巅峰极客——flag{4b0b4c8a-82f3-4d80-902b-8e7a5706f8fe}
描述
有两个公钥，两个密文
分析
- 用RsaCtfTool.py分析公钥信息，发现n相同，e不同
root@ben-PC:/mnt/d/security/misc-tool/RSA/rsatools# python RsaCtfTool.py --pkey pubkey1.pem --v
"n" is:17362520124149736059291605717839814089431261833972408175766504894876091272021197374480215582589878198406028065354454242540322618614670160317701698407729515781811530180885334265851364490357884909336085410775168953942120359215038925025305363480538685487988827339463890539279008285241711326041868183805848503077373967082910932422798165242481154593794712639251157856102009630894845049984346776659339380886766804814959778048440996937820138560802077375885700500737699904011032451007341777160586467318264288370080315519305800247682611802774996999330812534723806925426052547128371180683265963525581842037399869323246530085399
************************************************************
"e" is:2333
************************************************************
Try weak key attack
Try Wiener's attack
root@ben-PC:/mnt/d/security/misc-tool/RSA/rsatools# python RsaCtfTool.py --pkey pubkey2.pem --v
"n" is:17362520124149736059291605717839814089431261833972408175766504894876091272021197374480215582589878198406028065354454242540322618614670160317701698407729515781811530180885334265851364490357884909336085410775168953942120359215038925025305363480538685487988827339463890539279008285241711326041868183805848503077373967082910932422798165242481154593794712639251157856102009630894845049984346776659339380886766804814959778048440996937820138560802077375885700500737699904011032451007341777160586467318264288370080315519305800247682611802774996999330812534723806925426052547128371180683265963525581842037399869323246530085399
************************************************************
"e" is:23333
************************************************************
Try weak key attack
Try Wiener's attack
密文内容如下：
[root RsaCtfTool]$ cat flag1.enc
XSKBJ2biS6brC5iGwU0GZitHdVM3HXAiwtFnVf2+HTaUqFahxL+BxBi2QDcx7gLxcjEWCMwFP6DS92nMAU4r0gPWSEUIoY57sgNZsjDIDAukiYeLNDUgYz+1P+nF4fk7gwPdozrIvAXGDBvMBjuviqsC8vmVP3I6eLLkt9C46HFt0SBw5ycfAjVoDF2r7/4B1UDs4G0dpIDUCk4khezzgqspn6tqtwOGB27vrKegoL/FlwmutFYIuRKKCBKx3yc/qfWXZ84Oo8nPqgaxgDlxWeLtGM9ZouwFKnagmjbnH+58Pescw4XYafXKqFjQz3XrK/uUESE8jIEIPeL1+8yUpw==
[root RsaCtfTool]$ cat flag2.enc
EruzwVAXSVLC3rldjcsx6HO0UUICdR9xxgr9eWNhIW0T8l2O3yT/LlFLK2+YU0HB97xr5HaiZesk4T6IuJ9+iOzB8YSkWMfYvOSDKn7Jng/1Q3wQuoldm+UurmZkiEs9kFi+EhsCNAbVAnLzLXLwzYm3emamueDqru4Doo/lSMz8p0+jqz24HscJN9shU85WX4JngW92REHHV8rPHaisCdxeAs+uPyTNzO4IbwDaJvw3ZR/Lo4m1K2Qw8PbYnOcgVr9CWR7mVyxofoWk6qWpQf3d0fX6wbbPcQkXxnnqLWy5S3PZcNQa1wkfRTJJO03QmNVsOivXGb3GzmeZbxmVhQ==
[root RsaCtfTool]$
- 对于同一明文使用同样的N不同的E分别进行加密，满足共模攻击条件
解密代码如下：
#!/usr/bin/env python
# -*- coding: utf-8 -*-
import sys,gmpy,base64
def egcd(a, b):
if a == 0:
return (b, 0, 1)
else:
g, y, x = egcd(b % a, a)
return (g, x - (b // a) * y, y)
def modinv(a, m):
g, x, y = egcd(a, m)
if g != 1:
raise Exception('modular inverse does not exist')
else:
return x % m
def pad_even(x):#重要！凑齐2位，将0x1 变成 0x01
return ('', '0')[len(x)%2] + x
def CipherB2n(c):#将base64编码后的密文转成数字
c2 = base64.b64decode(c)
temp = ''
for i in c2:
temp += pad_even(str(hex(ord(i)))[2:])
temp = eval('0x'+temp)
return (temp)
def CipherN2b(m):#将数字转换成ascii
hex_m=hex(m)[2:]
if hex_m[-1] == 'L' :
hex_m=hex_m[:-1]
return hex_m.decode('hex')
if __name__ == '__main__':
sys.setrecursionlimit(1000000)
e1 = 2333 #根据分解结果
e2 = 23333 #根据分解结果
s = egcd(e1, e2)
s1 = s[1]
s2 = s[2]
c1 = 'XSKBJ2biS6brC5iGwU0GZitHdVM3HXAiwtFnVf2+HTaUqFahxL+BxBi2QDcx7gLxcjEWCMwFP6DS92nMAU4r0gPWSEUIoY57sgNZsjDIDAukiYeLNDUgYz+1P+nF4fk7gwPdozrIvAXGDBvMBjuviqsC8vmVP3I6eLLkt9C46HFt0SBw5ycfAjVoDF2r7/4B1UDs4G0dpIDUCk4khezzgqspn6tqtwOGB27vrKegoL/FlwmutFYIuRKKCBKx3yc/qfWXZ84Oo8nPqgaxgDlxWeLtGM9ZouwFKnagmjbnH+58Pescw4XYafXKqFjQz3XrK/uUESE8jIEIPeL1+8yUpw=='
c2 ='EruzwVAXSVLC3rldjcsx6HO0UUICdR9xxgr9eWNhIW0T8l2O3yT/LlFLK2+YU0HB97xr5HaiZesk4T6IuJ9+iOzB8YSkWMfYvOSDKn7Jng/1Q3wQuoldm+UurmZkiEs9kFi+EhsCNAbVAnLzLXLwzYm3emamueDqru4Doo/lSMz8p0+jqz24HscJN9shU85WX4JngW92REHHV8rPHaisCdxeAs+uPyTNzO4IbwDaJvw3ZR/Lo4m1K2Qw8PbYnOcgVr9CWR7mVyxofoWk6qWpQf3d0fX6wbbPcQkXxnnqLWy5S3PZcNQa1wkfRTJJO03QmNVsOivXGb3GzmeZbxmVhQ=='
c1 = CipherB2n(c1)
c2 = CipherB2n(c2)
#print hex(c1)
n = 17362520124149736059291605717839814089431261833972408175766504894876091272021197374480215582589878198406028065354454242540322618614670160317701698407729515781811530180885334265851364490357884909336085410775168953942120359215038925025305363480538685487988827339463890539279008285241711326041868183805848503077373967082910932422798165242481154593794712639251157856102009630894845049984346776659339380886766804814959778048440996937820138560802077375885700500737699904011032451007341777160586467318264288370080315519305800247682611802774996999330812534723806925426052547128371180683265963525581842037399869323246530085399 #共n
if s1<0:
s1 = - s1
c1 = modinv(c1, n)
elif s2<0:
s2 = - s2
c2 = modinv(c2, n)
m=(pow(c1,s1,n)*pow(c2,s2,n)) % n
print m
print CipherN2b(m)
运行结果
flag{4b0b4c8a-82f3-4d80-902b-8e7a5706f8fe}
总结
串了一下rsa的明文和密文处理方式。计算过程中明文和密文都作为一个大数字，于是有：
- 明文是字符串。
1.1. 将字符串转hex编码（py3里只能逐字，py2里可以直接转），拼上0x头和L尾，成为大数字。
hex_m.decode('hex')
1.2. 将字符串分成每行一个字，ord(i)。（解密时chr(i)再拼接） - 算出密文是大数字。
2.1. 转成16进制，去掉0x和L后变成一串16进制。此时可能出现：
2.1.1 将十六进制直接转存为文件。（解密时提取十六进制值）
2.1.2 将十六进制进行base64编码，变为可见字符。（解密时进行base64解码,由于二位一组，需注意对0x1这种补成0x01）
def pad_even(x): return ('', '0')[len(x)%2] + x
for i in c2:temp += pad_even(str(hex(ord(i)))[2:])
2.2. 直接是10进制数字，无需处理
m = chr(pow(int(i),d,n))
