在此python主要因子程序中寻找反例

Question

I cobbled together this script to spit out the prime factorization of an integer n . At the very end of the script, I multiply the current list of prime factors, including multiplicity, and check if its equal to n ; if not, I divide out that product and append the quotient if it's prime.

I've tested a ton of numbers, and it's worked each time, but mathematically I'm not sure why. Shouldn't I have to loop that bit about checking for leftover primes in the range (sqrt(n), n) ? Couldn't there be an integer with two primes in that range? If there is, I haven't been able to find an example yet. I'm new to python, so another explanation is that I don't understand my own code, and its looping for me somehow.

from math import sqrt
n = int(input("Let's factor a number:"))

def isPrime(x):
    for i in range(2,int(sqrt(x)+1)):
        if x%i==0:
            return False
    return True

m = int(sqrt(n)+1)
i = 1
factors = []
while(1 < m):
    if (n % (m**i) == 0 and isPrime(m) == True):
        factors.append(m)
    else:
        m -= 1
        i = 1
        continue
    i += 1

prod = 1
for i in factors:
    prod *= i

if prod == n:
    print(factors)
elif isPrime(n/prod) == True:
    factors.append(int(n/prod))
    print(factors)

Answer 1

No, you don't need such a loop. n can have only one prime factor > sqrt(n) . Assume the opposite: there are at least two. i, j > sqrt(n). In this case, i*j must be > sqrt(n)*sqrt(n), and there's your contradiction.

Once you've found all factors less than sqrt(n) , any remaining amount is prime, and would be the only factor larger than sqrt(n) .