#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?