Here’s a link to the room: https://tryhackme.com/room/flip

TASK [1] Source Code

First, go ahead and review the source code before moving on to Task 2.

You can review the source code by clicking on the Download Task Files button at the top of this task to download the required file.

import socketserver 
import socket, os
from Crypto.Cipher import AES
from Crypto.Util.Padding import pad,unpad
from Crypto.Random import get_random_bytes
from binascii import unhexlify
flag = open('flag','r').read().strip()
def encrypt_data(data,key,iv):
    padded = pad(data.encode(),16,style='pkcs7')
    cipher = AES.new(key, AES.MODE_CBC,iv)
    enc = cipher.encrypt(padded)
    return enc.hex()
def decrypt_data(encryptedParams,key,iv):
    cipher = AES.new(key, AES.MODE_CBC,iv)
    paddedParams = cipher.decrypt( unhexlify(encryptedParams))
    if b'admin&password=sUp3rPaSs1' in unpad(paddedParams,16,style='pkcs7'):
        return 1
    else:
        return 0
def send_message(server, message):
    enc = message.encode()
    server.send(enc)
def setup(server,username,password,key,iv):
        message = 'access_username=' + username +'&password=' + password
        send_message(server, "Leaked ciphertext: " + encrypt_data(message,key,iv)+'\n')
        send_message(server,"enter ciphertext: ")
        enc_message = server.recv(4096).decode().strip()
        try:
                check = decrypt_data(enc_message,key,iv)
        except Exception as e:
                send_message(server, str(e) + '\n')
                server.close()
        if check:
                send_message(server, 'No way! You got it!\nA nice flag for you: '+ flag)
                server.close()
        else:
                send_message(server, 'Flip off!')
                server.close()
def start(server):
        key = get_random_bytes(16)
        iv = get_random_bytes(16)
        send_message(server, 'Welcome! Please login as the admin!\n')
        send_message(server, 'username: ')
        username = server.recv(4096).decode().strip()
        send_message(server, username +"'s password: ")
        password = server.recv(4096).decode().strip()
        message = 'access_username=' + username +'&password=' + password
        if "admin&password=sUp3rPaSs1" in message:
            send_message(server, 'Not that easy :)\nGoodbye!\n')
        else:
            setup(server,username,password,key,iv)
class RequestHandler(socketserver.BaseRequestHandler):
    def handle(self):
        start(self.request)
if __name__ == '__main__':
    socketserver.ThreadingTCPServer.allow_reuse_address = True
    server = socketserver.ThreadingTCPServer(('0.0.0.0', 1337), RequestHandler)
    server.serve_forever()

~ In the source code you could find the password : sUp3rPaSs1

TASK [2] What is the flag?

Log in as the admin and capture the flag!

If you can…

Whenever you are ready, click on the Start Machine button to fire up the Virtual Machine. Please allow 3–5 minutes for the VM to fully start.

The server is listening on port 1337 via TCP. You can connect to it using Netcat or any other tool you prefer.

For better use of Netcat do use rlwrap
syntax : rlwrap nc <ip> 1337

So when i login using the credential given
Username : admin ,Password : sUp3rPaSs1

I got the output

When I attempted to log in with changes in username , the output I received was…

As it posed some challenges, I took the initiative to comprehend the source code in order to determine the encryption method employed and identify the specific values involved in the encryption process.

so as in the source code said the cipher text is decrypt using CBC(cipher-block-chaining)

i got the leaked ciphertext : f395a700538e092c2d4a75f2a47d408028e8e6876fcc60879258be9be53abf95da70307fef52558fdd8ddc4aae7e4a33

as shown in the source the leaked ciphertext is in the form of
message = ‘access_username=’ + username +’&password=’ + password

make it more simple using using the above picture
message = access_username=ddmin&password=sUp3rPaSs1
the ciphertext is a combination of this 3 thing highlighted

By observing the AES Encryption and Decryption process, we can determine that performing an XOR operation between two texts allows us to derive the third text.

images source: https://en.wikipedia.org/wiki/Block_cipher_mode_of_operation

The Process

Leaked ciphertext = f395a700538e092c2d4a75f2a47d408028e8e6876fcc60879258be9be53abf95da70307fef52558fdd8ddc4aae7e4a33

Password = sUp3rPaSs1 admin&password=sUp3rPaSs1' in unpad(paddedParams,16,style=’pkcs7'):

access_username=
ddmin&password=s
Up3rPaSs1@@@@@@@

f395a700538e092c2d4a75f2a47d4080
28e8e6876fcc60879258be9be53abf95
da70307fef52558fdd8ddc4aae7e4a33

code = message = ‘access_username=’ + username +’&password=’ + password

f395a700538e092c2d4a75f2a47d4080
xor
28e8e6876fcc60879258be9be53abf95 -> dec aes
xor
da70307fef52558fdd8ddc4aae7e4a33 -> dec aes

a=f3 from acces_username=
d=28 from ddmin&password=s

f3 xor dec(28) -> d
f3 xor dec(28) -> 64 {d hex value is 64}

f3 xor 64 = dec(28)
f3 xor 64 = 97
f3 xor 97 -> 64 d

to convert (d) into (a) we have to convert old cipher

? xor 97 = 61 {a hex value is 61}
97 xor 61 = f6
f6 — — -> you have to just change the starting two alphabet of the leaked ciphertext

enter the ciphertext — f695a700538e092c2d4a75f2a47d408028e8e6876fcc60879258be9be53abf95da70307fef52558fdd8ddc4aae7e4a33

Result :