["python","math","integer","sqrt"]
Integer square root in python

Is there an integer square root somewhere in python, or in standard libraries? I want it to be exact (i.e. return an integer), and raise an exception if the input isn't a perfect square. I tried using this code: def isqrt(n): i = int(math.sqrt(n) + 0.5) if i**2 == n: return i raise ValueError('input was not a perfect square') But it's ugly and I don't really trust it for large integers. I could iterate through the squares and give up if I've exceeded the value, but I assume it would be kinda slow to do something like that. Also, surely this is already implemented somewhere? See also: Check if a number is a perfect square.
Note: There is now math.isqrt in stdlib, available since Python 3.8. Newton's method works perfectly well on integers: def isqrt(n): x = n y = (x + 1) // 2 while y < x: x = y y = (x + n // x) // 2 return x This returns the largest integer x for which x * x does not exceed n. If you want to check if the result is exactly the square root, simply perform the multiplication to check if n is a perfect square. I discuss this algorithm, and three other algorithms for calculating square roots, at my blog.