Mark Dickinson <dicki...@gmail.com> added the comment:
> the only thing I'm not sure about is whether the final correction in the > original `isqrt` is needed Well, *some* part of the algorithm has to make use of the low-order bits of n. Otherwise we won't be able to distinguish n = 4a**2 + 4a + 1 (whose isqrt is 2a + 1) from 4a**2 + 4a (whose isqrt is 2a). ---------- _______________________________________ Python tracker <rep...@bugs.python.org> <https://bugs.python.org/issue43053> _______________________________________ _______________________________________________ Python-bugs-list mailing list Unsubscribe: https://mail.python.org/mailman/options/python-bugs-list/archive%40mail-archive.com
