## Salt and Pepper
### Challenge

#!/usr/bin/env python3

from hashlib import md5, sha1
import sys
from secret import salt, pepper
from flag import flag

assert len(salt) == len(pepper) == 19
assert md5(salt).hexdigest() == '5f72c4360a2287bc269e0ccba6fc24ba'
assert sha1(pepper).hexdigest() == '3e0d000a4b0bd712999d730bc331f400221008e0'

def auth_check(salt, pepper, username, password, h):
return sha1(pepper + password + md5(salt + username).hexdigest().encode('utf-8')).hexdigest() == h

def die(*args):

def pr(*args):
s = " ".join(map(str, args))
sys.stdout.write(s + "\n")

def sc():
return sys.stdin.readline().strip()

def main():
border = "+"
pr(border, " welcome to hash killers battle, your mission is to login into the ", border)
pr(border, " ultra secure authentication server with provided information!! ", border)

USERNAME = b'n3T4Dm1n'
PASSWORD = b'P4s5W0rd'

while True:
pr("| Options: \n|\t[L]ogin to server \n|\t[Q]uit")
ans = sc().lower()
if ans == 'l':
pr('| send your username, password as hex string separated with comma: ')
inp = sc()
inp_username, inp_password = [bytes.fromhex(s) for s in inp.split(',')]
die('| your input is not valid, bye!!')
pr('| send your authentication hash: ')
inp_hash = sc()
if USERNAME in inp_username and PASSWORD in inp_password:
if auth_check(salt, pepper, inp_username, inp_password, inp_hash):
die(f'| Congrats, you are master in hash killing, and it is the flag: {flag}')
die('| your credential is not valid, Bye!!!')
die('| Kidding me?! Bye!!!')
elif ans == 'q':
die("Quitting ...")
die("Bye ...")

if __name__ == '__main__':

We send a username and password to the server, along with an authentication hash. These are all passed as parameters to the `auth_check` function, and the username contains`n3T4Dm1n`, the password contains `P4s5W0rd`, and the function returns true, we get the flag.

### Solution

The `check_auth` function uses two secrets, `salt` and `pepper`, which we know the length of, however we don't know the value of.

The `check_auth` function calculates the authentication hash using the following line

sha1(pepper + password + md5(salt + username).hexdigest().encode('utf-8')).hexdigest()

Since these two secrets are hashed as well as our username and password, we cannot directly work out the authentication hash. However, we get given the MD5 hash of `salt`, and the SHA1 hash of `pepper`. Since both of the secret values are put as prefixes to our input, we can perform a hash length extension attack.

[HashPump](https://github.com/bwall/HashPump) is a useful tool to do this, as all we need to do is provide the parameters and the tool does most of the work for us. One thing that needed to be changed however is that since we get the raw hashes, we don't have any data to give to the tool, and Hashpump complains when we do that.

To get around this, I simply removed this check in the `main.cpp` file (line 255) and recompiled it.

First, we will create a MD5 of (`salt` + `padding` + `n3T4Dm1n`) using the tool:

hashpump -s "5f72c4360a2287bc269e0ccba6fc24ba" -d "" -a "n3T4Dm1n" -k 19

giving an output of


Then, we will create our authentication hash by creating a SHA1 of (`pepper` + `padding` + `P4s5W0rd` + `95623660d3d04c7680a52679e35f041c`)

hashpump -s "3e0d000a4b0bd712999d730bc331f400221008e0" -d "" -a "P4s5W0rd95623660d3d04c7680a52679e35f041c" -k 19

giving an output of


`83875efbe020ced3e2c5ecc908edc98481eba47f` should now be our authentication hash when we use `\x80\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x98\x00\x00\x00\x00\x00\x00\x00n3T4Dm1n` as our username and `\x80\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x98P4s5W0rd` as our password (note that we remove the MD5 hash at the end as it gets added when the `auth_check` function is called).

Submitting these to the server gives us the flag.

##### Flag


Original writeup (https://blog.cryptohack.org/cryptoctf2021-easy#salt-and-pepper).