[issue43053] Speed up math.isqrt, again

[Mark Dickinson]

Mark Dickinson <dickinsm at 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).

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<https://bugs.python.org/issue43053>
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