🟩The Last Dance🕺

First you MUST understand how chacha20 works before solving this challenge, you can check this or any other source

Source code

from Crypto.Cipher import ChaCha20
from secret import FLAG
import os


def encryptMessage(message, key, nonce):
    cipher = ChaCha20.new(key=key, nonce=iv)
    ciphertext = cipher.encrypt(message)
    return ciphertext


def writeData(data):
    with open("out.txt", "w") as f:
        f.write(data)


if __name__ == "__main__":
    message = b"Our counter agencies have intercepted your messages and a lot "
    message += b"of your agent's identities have been exposed. In a matter of "
    message += b"days all of them will be captured"

    key, iv = os.urandom(32), os.urandom(12)

    encrypted_message = encryptMessage(message, key, iv)
    encrypted_flag = encryptMessage(FLAG, key, iv)

    data = iv.hex() + "\n" + encrypted_message.hex() + "\n" + encrypted_flag.hex()
    writeData(data)

output 👇

c4a66edfe80227b4fa24d431
7aa34395a258f5893e3db1822139b8c1f04cfab9d757b9b9cca57e1df33d093f07c7f06e06bb6293676f9060a838ea138b6bc9f20b08afeb73120506e2ce7b9b9dcd9e4a421584cfaba2481132dfbdf4216e98e3facec9ba199ca3a97641e9ca9782868d0222a1d7c0d3119b867edaf2e72e2a6f7d344df39a14edc39cb6f960944ddac2aaef324827c36cba67dcb76b22119b43881a3f1262752990
7d8273ceb459e4d4386df4e32e1aecc1aa7aaafda50cb982f6c62623cf6b29693d86b15457aa76ac7e2eef6cf814ae3a8d39c7

---

Understanding the code

this is a chacha20 encrypting which is a stream key encrypting cipher and as you see the code there is a msg and a flag encrypted using the same initial key and vector which means that both flag and msg are encrypted using the same key stream, chacha20 works by XORing the key with the text so if we XORed the encrypted msg and flag we will get the plain msg xored with the plain flag, by xoring the result with the plain msg we cancel its bytes and get our original flag :D

msg ⊕ key-stream = enc-msg

flag ⊕ key stream = enc-flag

enc-msg ⊕ enc-flag = flag ⊕ msg

or in other words its:

msg ⊕ key-stream ⊕ flag ⊕ key stream = flag ⊕ msg

then

flag ⊕ msg ⊕ msg = flag

---

solution code

enc_msg_hex = "7aa34395a258f5893e3db1822139b8c1f04cfab9d757b9b9cca57e1df33d093f07c7f06e06bb6293676f9060a838ea138b6bc9f20b08afeb73120506e2ce7b9b9dcd9e4a421584cfaba2481132dfbdf4216e98e3facec9ba199ca3a97641e9ca9782868d0222a1d7c0d3119b867edaf2e72e2a6f7d344df39a14edc39cb6f960944ddac2aaef324827c36cba67dcb76b22119b43881a3f1262752990"
enc_flag_hex = "7d8273ceb459e4d4386df4e32e1aecc1aa7aaafda50cb982f6c62623cf6b29693d86b15457aa76ac7e2eef6cf814ae3a8d39c7"
enc_flag = bytes.fromhex(enc_flag_hex)
enc_msg = bytes.fromhex(enc_msg_hex)
temp = b""
flag = b""
msg = b"Our counter agencies have intercepted your messages and a lot "
msg += b"of your agent's identities have been exposed. In a matter of "
msg += b"days all of them will be captured"
for i in range(len(enc_flag)):
    temp = enc_flag[i] ^ enc_msg[i]
    flag += bytes([temp ^ msg[i]])
print(flag)

this will give you the flag HTB{und3r57AnD1n9_... That's it

Happy hacking