Square free number

Square free no: It is an integer which is divisible by no other
perfect square than 1. For example, 10 is square-free but 18 is not,
as 18 is divisible by 9 = 3^2.


The smallest positive square-free numbers are 1, 2, 3, 5, 6, 7, 10,
11, 13, 14, 15, 17, 19, 21, 22, 23, 26, 29, 30, 31, 33, 34, 35, 37,
38, 39

Is it possible to reduce the time execution for this code, without importing any Python module?

What should I use in place of a for loop body containing an if condition? Will using a lambda function be helpful?

def squarefree(n): 

 for i in range (2, round(n**0.5)): 
    if n % (i**2) == 0: 
        return False 
 return n

Review

Python has an official style-guide, PEP8. It recommends using white-space only where necessary (so the blank line after def squarefree(n): could go, as could the trailing space) and using four spaces as indentation.

Normally I would consider it best practice for a function like this to return actually True or False. Your function returns False or a truthy value, which also works, but is not as clean. Consider changing it from return n to return True.

Alternate implementation

As @Dair mentioned in the comments, depending on your use-case for this, it would be faster to write a sieve that gets all square-free numbers up to some limit.

Starting with the Sieve of Eratosthenes, taken from another of my answers, this could be implemented like this:

def squarefree(n):
    """
    Check if n is a square-free number, i.e. is divisible by no other perfect square than 1.

    Args:
        n     positive integer to check
    Returns:
        n     if n is a square-free number
        False else
    """
    for i in range(2, round(n**0.5) + 1):
        if n % (i**2) == 0:
            return False
    return n

def square_free_sieve(limit):
    """Generator that yields all square free numbers less than limit"""
    a = [True] * limit
    # Needed so we don't mark off multiples of 1^2
    yield 1
    a[0] = a[1] = False
    for i, is_square_free in enumerate(a):
        if is_square_free:
            yield i
            i2 = i * i
            for n in range(i2, limit, i2):
                a[n] = False

test1 = [n for n in range(100) if squarefree(n)]
test2 = list(square_free_sieve(100))
assert test1 == test2

When generating all square free numbers up to 100,000, your code needs about 3.5s, while the sieve only takes 0.05s, which is about 70 times faster.

StackExchange Code Review Q#151949, answer score: 6