#Ransomware CURE
#RSA#Cryptography#hacker#cybersecurity
to factor N = (6 * a + 1) * (6 * b + 1)
it takes 9 cases
I'll show you 1 of 9 when a=3*v+1 and
[2*((N-1)/6)^2+(N-1)/6] mod N] mod 8 =3
In particular to find the general solution to case 1 of 9
you have to study this
[[[2*((N-1)/6)^2-(N-1)/6] mod N]+5]/8-(1/3*(8*a+1)+5)/8+c*(6*a+1)=X , a=3*n+1 -> X=(c*18-1)*n+(7*c+M-1) where M=[[[2*((N-1)/6)^2-(N-1)/6] mod N]+5]/8
and
(c*18-1) and (7*c+M-1) they must have one factor in common
so i thought of a little bruteforce [but ....]
(c*18-1)=s*z
,
(7*c+M-1)=t*z
Example
N=25*55
77-(1/3*(8*a+1)+5)/8+2*(6*a+1)=X , a=3*n+1
X=(35*n+90)-> 7*n+18
18*(7*n+18)-7*(6*(3*n+1)+1)=275
GCD(275,1375)=25
WOW
studying studying I discovered that bruteforce is not needed, in fact
(c*18-1)+(7*c+77-1)=t*z
G=t*z
25*c-G+75=0
Generalized Euclid's algorithm
G=25u
c=u-3
GCD(G,N)=factor of N
what do you think about it?