RSA factorization conjecture
This conjecture is valid for N = a * b with a in the form a = 4 * G + 3
The part (h-k)/(a-3)=Z is the conjecture the rest is shown
Suppose N = 5369
[34+34+24k](k+1)/2=h
,
k=(a-7)/4
,
(6+6+8(w-1))(w)/2=z
,
w=(a-3)/4
,
5369-n*a-h-z=9
,
(h-k)/(a-3)=Z
solving Z with variable a we have:
Z=(3a^2+3a-32)/(4*(a-3))
therefore (3a^2+3a-32)/(a-3)
3(a-3)^2=3(a^2-6a+9)=3a^2-18a+27 -> (3a^2-18a+27)-(3a^2+3a-32)=-21a+59
therefore (-21*a+59)/(a-3)
21(a-3)=21a-63 -> (-21a+59)-(21a-63)=112
then 112/(a-3)
so just factor 112 = 2 * 2 * 23
follows a-3 = 56 -> a = 59
what do you think?