Given $N=pq$, would $\frac{p-1}{2}$ steps be fast compared with extant factoring methods?
Tell me more
×
Mathematics Stack Exchange is a question and answer site for
people studying math at any level and professionals in related fields. It's 100% free, no registration required.
|
|
See for example Wikipedia, where sub-exponential methods are discussed (note that this is measured in terms of the bits needed to represent $N$, not of $N$ itself). |
|||
|
|
