posts Quote:
 Originally Posted by Alberico Lepore 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.   2020-05-17, 10:33   #22
Alberico Lepore

May 2017
ITALY

22·97 Posts Quote:
 Originally Posted by retina 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



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



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



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



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   Thread Tools Show Printable Version Email this Page Similar Threads Thread Thread Starter Forum Replies Last Post Alberico Lepore Alberico Lepore 9 2020-05-01 09:41 jvang jvang 10 2019-02-03 23:17 wildrabbitt Hardware 0 2016-05-22 17:22 ixfd64 Lounge 7 2005-11-30 17:39 JHansen Factoring 34 2005-05-27 19:24

All times are UTC. The time now is 01:45.

Thu Jul 9 01:45:19 UTC 2020 up 105 days, 23:18, 0 users, load averages: 0.93, 1.21, 1.37 