Calculating Odd/Even variety of divisors in Python

The problem

Given an integer n return "odd" if the variety of its divisors is odd. In any other case, return "even".

Be aware: large inputs will probably be examined.

Examples:

All prime numbers have precisely two divisors (therefore "even").

For n = 12 the divisors are [1, 2, 3, 4, 6, 12] – "even".

For n = 4 the divisors are [1, 2, 4] – "odd".

The answer in Python code

Choice 1:

def oddity(n):
    #your code right here
    return 'odd' if n**0.5 == int(n**0.5) else 'even'

Choice 2:

import math
def oddity(n):
    return math.sqrt(n) % 1 == 0 and 'odd' or 'even'

Choice 3:

oddity=lambda n: ["odd","even"][n**.5%1!=0]

Take a look at instances to validate our answer

import take a look at
from answer import oddity

@take a look at.describe("Pattern exams")
def exams():
    @take a look at.it("Some examples")
    def exams():
        take a look at.assert_equals(oddity(1), 'odd')
        take a look at.assert_equals(oddity(5), 'even')
        take a look at.assert_equals(oddity(16), 'odd')