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There are some good reasons why this hasn't been tried. Firstly, it's not "discrete boundaries" but all points with integer coordinates $(x,y)$ on the hyperbola $xy=N$ which are candidates for factorization. The only information that matters is in those points, it doesn't matter if one uses your $$ (N-xy)^2 $$ and look for zeroes, or if one uses $$ ...


0

Mathematically, yes, it will work. Practically, you will require an extremely very long time and an incredible amount of energy, considering the sizes of the primes involved in RSA (usually around 1024-bit prime numbers). It is about billion and billions of years and billions and billions times the energy of the whole universe (RSA: How effective is this ...


4

The fancy name for this is factorization with an implicit hint. If the primes are unbalanced, i.e., if $\log_2 p_i > 2 \log_2 q_i$, we know how to factor $n_1$ and $n_2$ quite easily. Let $k$ be the number of bits $p_1$ and $p_2$ differ by (a very small $k$, usually, for $(p, p+2)$); reduce the following lattice $$ \begin{pmatrix} 2^k & 0 & n_2 ...



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