返回x的除数的最小质数

您陷入无限循环，因为如果不满足返回条件，则不更改n的值，因为您可以看到返回条件仅在您的数字x是2的倍数时得到满足改变 ：

if x % n == 0:
    x += 1

与：

while n <= x:
    if x % n == 0:
        return x
    n += 1

为了优化代码，您可以搜索一个以质数n来除以int（math.sqrt（x）+ 1）的x：

import math

def smallest_prime_factor(x):
    """Returns the smallest prime number that is a divisor of x"""
    # Start checking with 2, then move up one by one
    n = 2
    max_n = int(math.sqrt(x) + 1)
    while n < max_n:
        if x % n == 0:
            return n
        n += 1

    return x

甚至更好的是，您可以使用Eratosthenes筛子快速生成质数并针对x进行测试：

# Sieve of Eratosthenes
# Code by David Eppstein, UC Irvine, 28 Feb 2002
# http://code.activestate.com/recipes/117119/

def gen_primes(y):
    """ Generate an infinite sequence of prime numbers.
    """
    # Maps composites to primes witnessing their compositeness.
    # This is memory efficient, as the sieve is not "run forward"
    # indefinitely, but only as long as required by the current
    # number being tested.
    #
    D = {}

    # The running integer that's checked for primeness
    q = 2

    while q < y:
        if q not in D:
            # q is a new prime.
            # Yield it and mark its first multiple that isn't
            # already marked in previous iterations
            # 
            yield q
            D[q * q] = [q]
        else:
            # q is composite. D[q] is the list of primes that
            # divide it. Since we've reached q, we no longer
            # need it in the map, but we'll mark the next 
            # multiples of its witnesses to prepare for larger
            # numbers
            # 
            for p in D[q]:
                D.setdefault(p + q, []).append(p)
            del D[q]


def smallest_prime_factor(x):
    """Returns the smallest prime number that is a divisor of x"""
    # Start checking with 2, then move up one by one

    return next((i for i in gen_primes(int(math.sqrt(x) + 1)) if x % i == 0), x)

回答来源：stackoverflow