2022 | LITCTF | crypto

Dynamic RSA

#rsa #gcd #crt

题目¶

Nowadays, clients just keep changing their requirements! They said e=65537 is too boring for RSA and that they wanted a dynamic encryption system instead. Oh, I'll give it to them!

Connect with nc litctf.live 31792

https://drive.google.com/uc?export=download&id=1JY3LfzcoIWUEhC0C8NuwO-fOwZnp3tKt

解题思路¶

已知经过 RSA 加密的 Flag 的密文及随机数种子/加密指数 \(e\)

flag = open('flag.txt','rb').read();

m = bytes_to_long(flag);
e = 65537;
p = getPrime(200);
q = getPrime(200);
random.seed(e);
phi = (p-1) * (q-1);
n = p * q;
ct = pow(m, e, n)
print("My secret flag is " + str(ct));

可以与服务器进行两种交互

提供私钥，获得 Flag 密文的解密结果
提供任意消息，获得「消息+盐值」的加密结果

while True:
    inp = input("Guess Private Key (1) or Encrypt Message (2): ");
    if (inp == "1"):
        d = int(input("Enter Private Key: "));
        print(long_to_bytes(pow(ct, d, n)));
        exit()

    elif (inp == "2"):
        test_e = e_gen()
        # Oh great they made me change the algorithm again
        # They said its "TOO BLAND"
        # FINE I'll add more seasoning
        salt = get_random_bytes(8).hex();
        inp = bytes_to_long((input("Enter Message: ") + salt).encode());
        test_ct = pow(inp, test_e, n);
        print("Your Message (remember to convert): " + str(test_ct));

    else:
        print("BAD OPTION");
        exit();

不过重点在函数 e_gen 以及 gcd。随机数种子已知，则可以求出 test_e。因为 test_e 是质数，所以 gcd(test_e, phi) 只有两种结果，要么是 \(1\)，要么是 test_e。当 gcd(test_e, phi) 的结果为 test_e 时，new_e = 1，此时密文与明文相同，可以确定 phi 的其中一个因数。但显然，结果 gcd(test_e, phi) 为 \(1\) 的情况更多

def gcd(a,b):
    # Client said the loading screen is too boring
    # So they want something with more flair and movement while they wait
    if(a == 0):
        return b;
    if(b == 0):
        return a;
    print(".,"[(b // a) & 1], end = "");

    return gcd(b % a,a);

# Clients keep complaining that making e always 65537 is too boring
# So they changed their requirements and wanted a "dynamic encryption system"
# I literally can't
def e_gen():
    print("Loading", end = "")
    test_e = nextprime(random.randint(1, 100000));
    # Okay but I literally can't use a random e if gcd is not 1
    # It's like most fundamental part of RSA!!!
    new_e = test_e // gcd(test_e, phi);
    print()
    return new_e

重写的函数 gcd 给出了每一步 b // a 结果的奇偶性，因为第一步 phi 的值未知，所以从第二步开始考虑，即 gcd(phi % test_e, test_e)。test_e 已知，phi % test_e 是小于 test_e 的自然数，仅根据每一步 b // a 结果的奇偶性是有一定概率能够确定 phi % test_e 的。以 \(13\) 为例 gcd 的计算结果如下，可见部分 b // a 奇偶性结果序列与 phi % test_e 有一对一映射关系
```
gcd(1, 13) ,
gcd(2, 13) ..
gcd(3, 13) .,
gcd(4, 13) ,.
gcd(5, 13) .,,.
gcd(6, 13) ..
gcd(7, 13) ,,.
gcd(8, 13) ,,,,.
gcd(9, 13) ,..
gcd(10, 13) ,,,
gcd(11, 13) ,,.
gcd(12, 13) ,.
```

由此可利用中国剩余定理计算出 phi，从而得出私钥 d

import random, pwn
from sympy import nextprime
from sage.all import crt, Integer
from Crypto.Util.number import inverse

def gcd(a, b):
    if a == 0 or b == 0:
        return ''
    return '.,'[(b // a) & 1] + gcd(b % a, a)

e = 65537

random.seed(e)
state = random.getstate()

gcd_res = dict()
for i in set(nextprime(random.randint(1, 100000)) for _ in range(750)):
    gcd_res[i] = ['']
    for j in range(1, i):
        gcd_res[i].append(gcd(j, i))

conn = pwn.remote('litctf.live', 31792)
random.setstate(state)

mo, rem = [], []
pre_crt = 0
while True:
    conn.sendafter('Encrypt Message (2): ', '2\n')
    test_e = nextprime(random.randint(1, 100000))
    res = conn.recvline().decode().strip()[7:]
    conn.sendafter('Enter Message:', '0\n')
    if len(res) == 1 and test_e not in mo:
        mo.append(Integer(test_e))
        rem.append(Integer(0))
        cur = crt(rem, mo)
        if cur == pre_crt:
            break
        pre_crt = cur
    if test_e in gcd_res and test_e not in mo and gcd_res[test_e].count(res[1:]) == 1:
        mo.append(Integer(test_e))
        rem.append(Integer(gcd_res[test_e].index(res[1:])))
        cur = crt(rem, mo)
        if cur == pre_crt:
            break
        pre_crt = cur
    print(pre_crt)

d = inverse(e, pre_crt)
conn.sendafter('Encrypt Message (2): ', '1\n')
conn.sendafter('Enter Private Key:', f'{d}\n')
conn.interactive()

Flag¶

LITCTF{0op5i3_dyn4m1c_n0t_gr3at_1t_s33m5}

最后更新: 2022年8月2日 16:15:12

Contributors: YanhuiJessica
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Dynamic RSA

题目¶

解题思路¶

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