salt writeup

We have a task:

We managed to capture the following traffic from a user who retrieved a note from that new secure notebook service:                                                         
request: b'\xa0f?u\xfb;AZf\xfc@{M!\xcdP\x92\xf6\x0f\xea\x1d\xad@\xc5\x8c\xd0R\xd8\xfdX81\x01d\xc8\x9b\xc4\xfd\x04\x9b\x843a\x940U\xc4\x7fa\x11W\xa9uf\xa9\xf4%w;`s[\xad\xa8V\x90\xe0w,\xb6<\xbd\xb1\xcbh=\x0b\x80\xba\xd8\x9bM\x17\xc6\x1f\x83<G\xcfV\x93\x00E\xe97\xcc\x9a.\xa1\xe6\x13\x11\xe9<\xae'

response: b"e\xc8\x9b\xc4\xfd\x04\x9b\x843a\x940U\xc4\x7fa\x11W\xa9uf\xa9\xf4%\r6C\x8d\xe1Z\x95\xb1^\x92\xddF\xa7\xbb\x86\x19\xbaCW\xde\x9bo\xd3Z\x8d\x85kx\x81a\xb0\x0b\xc9\x14'L\xc6i\xc4V\x86=\xba\x11~\xcc\x9bw#i\xc7\xb2\xc0Z\x9d\x1d\xb3\x96\\\xf9\xffG\x8a\xa2"

That crypto is pretty much invincible - after all, they're using NaCl - but maybe you can figure out some crazy attack anyway? I count on you!!!

You need to install the "pynacl" python module for python 3 to make this work.

connect to school.fluxfingers.net:1512

You can download sources here.

Let’s see what do we have here. Client generates an assymetric keypair (Box NaCl is used) and sends an encrypted and signed message to a server:

  req = my_sk.public_key.encode() + bytes([command]) + box.encrypt(data.encode('utf8'), nonce)

Because Box signs the request, there is no reason to change public key or encrypted data. Hovewer, we can change the command (it’s just 0x00 for CMD_STORE and 0x01 for CMD_LOOKUP request).

These commands have different numbers of arguments which are separated by a ‘#’ symbol. If server recieves an incorrect number of arguments for a command, it returns an error. Very large encrypted error.

if cmd == CMD_STORE:
  if len(params) < 2:
        return reply("you need to send more parameters! seriously, why would you just send one parameter? I should just let this fail with a silent error. sending just one parameter is stupid.")

How can we use it? Let see, how server encrypts response. It takes client’s nonce and increments it:

# the client already picked a random nonce for the request, so we can just increment it by one for the reply
answer_nonce = bytes([nonce[0] + 1]) + nonce[1:]
  def reply(str):
      s.sendall(my_box.encrypt(str.encode('utf8'), answer_nonce))

Great. It means that we can get that known response encrypted with the same nonce and key. Now we need to find out how Box encryption works.

According to the Internet, “Box uses Curve25519, XSalsa20 and Poly1305 to encrypt and authenticate messages”. Curve25519 is some weird elliptic shit, and Poly1305 is used for signing, so the encryption algorithm is XSalsa20. The first 32 bytes of the output of XSalsa20 are used for the MAC, the rest are xored into the plaintext to encrypt it.

Wait, is that just a XOR? Solution becomes super-easy:

PLAINTEXT=b'you need to send more parameters! seriously, why would you just send one parameter? I should just let this fail with a silent error. sending just one parameter is stupid.'
CIPHERTEXT=b"e\xc8\x9b\xc4\xfd\x04\x9b\x843a\x940U\xc4\x7fa\x11W\xa9uf\xa9\xf4%\r6C\x8d\xe1Z\x95\xb1^\x92\xddF\xa7\xbb\x86\x19\xbaCW\xde\x9bo\xd3Z\x8d\x85kx\x81a\xb0\x0b\xc9\x14'L\xc6i\xc4V\x86=\xba\x11~\xcc\x9bw#i\xc7\xb2\xc0Z\x9d\x1d\xb3\x96\\\xf9\xffG\x8a\xa2"

# that is what we get if we re-send captured request with a 0x00 changed to 0x01
# only PLAINTEXT is encrypted here
RECV=b"e\xc8\x9b\xc4\xfd\x04\x9b\x843a\x940U\xc4\x7fa\x11W\xa9uf\xa9\xf4%\xcco\x95\xb8i\x10`Tz\xd6`\x8bJ&<i\xa5@C\x99\x8ed\xd9a\xce\x83}(\x86k\x81\x06\x9a&!P\xd5 \xdaT\x9d>\xbb\x11U\xc8\x8eek'\xc7\xa3\xed@\x86\x1d\xb0\x93A\xbc\xabI\x9f\xd1\xbcp\xf8\x1a\xba2u\xf8\x88\x05H\x16\xfe\xafX8K)\x03\xbb'&\x01\xcc\xaf\x1f\x1a{\x9dB\xb1\xb7\xc7\x17\x9ahe\x80'\x13\xdbG`\xf8~L\xcf@\x04Q\x97\xac\xf5\x157\x83b\xd4\x086\x08\xb1\xeb7\xf8*\xcf\x8a]q\xb1=\xc8\xd2\xb6\xc8\xdc\xf2)\x8e\x13\xd2\x16`\x84&Wn^\xce\xa8\x84\x9e\xc0\xeb\xb2\xc2\xcc\x92\xaf\x10\xad$c\x13\x92\xae]\x87\x83\x0c\x8a.D\xdf\xe0\xbf\x12\x87 \x11:"
XOR=lambda s1,s2: bytes([(a ^ b) for a,b in zip(s1,s2)])

#I was too lazy to calculate the correct shift, so let's just iterate
for i in range(23,42): 
    print( XOR(CIPHERTEXT[i:], XOR(RECV[i:], PLAINTEXT)) )
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