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For the sake of argument, let's use your code above as our basic requirements. Let's say we have access to sympy and basic Python, but nothing else. As a baseline, let's consider a trivial prime number finder that use nothing but pure Python. We'll just count up from 3 up to sqrt(n): it's naive, it's dead simple to write, and it's actually reasonably fast just because Python is a decent language (and it even has an O(sqrt(N)) runtime, which is not a bad place to start):

For the sake of argument, let's use your code above as our basic requirements. Let's say we have access to sympy and basic Python, but nothing else. As a baseline, let's consider a trivial prime number finder that use nothing but pure Python. We'll just count up from 3 up to \$\sqrt{n}\$: it's naive, it's dead simple to write, and it's actually reasonably fast just because Python is a decent language (and it even has an \$O(\sqrt{N})\$ runtime, which is not a bad place to start):

def naive_is_prime(n):
    if n <= 2 or n % 2 == 0:
        return False

    # Consolidated code thanks to GZ0's comment
    return all(n % i != 0 for i in range(3, int(n ** 0.5 + 1), 2)) 

def naive_is_prime(n):
    if n == 2:
        return True
    if n < 2 or n % 2 == 0:
        return False

    # Consolidated code thanks to GZ0's comment
    return all(n % i != 0 for i in range(3, int(n ** 0.5 + 1), 2)) 

import math

def naive_is_prime(n):
    if n <= 2 or n % 2 == 0:
        return False

    for i in range(3, int(math.ceil(math.sqrt(n))), 2):
        if n % i == 0:
            return False

    return True

def naive_is_prime(n):
    if n <= 2 or n % 2 == 0:
        return False

    # Consolidated code thanks to GZ0's comment
    return all(n % i != 0 for i in range(3, int(n ** 0.5 + 1), 2))