RSA and number factoring, can be used for solving CTF challenges. It is fast enough for up to 64 bit numbers.
import math
from typing import Generator
def rsa(n:int)->tuple[int,int]:
if n==0: return None
if n % 2 ==0: return (2,n//2)
i=math.isqrt(n)
i=i+1+(i%2)
while(i:=i-2):
if n%i==0:
return (n//i,i)
def factor(n:int)->Generator[int,None,None]:
factor1,factor2=rsa(n)
if factor2 ==1:
yield n
else:
yield from factor(factor1)
yield from factor(factor2)Usage:
print(list(factor(500)))