what do you think of this?

Lost Factorization


Each number H=x*y && (y-x) mod 8 == 0

K=(H-1)/8 ed n=y-x

then one of these is true

1
solve (H-1)/8=3*x*(x+1)/2-3*y*(y-1)/2+(3*x+1)*(3*x+2)/2
2
solve (H-1)/8=3*x*(x+1)/2-3*y*(y-1)/2+(3*x-1+1)*(3*x-1+2)/2-(x-y+1)
3
solve (H-1)/8=3*x*(x+1)/2-3*y*(y-1)/2+(3*x-2+1)*(3*x-2+2)/2-2*(x-y+1)

We observe in particular that in 1 :

H =0 mod 3

therefore H/3=N=p*q

when p=[2*(3*x+1-(x-y+1))+1-(4*y-2)] and q=[2*(3*x+1-(x-y+1))+1]

and p+q = 4 mod 8

H has the characteristic that

H-1 = 0 mod 8

and

((H-1)/8-1)=0 mod 3

Let's rewrite p as a function of only x where M=(H-1)/8

solve
p=[2*(3*x+1-(x-y+1))+1-(4*y-2)]
,
3*x*(x+1)/2-3*y*(y-1)/2+(3*x+1)*(3*x+2)/2=M

and we get

p=[12*x+6-sqrt[3*(48*x^2+48*x+11-8*M)]]/3

A=((H-1)/8-1)/3

has the characteristic

A-(x-1)*p=((p*3*p*3-1)/8-1)/3-p*(p-3)/2

and in a about c of p, x will be very near (d) to x0 therefore


If N=p*q e (p+q-4) mod 8 =0 e p >= (p+q)/4

N=66390187

(3*N-1)/8=24896320

((3*N-1)/8-1)/3=8298773

8298773-((x+d)-1)*(p-c)=(((p-c)*3*(p-c)*3-1)/8-1)/3-(p-c)*((p-c)-3)/2
,
c=1/100000000
,
p=[12*x+6-sqrt[3*(48*x^2+48*x+11-8*24896320)]]/3

to vary d in
d=1/10000000000
or
d=2/10000000000
or
d=3/10000000000
or
d=4/10000000000
or
d=5/10000000000
or
d=6/10000000000
or
d=7/10000000000
or
d=8/10000000000
or
d=9/10000000000
or
d=10/10000000000

for d=6/10000000000 -> x0=2078 and x=2075,.....
What size order should c and d have for p and N and for each other?