题目描述：
给了4个文件
pubkey2.pem:
-----BEGIN PUBLIC KEY-----
MIIBIDANBgkqhkiG9w0BAQEFAAOCAQ0AMIIBCAKCAQB1qLiqKtKVDprtS+NGGN++
q7jLqDJoXMlPRRczMBAGJIRsz5Dzwtt1ulr0s5yu8RdaufiYeU6sYIKk92b3yygL
FvaYCzjdqBF2EyTWGVE7PL5lh3rPUfxwQFqDR8EhIH5x+Ob8rjlkftIjHTBt1ThJ
JXvDBumXpQKGcBIknRaR9dwR1q8GU58/gIk5ND3eCTAadhrhLByWkHbFArxalx4Q
q8s2ZUe8lDc/N6V93EOFjbKbqqqtDmhniF6jdXQDAIwWTpx6+jmzxlCJoVHd2MBs
ZCcQhvklWtuKz4IYL4+iUpMKGHlhY1vCqFx2EzD4XIljFLP9rk7+9+CoyTuIVL/D
AgJbJQ==
-----END PUBLIC KEY-----
pubkey1.pem:
-----BEGIN PUBLIC KEY-----
MIIBITANBgkqhkiG9w0BAQEFAAOCAQ4AMIIBCQKCAQB1qLiqKtKVDprtS+NGGN++
q7jLqDJoXMlPRRczMBAGJIRsz5Dzwtt1ulr0s5yu8RdaufiYeU6sYIKk92b3yygL
FvaYCzjdqBF2EyTWGVE7PL5lh3rPUfxwQFqDR8EhIH5x+Ob8rjlkftIjHTBt1ThJ
JXvDBumXpQKGcBIknRaR9dwR1q8GU58/gIk5ND3eCTAadhrhLByWkHbFArxalx4Q
q8s2ZUe8lDc/N6V93EOFjbKbqqqtDmhniF6jdXQDAIwWTpx6+jmzxlCJoVHd2MBs
ZCcQhvklWtuKz4IYL4+iUpMKGHlhY1vCqFx2EzD4XIljFLP9rk7+9+CoyTuIVL/D
AgMACR0=
-----END PUBLIC KEY-----
myflag2:
O+rRCXI3aTB6P1rYIOPUdalUp6ujpwEq4I20CoWA+HIL8xxGtqY6N5gpr0guZv9ZgOEAMFnBxOqMdVNnB9GgnhmXtt1ZWydPqIcHvlfwpd/Lyd0XSjXnjaz3P3vOQvR71cD/uXyBA0XPzmnTIMgEhuGJVFm8min0L/2qI7wg/Z7w1+4mOmi655JIXeCiG23ukDv6l9bZuqfGvWCa1KKXWDP31nLbp0ZN2obUs6jEAa1qVTaX6M4My+sks+0VvHATrAUuCrmMwVEivqIJ/nS6ymGVERN6Ohnzyr168knEBKOVj0FAOx3YLfppMM+XbOGHeqdKJRLpMvqFXDMGQInT3w==
myflag1:
R3Noy6r3WLItytAmb4FmHEygoilucEEZbO9ZYXx5JN03HNpBLDx7fXd2fl+UL5+11RCs/y0qlTGURWWDtG66eNLzGwNpAKiVj6I7RtUJl2Pcm3NvFeAFwI9UsVREyh7zIV6sI9ZP8l/2GVDorLAz5ULW+f0OINGhJmZm8FL/aDnlfTElhQ87LPicWpXYoMtyr6WrxjK6Ontn8BqCt0EjQ7TeXZhxIH9VTPWjDmFdmOqaqdVIT+LZemTgLNESwM5nn4g5S3aFDFwj1YiDYl0/+8etvKfOrfoKOwR0CxsRHagwdUUTES8EcHLmMGCxCkDZn3SzmmA6Nb3lgLeSgG8P1A==
题目分析：
- 首先这种出现pubkey的题都很熟悉，那么可以用惯用的套路来求解其中的n,e
- 第一步：在kali中分别使用命令：
openssl rsa -pubin -text -modulus -in warmup -in pubkey1.pem
openssl rsa -pubin -text -modulus -in warmup -in pubkey2.pem
-
分别得到：
-
-
由此可以得出：e不同，n相同，想到了共模攻击
e1 = 2333
e2 = 23333
n = 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
- 然而共模攻击还要知道密文c的值，文件中myflag便是对应的密文（这种应该可以说是约定俗成了：公钥对应n和e ，私钥对应n和d ，flag对应密文c或明文m）
- 在rsa中对于类似base64编码的密文，我们可以用以下方式进行解码：
f1="R3Noy6r3WLItytAmb4FmHEygoilucEEZbO9ZYXx5JN03HNpBLDx7fXd2fl+UL5+11RCs/y0qlTGURWWDtG66eNLzGwNpAKiVj6I7RtUJl2Pcm3NvFeAFwI9UsVREyh7zIV6sI9ZP8l/2GVDorLAz5ULW+f0OINGhJmZm8FL/aDnlfTElhQ87LPicWpXYoMtyr6WrxjK6Ontn8BqCt0EjQ7TeXZhxIH9VTPWjDmFdmOqaqdVIT+LZemTgLNESwM5nn4g5S3aFDFwj1YiDYl0/+8etvKfOrfoKOwR0CxsRHagwdUUTES8EcHLmMGCxCkDZn3SzmmA6Nb3lgLeSgG8P1A=="
f2="O+rRCXI3aTB6P1rYIOPUdalUp6ujpwEq4I20CoWA+HIL8xxGtqY6N5gpr0guZv9ZgOEAMFnBxOqMdVNnB9GgnhmXtt1ZWydPqIcHvlfwpd/Lyd0XSjXnjaz3P3vOQvR71cD/uXyBA0XPzmnTIMgEhuGJVFm8min0L/2qI7wg/Z7w1+4mOmi655JIXeCiG23ukDv6l9bZuqfGvWCa1KKXWDP31nLbp0ZN2obUs6jEAa1qVTaX6M4My+sks+0VvHATrAUuCrmMwVEivqIJ/nS6ymGVERN6Ohnzyr168knEBKOVj0FAOx3YLfppMM+XbOGHeqdKJRLpMvqFXDMGQInT3w=="
c1 = bytes_to_long(base64.b64decode(f1)) # base64转字节后再转整数
c2 = bytes_to_long(base64.b64decode(f2)) # base64转字节后再转整数
- 由此可写出完整代码：
import gmpy2
import base64
from Crypto.Util.number import *
f1="R3Noy6r3WLItytAmb4FmHEygoilucEEZbO9ZYXx5JN03HNpBLDx7fXd2fl+UL5+11RCs/y0qlTGURWWDtG66eNLzGwNpAKiVj6I7RtUJl2Pcm3NvFeAFwI9UsVREyh7zIV6sI9ZP8l/2GVDorLAz5ULW+f0OINGhJmZm8FL/aDnlfTElhQ87LPicWpXYoMtyr6WrxjK6Ontn8BqCt0EjQ7TeXZhxIH9VTPWjDmFdmOqaqdVIT+LZemTgLNESwM5nn4g5S3aFDFwj1YiDYl0/+8etvKfOrfoKOwR0CxsRHagwdUUTES8EcHLmMGCxCkDZn3SzmmA6Nb3lgLeSgG8P1A=="
f2="O+rRCXI3aTB6P1rYIOPUdalUp6ujpwEq4I20CoWA+HIL8xxGtqY6N5gpr0guZv9ZgOEAMFnBxOqMdVNnB9GgnhmXtt1ZWydPqIcHvlfwpd/Lyd0XSjXnjaz3P3vOQvR71cD/uXyBA0XPzmnTIMgEhuGJVFm8min0L/2qI7wg/Z7w1+4mOmi655JIXeCiG23ukDv6l9bZuqfGvWCa1KKXWDP31nLbp0ZN2obUs6jEAa1qVTaX6M4My+sks+0VvHATrAUuCrmMwVEivqIJ/nS6ymGVERN6Ohnzyr168knEBKOVj0FAOx3YLfppMM+XbOGHeqdKJRLpMvqFXDMGQInT3w=="
c1 = bytes_to_long(base64.b64decode(f1)) # base64转字节后再转
c2 = bytes_to_long(base64.b64decode(f2))
e1 = 2333
n = 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
e2 = 23333
_,s1,s2 = gmpy2.gcdext(e1,e2)
m = (pow(c1,s1,n)*pow(c2,s2,n)%n)
print(long_to_bytes(m))
- 得到 flag{23re_SDxF_y78hu_5rFgS}
收获与体会：
- 在kali中打开公钥或私钥文件用以下命令：
openssl rsa -pubin -text -modulus -in warmup -in pubkey1.pem
- RSA中遇到类似base46的字符串用以下代码进行转码：
c1 = bytes_to_long(base64.b64decode(f1))
