202207_QQ群_KCS1_OAEP模式RSA解密
0x00. 题目
题目给了两个文件。RSA公钥文件public.key
,密文文件flag.enc
public.key
-----BEGIN PUBLIC KEY-----
MIIBJDANBgkqhkiG9w0BAQEFAAOCAREAMIIBDAKCAQMlsYv184kJfRcjeGa7Uc/4
3pIkU3SevEA7CZXJfA44bUbBYcrf93xphg2uR5HCFM+Eh6qqnybpIKl3g0kGA4rv
tcMIJ9/PP8npdpVE+U4Hzf4IcgOaOmJiEWZ4smH7LWudMlOekqFTs2dWKbqzlC59
NeMPfu9avxxQ15fQzIjhvcz9GhLqb373XDcn298ueA80KK6Pek+3qJ8YSjZQMrFT
+EJehFdQ6yt6vALcFc4CB1B6qVCGO7hICngCjdYpeZRNbGM/r6ED5Nsozof1oMbt
Si8mZEJ/Vlx3gathkUVtlxx/+jlScjdM7AFV5fkRidt0LkwosDoPoRz/sDFz0qTM
5q5TAgMBAAE=
-----END PUBLIC KEY-----
flag-enc
CQGd9sC/h9lnLpua50/071knSsP4N8WdmRsjoNIdfclrBhMjp7NoM5xy2SlNLLC2
yh7wbRw08nwjo6UF4tmGKKfcjPcb4l4bFa5uvyMY1nJBvmqQylDbiCnsODjhpB1B
JfdpU1LUKtwsCxbc7fPL/zzUdWgO+of/R9WmM+QOBPagTANbJo0mpDYxvNKRjvac
9Bw4CQTTh87moqsNRSE/Ik5tV2pkFRZfQxAZWuVePsHp0RXVitHwvKzwmN9vMqGm
57Wb2Sto64db4gLJDh9GROQN+EQh3yLoSS8NNtBrZCDddzfKHa8wv6zN/5znvBst
sDBkGyi88NzQxw9kOGjCWtwpRw==
0x01. WP
可以通过openssl
可以得到n和e或者利用python
中的RSA模块获取n和e
openssl rsa -pubin -text -modulus -in warmup -in public.key
使用yafu分解n,得到p和q
由于加密脚本采用了PKCS1_OAEP
模式下的RSA加密,所以我们需要通过手动构造私钥进而才可以去解密密文。采用原始的pow(c,d,n)
是无法正确的解密密文的。
因此要先采用PKCS1_OAEP
模式构造私钥,然后利用这个私钥来解密密文文件。
exp.py
# coding=utf-8
from Crypto.PublicKey import RSA
from Crypto.Cipher import PKCS1_OAEP
import base64
from Crypto.Util.number import *
fPK = "public.key"
fFlag = "flag.enc"
with open(fPK,'rb') as f:
key=RSA.importKey(f.read().decode("utf-8"))
n=key.n
e=key.e
with open(fFlag,'r') as f:
c = base64.b64decode(f.read())
# 利用factordb对n进行分解,得到pq
p = 3133337
q = 25478326064937419292200172136399497719081842914528228316455906211693118321971399936004729134841162974144246271486439695786036588117424611881955950996219646807378822278285638261582099108339438949573034101215141156156408742843820048066830863814362379885720395082318462850002901605689761876319151147352730090957556940842144299887394678743607766937828094478336401159449035878306853716216548374273462386508307367713112073004011383418967894930554067582453248981022011922883374442736848045920676341361871231787163441467533076890081721882179369168787287724769642665399992556052144845878600126283968890273067575342061776244939
phi_n = (p-1) * (q - 1)
d = inverse(e, phi_n)
private_key = RSA.construct((n, e, d, p, q))
decipher = PKCS1_OAEP.new(private_key)
print(decipher.decrypt(c).decode())
# EKO{classic_rsa_challenge_is_boring_but_necessary}