Python: Function to determine if number is square, cube, etc

Question

My simple problem is to create a function that determines if a number N can be written as a^n for some given n, i.e. I need to check if N^(1/n) is a whole number. Somehow this function yields wrong results:

def is_power(N, n):
    r = float(N) ** ( 1. / float(n) )
    return r.is_integer()

For n=2 it works.

For n=3 and N=1,8,27 the function yields True, which is correct. But from then on False, e.g. for 4*4*4=64 or 5*5*5=125. How can I create a working function that finds numbers that are squares/cubes/etc.?

Thank you. How can I create a function that does not fail due to floating point arithmetic issues? — Fabian Schn., Jul 27 '17 at 16:02
Would either of http://mpmath.org/ or http://pythonhosted.org/bigfloat/ be helpful? — , Jul 28 '17 at 02:51

Rory Daulton · Accepted Answer · 2017-07-27T16:08:00.063

Floating point arithmetic is not exact--see Is floating point math broken?.

So check your answer using exact-integer math. Round r to the nearest integer then see if the power works. This Python 3 code removes some of your redundant type-casting. For Python 2, wrap the calculation of r into an int() typecast (which is not needed for Python 3).

def is_power(N, n):
    r = round(N ** (1.0 / n))
    return r**n == N

Python: Function to determine if number is square, cube, etc

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