Tim Peters <[email protected]> added the comment:
Suppose we added isqrt_rem(n), returning the integer pair (i, rem) such that
n == i**2 + rem
0 <= rem <= 2*i
Then:
- Want the floor of sqrt(n)? i.
- The ceiling? i + (rem != 0).
- Rounded? i + (rem > i).
- Is n a perfect square? not rem.
That's how mpz addresses these, although it has a different function to compute
the floor without returning the remainder too.
I wouldn't object to that - just +0, though. Depending on implementation
details, which I haven't investigated, it may or may not be materially faster
than doing:
def isqrt_rem(n):
return (i := isqrt(n)), n - i*i
myself.
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<https://bugs.python.org/issue46187>
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