---Category: CRY | 100 pts - 170 solves---
# CryMePlx
> Description: Awesome service. Now I don't need to encrypt anything myself! \> Connect via: `nc chall.rumble.host 2734`
We had a service to netcat with: `nc chall.rumble.host 2734` and the source code to download & unzip. \Looking at the source code, the program ask for input and then encrypt the flag and the input with AES-CTR-128 mode.
```pythonfrom Crypto.Cipher import AESfrom secret import flagimport os
kwargs = {"nonce": os.urandom(8)}key = os.urandom(16)
def encrypt(msg):    aes = AES.new(key, AES.MODE_CTR, **kwargs)    return aes.encrypt(msg).hex()
print(encrypt(flag))q = input("Encrypt this string:").encode()print(encrypt(q))```
The vulnerability here is that the Nonce is reused for all the encryptions, and as the name says, it should be used only (n)once. \So, the encryption of flag is done with: 
`Ciphertext1 = flag ⊕ AES(Key, Nonce)`
and the encryption of user input:
`Ciphertext2 = input ⊕ AES(Key, Nonce)`
Note that we used the same Nonce and Key pair, so we can say that:
`AES(Key, Nonce) = flag ⊕ Ciphertext1`\`AES(Key, Nonce) = input ⊕ Ciphertext2`
Therefore:
`flag ⊕ Ciphertext1 = input ⊕ Ciphertext2`\`flag = input ⊕ Ciphertext2 ⊕ Ciphertext1`\\Let's write our exploit:
```pythonfrom pwn import *import binascii
context.log_level = 'debug'p = remote('chall.rumble.host', 2734)
enc_flag = binascii.unhexlify(p.recvline().strip())input = 'A' * len(enc_flag)p.sendlineafter('Encrypt this string:', input)enc_input = binascii.unhexlify(p.recvline().strip())
flag = xor(input, enc_input, enc_flag)log.success(f'{flag=}')```