CTF：HITB 2015加密解密部分Write-up

15-11-14 来源：[db:作者]

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介绍

加密类300分这道挑战题目是有关于RSA生成的质数p和q。我们现目前获得了一份RSA加密过的邮件mail.msg，以及隐藏有公钥的证书hitbctf.crt。所有文件会在文尾分享！

RSA参数结构

模N是由第一个质数P随机选择，第二个质数是由以下方式构成：

k是一个正整数，q是一个质数，e是一个公用的指数且还是一个质数。

下面的代码是由Python实现：

def gen_rsa_parameters():
    r = os.urandom(63)
    e = int(r.encode('hex'), 16)
    e = next_prime(e)
    r = os.urandom(64)
    p = int(r.encode('hex'), 16)
    p = next_prime(p)
    q = (p*modinv(p-1, e)%e)    
    while not is_prime(q):
        q += e
    N = p*q
    phi = (p-1)*(q-1)
    d = modinv(e,phi)    
    return N,e,d,p,q

攻击假想

令N′=N mod e，我们可以得到：

由于方程中存在α，所以我们引入p−1来替换

现在我们获得了一个取决于p的二次方程。

令X = p−1并假设N′−2位偶数且N′−2 = 2b

通过使用二次剩余[百度百科,维基百科中文,维基百科英文]，我们确定并找到了解决方案。我们还可以使用SAGE以及sqrt()函数：

Np = N % e
b = (Np - 2) / 2
p = Mod(pow(b, 2) - 1, e).sqrt() + b + 1

我们获得的是p mod e而不是p！

我尝试增加一些次方来找到p，但是p和 p mod e之间的距离实在太大了。

所以我们必须放弃这条道，选择另外的方法了。

通过了解p mod e我们还可以计算α

拥有了α和p mod e我们就可以通过递增e直到(pα mod e)+k⋅e为质数且能整除N。

我们假设N′−2为偶数(所以N′必须是偶数)，如果在证书中N′为偶数那么这一切猜想也就成立了。

实际攻击

首先我们从证书中提取模N和公钥e：

openssl x509 -in hitbctf.crt -text -noout

Certificate:    
    Data:        
        Version: 1 (0x0)        
        Serial Number: 18379438180976429416 (0xff10e1a5ac5a0968)    
    Signature Algorithm: sha1WithRSAEncryption        
        Issuer: C=NL, ST=Noord-Holland, L=Amsterdam, O=HITB, OU=CTF        
        Validity
            Not Before: May 24 09:58:26 2015 GMT            
            Not After : May 23 09:58:26 2016 GMT        
          Subject: C=NL, ST=Noord-Holland, L=Amsterdam, O=HITB, OU=CTF        
          Subject Public Key Info:            
         Public Key Algorithm: rsaEncryption                
             Public-Key: (1024 bit)                
             Modulus:                    
                 00:e6:eb:89:c1:8d:49:c9:a2:02:2b:e0:b4:65:14:           
                 6e:0f:90:45:1e:a3:4c:6b:60:56:00:4e:bd:15:59:            
                 55:b1:35:96:c2:d6:83:ad:2f:23:6b:0b:2c:0e:0b:            
                 88:83:b5:d6:cb:8a:0b:4f:f9:b7:eb:64:8c:95:2b:            
                 6b:ef:5a:6f:04:f5:64:17:f5:1c:a9:14:d9:ea:73:            
                 e7:dd:c5:f2:0d:ce:c3:9c:e8:4b:72:2a:0c:f3:d8:            
                 5e:80:ce:78:64:63:e1:44:f6:1d:b5:9c:cf:45:ff:            
                 0e:d3:7f:d0:ce:bd:37:a5:8d:8a:4b:08:33:9e:a3:            
                 2c:bc:ab:61:64:03:fd:2c:c5                
             Exponent:                    
                 69:60:2d:93:8a:81:5f:14:cf:9f:b8:36:c2:e0:4d:                                      4d:de:82:ba:fc:8d:56:c2:6d:8c:89:ef:3c:40:69:            
                 5d:d5:d4:ef:a7:36:36:43:15:14:95:f3:8c:bf:24:            
                 ae:94:30:92:40:79:12:00:1b:17:f5:53:33:9e:92:            
                 70:70:49    
    Signature Algorithm: sha1WithRSAEncryption         
    17:2b:ea:be:90:ad:98:f2:2b:ff:f5:61:d3:ea:af:fb:35:3a:         
    67:10:91:13:db:60:55:d9:09:8b:c2:1a:cf:6b:c6:1f:f2:10:         
    7a:d1:7b:9d:ff:10:f2:f2:c0:a9:f5:aa:2e:09:93:40:88:92:         
    7d:98:ff:e1:cb:dc:db:35:8d:e0:4b:21:99:76:bf:db:04:a2:         
    62:a4:18:4e:fc:bb:a7:53:be:6a:a1:ef:ec:15:86:c1:f1:1e:         
    87:6a:e9:af:fe:d1:08:eb:de:22:28:c4:5e:be:f1:41:0a:ca:         
    cf:cf:da:63:b1:c1:56:e8:0c:8e:56:7f:08:94:0d:2b:2a:08:

N = 162157588231432175750266419709084494256738149198416702818838192688585199555839792754739411546929869488574731499231574687207152393171517768019327338646577588312972543620665360591059281057979460340279244616489314862289312195704820435867259965443285749719682327313893490163672147378911558526315013166594183212229
e = 215584882345398898379387706359713309034898391467668952244901790414655161931455822669631548838317075220811407344210520390992334648372016602816069805 30249
SAGE:

sage: Np = N % e
sage: b = (Np - 2) / 2sage: pp = int(Mod(pow(b, 2) - 1, e).sqrt()) + b + 1sage: alpha = inverse_mod(int(X), int(e))
sage: q = (pp * alpha) % e
sage: while not is_prime(q) and N % q != 0:
....:         q += e
sage: p = N / q
sage: p13317713478157317654574552532079837937895228108820477140030796245493222349714497856652987583926206280627498615972491072112647669795345566943409669535038641sage: q12176083266650126897170100375931110708350668494730113414987801764299563774952801449439933220072280766145748279998832962142839152786620322097065894585706069

rsatool.py：

#!/usr/bin/env python2
import base64, fractions, optparse, random
import gmpy
from pyasn1.codec.der import encoder
from pyasn1.type.univ import *
PEM_TEMPLATE = '-----BEGIN RSA PRIVATE KEY-----\n%s-----END RSA PRIVATE KEY-----\n'
DEFAULT_EXP = 65537
def factor_modulus(n, d, e):
    """
    Efficiently recover non-trivial factors of n
    See: Handbook of Applied Cryptography
    8.2.2 Security of RSA -> (i) Relation to factoring (p.287)
    http://www.cacr.math.uwaterloo.ca/hac/
    """
    t = (e * d - 1)
    s = 0
    while True:
        quotient, remainder = divmod(t, 2)
        if remainder != 0:
            break
        s += 1
        t = quotient
    found = False
    while not found:
        i = 1
        a = random.randint(1,n-1)
        while i <= s and not found:
            c1 = pow(a, pow(2, i-1, n) * t, n)
            c2 = pow(a, pow(2, i, n) * t, n)
            found = c1 != 1 and c1 != (-1 % n) and c2 == 1
            i += 1
    p = fractions.gcd(c1-1, n)
    q = (n / p)
    return p, q
class RSA:
    def __init__(self, p=None, q=None, n=None, d=None, e=DEFAULT_EXP):
        """
        Initialize RSA instance using primes (p, q)
        or modulus and private exponent (n, d)
        """
        self.e = e
        if p and q:
            assert gmpy.is_prime(p), 'p is not prime'
            assert gmpy.is_prime(q), 'q is not prime'
            self.p = p
            self.q = q
        elif n and d:   
            self.p, self.q = factor_modulus(n, d, e)
        else:
            raise ArgumentError('Either (p, q) or (n, d) must be provided')
        self._calc_values()
    def _calc_values(self):
        self.n = self.p * self.q
        phi = (self.p - 1) * (self.q - 1)
        self.d = gmpy.invert(self.e, phi)
        # CRT-RSA precomputation
        self.dP = self.d % (self.p - 1)
        self.dQ = self.d % (self.q - 1)
        self.qInv = gmpy.invert(self.q, self.p)
    def to_pem(self):
        """
        Return OpenSSL-compatible PEM encoded key
        """
        return PEM_TEMPLATE % base64.encodestring(self.to_der())
    def to_der(self):
        """
        Return parameters as OpenSSL compatible DER encoded key
        """
        seq = Sequence()
        for x in [0, self.n, self.e, self.d, self.p, self.q, self.dP, self.dQ, self.qInv]:
            seq.setComponentByPosition(len(seq), Integer(x))
        return encoder.encode(seq)
    def dump(self, verbose):
        vars = ['n', 'e', 'd', 'p', 'q']
        if verbose:
            vars += ['dP', 'dQ', 'qInv']
        for v in vars:
            self._dumpvar(v)
    def _dumpvar(self, var):
        val = getattr(self, var)
        parts = lambda s, l: '\n'.join([s[i:i+l] for i in xrange(0, len(s), l)])
        if len(str(val)) <= 40:
            print '%s = %d (%#x)\n' % (var, val, val)
        else:
            print '%s =' % var
            print parts('%x' % val, 80) + '\n'
if __name__ == '__main__':
    parser = optparse.OptionParser()
    parser.add_option('-p', dest='p', help='prime', type='int')
    parser.add_option('-q', dest='q', help='prime', type='int')
    parser.add_option('-n', dest='n', help='modulus', type='int')
    parser.add_option('-d', dest='d', help='private exponent', type='int')
    parser.add_option('-e', dest='e', help='public exponent (default: %d)' % DEFAULT_EXP, type='int', default=DEFAULT_EXP)
    parser.add_option('-o', dest='filename', help='output filename')
    parser.add_option('-f', dest='format', help='output format (DER, PEM) (default: PEM)', type='choice', choices=['DER', 'PEM'], default='PEM')
    parser.add_option('-v', dest='verbose', help='also display CRT-RSA representation', action='store_true', default=False)
    try:
        (options, args) = parser.parse_args()
        if options.p and options.q:
            print 'Using (p, q) to initialise RSA instance\n'
            rsa = RSA(p=options.p, q=options.q, e=options.e)
        elif options.n and options.d:
            print 'Using (n, d) to initialise RSA instance\n'
            rsa = RSA(n=options.n, d=options.d, e=options.e)
        else:
            parser.print_help()
            parser.error('Either (p, q) or (n, d) needs to be specified')
        rsa.dump(options.verbose)
        if options.filename:
            print 'Saving %s as %s' % (options.format, options.filename)
            if options.format == 'PEM':
                data = rsa.to_pem()
            elif options.format == 'DER':
                data = rsa.to_der()
            fp = open(options.filename, 'wb')
            fp.write(data)
            fp.close()
    except optparse.OptionValueError, e:
        parser.print_help()
        parser.error(e.msg)

接着使用rsatool.py生成一个私钥：

$ ./rsatools.py -o private.pem \
-e 21558488234539889837938770635971330903489839146766895224490179041465516193145582266963154883831707522081140734421052039099233464837201660281606980530249 \
-p 13317713478157317654574552532079837937895228108820477140030796245493222349714497856652987583926206280627498615972491072112647669795345566943409669535038641 \
-q 12176083266650126897170100375931110708350668494730113414987801764299563774952801449439933220072280766145748279998832962142839152786620322097065894585706069

然后解密信息：

openssl smime -decrypt -in mail.msg -inkey private.pem
hitb{0b21cc2025534dbd2965390d2bcef45d}

链接：http://pan.baidu.com/s/1kTAjeHT 密码：iy4v

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