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公钥指数相关攻击

小公钥指数攻击

攻击条件

e 特别小,比如 e 为 3。

攻击原理

假设用户使用的密钥 e=3。考虑到加密关系满足:

c\equiv m^3 \bmod N

则:

\begin{align*} m^3 &= c+k\times N\\ m &= \sqrt[3]{c+k\times n} \end{align*}

攻击者可以从小到大枚举 N,依次开三次根,直到开出整数为止。

范例

这里我们以 XMan 一期夏令营课堂练习为例进行介绍(Jarvis OJ 有复现),附件中有一个 flag.encpubkey.pem,很明显是密文和公钥了,先用 openssl 读一下公钥。

➜  Jarvis OJ-Extremely hard RSA git:(master) ✗ openssl rsa -pubin -in pubkey.pem -text -modulus       
Public-Key: (4096 bit)
Modulus:
    00:b0:be:e5:e3:e9:e5:a7:e8:d0:0b:49:33:55:c6:
    18:fc:8c:7d:7d:03:b8:2e:40:99:51:c1:82:f3:98:
    de:e3:10:45:80:e7:ba:70:d3:83:ae:53:11:47:56:
    56:e8:a9:64:d3:80:cb:15:7f:48:c9:51:ad:fa:65:
    db:0b:12:2c:a4:0e:42:fa:70:91:89:b7:19:a4:f0:
    d7:46:e2:f6:06:9b:af:11:ce:bd:65:0f:14:b9:3c:
    97:73:52:fd:13:b1:ee:a6:d6:e1:da:77:55:02:ab:
    ff:89:d3:a8:b3:61:5f:d0:db:49:b8:8a:97:6b:c2:
    05:68:48:92:84:e1:81:f6:f1:1e:27:08:91:c8:ef:
    80:01:7b:ad:23:8e:36:30:39:a4:58:47:0f:17:49:
    10:1b:c2:99:49:d3:a4:f4:03:8d:46:39:38:85:15:
    79:c7:52:5a:69:98:4f:15:b5:66:7f:34:20:9b:70:
    eb:26:11:36:94:7f:a1:23:e5:49:df:ff:00:60:18:
    83:af:d9:36:fe:41:1e:00:6e:4e:93:d1:a0:0b:0f:
    ea:54:1b:bf:c8:c5:18:6c:b6:22:05:03:a9:4b:24:
    13:11:0d:64:0c:77:ea:54:ba:32:20:fc:8f:4c:c6:
    ce:77:15:1e:29:b3:e0:65:78:c4:78:bd:1b:eb:e0:
    45:89:ef:9a:19:7f:6f:80:6d:b8:b3:ec:d8:26:ca:
    d2:4f:53:24:cc:de:c6:e8:fe:ad:2c:21:50:06:86:
    02:c8:dc:dc:59:40:2c:ca:c9:42:4b:79:00:48:cc:
    dd:93:27:06:80:95:ef:a0:10:b7:f1:96:c7:4b:a8:
    c3:7b:12:8f:9e:14:11:75:16:33:f7:8b:7b:9e:56:
    f7:1f:77:a1:b4:da:ad:3f:c5:4b:5e:7e:f9:35:d9:
    a7:2f:b1:76:75:97:65:52:2b:4b:bc:02:e3:14:d5:
    c0:6b:64:d5:05:4b:7b:09:6c:60:12:36:e6:cc:f4:
    5b:5e:61:1c:80:5d:33:5d:ba:b0:c3:5d:22:6c:c2:
    08:d8:ce:47:36:ba:39:a0:35:44:26:fa:e0:06:c7:
    fe:52:d5:26:7d:cf:b9:c3:88:4f:51:fd:df:df:4a:
    97:94:bc:fe:0e:15:57:11:37:49:e6:c8:ef:42:1d:
    ba:26:3a:ff:68:73:9c:e0:0e:d8:0f:d0:02:2e:f9:
    2d:34:88:f7:6d:eb:62:bd:ef:7b:ea:60:26:f2:2a:
    1d:25:aa:2a:92:d1:24:41:4a:80:21:fe:0c:17:4b:
    98:03:e6:bb:5f:ad:75:e1:86:a9:46:a1:72:80:77:
    0f:12:43:f4:38:74:46:cc:ce:b2:22:2a:96:5c:c3:
    0b:39:29
Exponent: 3 (0x3)
Modulus=B0BEE5E3E9E5A7E8D00B493355C618FC8C7D7D03B82E409951C182F398DEE3104580E7BA70D383AE5311475656E8A964D380CB157F48C951ADFA65DB0B122CA40E42FA709189B719A4F0D746E2F6069BAF11CEBD650F14B93C977352FD13B1EEA6D6E1DA775502ABFF89D3A8B3615FD0DB49B88A976BC20568489284E181F6F11E270891C8EF80017BAD238E363039A458470F1749101BC29949D3A4F4038D463938851579C7525A69984F15B5667F34209B70EB261136947FA123E549DFFF00601883AFD936FE411E006E4E93D1A00B0FEA541BBFC8C5186CB6220503A94B2413110D640C77EA54BA3220FC8F4CC6CE77151E29B3E06578C478BD1BEBE04589EF9A197F6F806DB8B3ECD826CAD24F5324CCDEC6E8FEAD2C2150068602C8DCDC59402CCAC9424B790048CCDD9327068095EFA010B7F196C74BA8C37B128F9E1411751633F78B7B9E56F71F77A1B4DAAD3FC54B5E7EF935D9A72FB176759765522B4BBC02E314D5C06B64D5054B7B096C601236E6CCF45B5E611C805D335DBAB0C35D226CC208D8CE4736BA39A0354426FAE006C7FE52D5267DCFB9C3884F51FDDFDF4A9794BCFE0E1557113749E6C8EF421DBA263AFF68739CE00ED80FD0022EF92D3488F76DEB62BDEF7BEA6026F22A1D25AA2A92D124414A8021FE0C174B9803E6BB5FAD75E186A946A17280770F1243F4387446CCCEB2222A965CC30B3929
writing RSA key
-----BEGIN PUBLIC KEY-----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-----END PUBLIC KEY-----

看到 e=3,很明显是小公钥指数攻击了。这里我们使用 Crypto 库来读取公钥,使用 multiprocessing 来加快破解速度。

#/usr/bin/python
# coding=utf-8
import gmpy2
from Crypto.PublicKey import RSA
from multiprocessing import Pool
pool = Pool(4)

with open('./pubkey.pem', 'r') as f:
    key = RSA.importKey(f)
    N = key.n
    e = key.e
with open('flag.enc', 'r') as f:
    cipher = f.read().encode('hex')
    cipher = int(cipher, 16)


def calc(j):
    print j
    a, b = gmpy2.iroot(cipher + j * N, 3)
    if b == 1:
        m = a
        print '{:x}'.format(int(m)).decode('hex')
        pool.terminate()
        exit()


def SmallE():
    inputs = range(0, 130000000)
    pool.map(calc, inputs)
    pool.close()
    pool.join()


if __name__ == '__main__':
    print 'start'
    SmallE()

爆破时间有点长,,拿到 flag

Didn't you know RSA padding is really important? Now you see a non-padding message is so dangerous. And you should notice this in future.Fl4g: flag{Sm4ll_3xpon3nt_i5_W3ak}

题目

RSA 衍生算法——Rabin 算法

攻击条件

Rabin 算法的特征在于 e=2

攻击原理

密文:

c = m^2\bmod n

解密:

  • 计算出 m_pm_q
\begin{align*} m_p &= \sqrt{c} \bmod p\\ m_q &= \sqrt{c} \bmod q \end{align*}
  • 用扩展欧几里得计算出 y_py_q
y_p \cdot p + y_q \cdot q = 1
  • 解出四个明文:
\begin{align*} a &= (y_p \cdot p \cdot m_q + y_q \cdot q \cdot m_p) \bmod n\\ b &= n - a\\ c &= (y_p \cdot p \cdot m_q - y_q \cdot q \cdot m_p) \bmod n\\ d &= n - c \end{align*}

注意:如果 p \equiv q \equiv 3 \pmod 4,则

\begin{align*} m_p &= c^{\frac{1}{4}(p + 1)} \bmod p\\ m_q &= c^{\frac{1}{4}(q + 1)} \bmod q \end{align*}

而一般情况下,p \equiv q \equiv 3 \pmod 4 是满足的,对于不满足的情况下,请参考相应的算法解决。

例子

这里我们以 XMan 一期夏令营课堂练习(Jarvis OJ 有复现)为例,读一下公钥。

➜  Jarvis OJ-hard RSA git:(master) ✗ openssl rsa -pubin -in pubkey.pem -text -modulus 
Public-Key: (256 bit)
Modulus:
    00:c2:63:6a:e5:c3:d8:e4:3f:fb:97:ab:09:02:8f:
    1a:ac:6c:0b:f6:cd:3d:70:eb:ca:28:1b:ff:e9:7f:
    be:30:dd
Exponent: 2 (0x2)
Modulus=C2636AE5C3D8E43FFB97AB09028F1AAC6C0BF6CD3D70EBCA281BFFE97FBE30DD
writing RSA key
-----BEGIN PUBLIC KEY-----
MDowDQYJKoZIhvcNAQEBBQADKQAwJgIhAMJjauXD2OQ/+5erCQKPGqxsC/bNPXDr
yigb/+l/vjDdAgEC
-----END PUBLIC KEY-----

e=2,考虑 Rabin 算法。首先我们先分解一下 p 和 q,得到

p=275127860351348928173285174381581152299
q=319576316814478949870590164193048041239

编写代码

#!/usr/bin/python
# coding=utf-8
import gmpy2
import string
from Crypto.PublicKey import RSA

# 读取公钥参数
with open('pubkey.pem', 'r') as f:
    key = RSA.importKey(f)
    N = key.n
    e = key.e
with open('flag.enc', 'r') as f:
    cipher = f.read().encode('hex')
    cipher = string.atoi(cipher, base=16)
    # print cipher
print "please input p"
p = int(raw_input(), 10)
print 'please input q'
q = int(raw_input(), 10)
# 计算yp和yq
inv_p = gmpy2.invert(p, q)
inv_q = gmpy2.invert(q, p)

# 计算mp和mq
mp = pow(cipher, (p + 1) / 4, p)
mq = pow(cipher, (q + 1) / 4, q)

# 计算a,b,c,d
a = (inv_p * p * mq + inv_q * q * mp) % N
b = N - int(a)
c = (inv_p * p * mq - inv_q * q * mp) % N
d = N - int(c)

for i in (a, b, c, d):
    s = '%x' % i
    if len(s) % 2 != 0:
        s = '0' + s
    print s.decode('hex')

拿到 flag,PCTF{sp3ci4l_rsa}

题目