RSA factorization conjecture

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^2-6*a+9)=3*a2-18a+27 -> (3a^{2-18*a+27)-(3*a}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?