# bfail *To 'B' secure or to 'b' fail? Strong passwords for admins are always great, right?* * *Autor do writeup:* [*@jackskelt*](https://github.com/jackskelt) * *Autor do desafio:* [*@gehaxelt*](https://github.com/gehaxelt) > Você pode acessar os arquivos do desafio no nosso repositório Ao acessa a página, nos deparamos com uma tela de login. ![Login](/files/owqrx5vtmIw3u0bk2gaI) Olhando o código da página renderizada, vemos que há um comentário dizendo que podemos acessa o código fonte. ![Código fonte da página renderizada](/files/iW6fP5cLR7jtU5PN0taY) {% code title="" overflow="wrap" lineNumbers="true" %} ```py from flask import Flask, request, redirect, render_template_string import sys import os import bcrypt import urllib.parse app = Flask(__name__) app.secret_key = os.urandom(16); # This is super strong! The password was generated quite securely. Here are the first 70 bytes, since you won't be able to brute-force the rest anyway... # >>> strongpw = bcrypt.hashpw(os.urandom(128),bcrypt.gensalt()) # >>> strongpw[:71] # b'\xec\x9f\xe0a\x978\xfc\xb6:T\xe2\xa0\xc9<\x9e\x1a\xa5\xfao\xb2\x15\x86\xe5$\x86Z\x1a\xd4\xca#\x15\xd2x\xa0\x0e0\xca\xbc\x89T\xc5V6\xf1\xa4\xa8S\x8a%I\xd8gI\x15\xe9\xe7$M\x15\xdc@\xa9\xa1@\x9c\xeee\xe0\xe0\xf76' app.ADMIN_PW_HASH = b'$2b$12$8bMrI6D9TMYXeMv8pq8RjemsZg.HekhkQUqLymBic/cRhiKRa3YPK' FLAG = open("flag.txt").read(); @app.route('/source') def source(): return open(__file__).read() @app.route('/', methods=["GET"]) def index(): username = request.form.get("username", None) password = request.form.get("password", None) if username and password: username = urllib.parse.unquote_to_bytes(username) password = urllib.parse.unquote_to_bytes(password) if username != b"admin": return "Wrong user!" if len(password) > 128: return "Password too long!" if not bcrypt.checkpw(password, app.ADMIN_PW_HASH): return "Wrong password!" return f"""Congrats! It appears you have successfully bf'ed the password. Here is your {FLAG}""" # Use f-string formatting within the template string template_string = """ Bfail

Login to get my secret, but 'B'-ware of the strong password!

""" return render_template_string(template_string) if __name__ == '__main__': app.run(debug=False, host="0.0.0.0", port="8080", threaded=True) ``` {% endcode %} O código nos revela 71 bytes da senha do usuário admin, que foi criada utilizando a função de hash do bcrypt. O hash do bcrypt tem 72 bytes, logo temos que descobrir somente o último byte da senha. Para isso, testamos as 256 possibilidades, passamos pela função de hash e comparamos com o hash da senha que temos. {% code title="exploit.py" overflow="wrap" lineNumbers="true" %} ```py import bcrypt import itertools import urllib.parse # Leaked first 71 bytes of the random password (from the challenge comment) leaked_prefix = b'\xec\x9f\xe0a\x978\xfc\xb6:T\xe2\xa0\xc9<\x9e\x1a\xa5\xfao\xb2\x15\x86\xe5$\x86Z\x1a\xd4\xca#\x15\xd2x\xa0\x0e0\xca\xbc\x89T\xc5V6\xf1\xa4\xa8S\x8a%I\xd8gI\x15\xe9\xe7$M\x15\xdc@\xa9\xa1@\x9c\xeee\xe0\xe0\xf76' print("Leaked prefix length:", len(leaked_prefix)) # The stored bcrypt hash for the admin user (only the first 72 bytes of the input matter) target_hash = b"$2b$12$8bMrI6D9TMYXeMv8pq8RjemsZg.HekhkQUqLymBic/cRhiKRa3YPK" found_password = None # Iterate over all possible combinations for the 1 missing byte for suffix in itertools.product(range(256)): candidate = leaked_prefix + bytes(suffix) if bcrypt.checkpw(candidate, target_hash): found_password = candidate print("Found password:", candidate) print("URL encoded: ", urllib.parse.quote(candidate)) break if not found_password: print("Password not found!") ``` {% endcode %} ```bash $ python exploit.py Leaked prefix length: 71 Found password: b'\xec\x9f\xe0a\x978\xfc\xb6:T\xe2\xa0\xc9<\x9e\x1a\xa5\xfao\xb2\x15\x86\xe5$\x86Z\x1a\xd4\xca#\x15\xd2x\xa0\x0e0\xca\xbc\x89T\xc5V6\xf1\xa4\xa8S\x8a%I\xd8gI\x15\xe9\xe7$M\x15\xdc@\xa9\xa1@\x9c\xeee\xe0\xe0\xf76\xaa' URL encoded: %EC%9F%E0a%978%FC%B6%3AT%E2%A0%C9%3C%9E%1A%A5%FAo%B2%15%86%E5%24%86Z%1A%D4%CA%23%15%D2x%A0%0E0%CA%BC%89T%C5V6%F1%A4%A8S%8A%25I%D8gI%15%E9%E7%24M%15%DC%40%A9%A1%40%9C%EEe%E0%E0%F76%AA ``` Agora podemos pegar a senha codificada para URL e fazer uma requisição form com `GET` passando `username=admin` e `password=%EC%9F%E0a%978%FC%B6%3AT%E2%A0%C9%3C%9E%1A%A5%FAo%B2%15%86%E5%24%86Z%1A%D4%CA%23%15%D2x%A0%0E0%CA%BC%89T%C5V6%F1%A4%A8S%8A%25I%D8gI%15%E9%E7%24M%15%DC%40%A9%A1%40%9C%EEe%E0%E0%F76%AA`, assim obtendo a flag ``` ENO{BCRYPT_FAILS_TO_B_COOL_IF_THE_PW_IS_TOO_LONG} ``` ![Requisição feita pelo BurpSuite](/files/NezL53m1JXOpkRqGK5KH) --- # Agent Instructions: Querying This Documentation If you need additional information that is not directly available in this page, you can query the documentation dynamically by asking a question. Perform an HTTP GET request on the current page URL with the `ask` query parameter: ``` GET https://hawksec.gitbook.io/pt/writeups/anos/2025/nullcon-hackim-ctf-goa-2025/bfail.md?ask= ``` The question should be specific, self-contained, and written in natural language. The response will contain a direct answer to the question and relevant excerpts and sources from the documentation. Use this mechanism when the answer is not explicitly present in the current page, you need clarification or additional context, or you want to retrieve related documentation sections.