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Given $N=pq$, would $\frac{p-1}{2}$ steps be fast compared with extant factoring methods?

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That is precisely as fast as trial division by all odd numbers (provided a single trial division has cost $O(1)$) – Hagen von Eitzen Feb 17 '13 at 16:12

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).

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thanks for the response. I see that now. – ratking Feb 19 '13 at 22:51

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