Question

我正在编写一个程序，该程序将为较弱的RSA公钥计算私钥。我想知道如何从值p确定q和n的值。到目前为止，这是Python代码：

from Crypto.PublicKey import RSA #PyCryptoDome
import .math as cm # My own module

with open(public_keyfile, 'rb') as key: # Public Keyfile Is in PEM format
    public_key = RSA.import_key(key)

n = public_key.n # N value of the public_key

e = public_key.e # E value of the public_key

p, q = get_factors_of(n) # This I don't know how to do, though there is a question that might help [see bottom]

t = cm.lcm(p-1, q-1) # Get the lowest common multiple of q and q

d = cm.mod_inverse(e, t) # Get d, the modular inverse of e % t

private_key = RSA.construct((n, e, d, p, q) # Construct the RSA private_key

上面引用的.math模块：

from math import gcd


def mod_inverse(a, b):
    a = a % b
    for x in range(1, b):
        if (a * x) % b == 1:
            return x
    return 1


def lcm(x, y):
    return x * y // gcd(x, y)

我需要做的事情似乎已被引用 here，但此代码是Java语言。

如果有人知道如何使用python从p获取q和n，将不胜感激。

非常感谢，Legolooj。

Answer 1

强制性警告：如果您追求性能，则需要亲自调查算法的详细信息。即便是“弱”的公钥，也将通过简单的算法（例如Erathostene的筛子）永久破解。

话虽这么说，sympy.ntheory.factorint()可能是您所需要的：

from sympy.ntheory import factorint

print(factorint(54))  # {2: 1, 3: 3} i.e. 54 == 2**1 * 3**3

Answer 2

经过大量的Google搜索和pdf阅读后，我发现了一种有效的算法。这是一个python实现：

import math
def get_factors_of(num):
    poss_p = math.floor(math.sqrt(num)) 

    if poss_p % 2 == 0: # Only checks odd numbers, it reduces time by orders of magnitude
        poss_p += 1
    while poss_p < num:
        if num % poss_p == 0:
            return poss_p
        poss_p += 2

此算法有效地找到了一个小的RSA密钥的P / Q因子。（我已经针对64位PEM公钥对其进行了测试）

如何使用python

2 个答案: