A probable attack for RSA (factorization): how to improve it?

Question

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?

Answer 1

This appears to be a special case of congruence of squares factoring. I don't see any particular reason why this form of congruence should be easier to solve than the generic form.

The congruence of squares method has of course led to many of the best factorisation algorithms including the best-known general purpose classical algorithm, the number field sieve.