Is cube root integer?

math, python, python-3.x

Solution

For small numbers (<~1013 or so), you can use the following approach:

def is_perfect_cube(n):
    c = int(n**(1/3.))
    return (c**3 == n) or ((c+1)**3 == n)

This truncates the floating-point cuberoot, then tests the two nearest integers.

For larger numbers, one way to do it is to do a binary search for the true cube root using integers only to preserve precision:

def find_cube_root(n):
    lo = 0
    hi = 1 << ((n.bit_length() + 2) // 3)
    while lo < hi:
        mid = (lo+hi)//2
        if mid**3 < n:
            lo = mid+1
        else:
            hi = mid
    return lo

def is_perfect_cube(n):
    return find_cube_root(n)**3 == n

Problem

This seems to be simple but I cannot find a way to do it. I need to show whether the cube root of an integer is integer or not. I used `is_integer()` float method in Python 3.4 but that wasn't successful. As ``` x = (3**3)**(1/3.0) is_integer(x) True ``` but ``` x = (4**3)**(1/3.0) is_integer(x) False ``` I tried `x%1 == 0`,`x == int(x)` and `isinstance(x,int)` with no success. I'd appreciate any comment.

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