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Oct 2, 2015 at 21:29 comment added Igor Rivin @johnmangual wikiwand.com/en/Integer_factorization#/…
Oct 2, 2015 at 21:20 comment added john mangual The factorization step takes $log(n)$ time or something ?
Oct 2, 2015 at 21:01 comment added Igor Rivin @johnmangual it's a lot better than that, see the edited version.
Oct 2, 2015 at 21:00 history edited Igor Rivin CC BY-SA 3.0
added complexity estimate
Oct 2, 2015 at 20:35 comment added john mangual For each of the numbers $k < \sqrt{n}$ I can factor $n - k^2$ which I can do in $\sqrt{n}$ time, so the total run time is $\sqrt{n} \times \sqrt{n} = n$.
Oct 2, 2015 at 20:15 history answered Igor Rivin CC BY-SA 3.0