A probable attack for RSA (factorization): how to improve it?
$N=8*G+3$ can be factored if there is a non-trivial negative $k$ such that
$\frac{(N*(9+24*k)-3)}{8}=-6*m^2 $
[to exclude the two trivial solutions $m=\frac{(N+1)}{4}-N*t$ and $m=N-\frac{(N+1)}{4}-N*t $]
given the system
$
\frac{[[8*[\frac{(N*(9+24*k)-3)}{8}+2*x^2*h^2-2*x^2+2*x*h-2*x]+3+6*n-(n*(n+4))]-4*n*y-3]}{8}-[4-\frac{(-(2*h-2)-7)*(-(2*h-2)-5)}{8}] = -(2*h-2)*(h*x-1)
$
,
$
h=-4*sqrt[\frac{-(N*(9+24*k)-3)}{48}]
$
,
$
n=h^2-1
$
,
$
(4*x+2)^2-(2*y-1)^2=N*(9+24*k)
$
you will have
$GCD(4*x+1-2*(y-1),N)= p || q || N$
Example
$
\frac{[[8*[\frac{(N*(9+24*k)-3)}{8}+2*x^2*h^2-2*x^2+2*x*h-2*x]+3+6*n-(n*(n+4))]-4*n*y-3]}{8}-[4-\frac{(-(2*h-2)-7)*(-(2*h-2)-5)}{8}] = -(2*h-2)*(h*x-1)
$
,
$
h=-4*sqrt[\frac{-(N*(9+24*k)-3)}{48}]
$
,
$
n=h^2-1
$
,
$
(4*x+2)^2-(2*y-1)^2=N*(9+24*k)
$
,
$k=-10$
,
$N=187$
$->x=30 ; y=121$
$GCD(4*30+1-2*(121-1),187)=17$
I had thought of this solution:
calculate $k$ as a function of $x$ or $y$ and see with the Coppersmith method if there is an $x <sqrt (N)$
if it exists and all the variables of the system are integers then we are done and we will find $p$ and $q$
Example as a function of $x$
$N*k=\frac{(-3*N-16*x^2+1)}{8}$
$N*k=\frac{(-3*N-16*x^2-32*x-15)}{8}$
how to improve it?
Do you have any ideas?
