Reversing : Secure transfer

Task

We are given 2 files : an executable and a packet capture of an exchange.

Process

We launch it in ghidra and read the pseudo code to find out what is happening. And we find out that the file transfer happenned this way :

1) Sender sends length of data
2) Receiver receives and verifies length
3) Sender encrypts the data using AES-CBC and then sends it
4) Receiver receives and verifies that the length corresponds to the real size of the data
5) Receiver decrypts the data using AES-CBC

Then we look for the values of variables that interests us the most which are the initialization value and the key.

Finding the initialization value is straight forward :

iv = "someinitialvalue"

The key is less straight forward but still easy to get. It's just a bunch of scrambled MOV instructions :

        001016d5 c6 45 d0 73     MOV        byte ptr [RBP + key], 's'
        001016d9 c6 45 d9 65     MOV        byte ptr [RBP + local_2f],'e'
        001016dd c6 45 d1 75     MOV        byte ptr [RBP + local_37],'u'
        001016e1 c6 45 d2 70     MOV        byte ptr [RBP + local_36],'p'
        001016e5 c6 45 e2 66     MOV        byte ptr [RBP + local_26],'f'
        001016e9 c6 45 e3 6f     MOV        byte ptr [RBP + local_25],'o'
        001016ed c6 45 e4 72     MOV        byte ptr [RBP + local_24],'r'
        001016f1 c6 45 e7 63     MOV        byte ptr [RBP + local_21],'c'
        001016f5 c6 45 da 74     MOV        byte ptr [RBP + local_2e],'t'
        001016f9 c6 45 db 6b     MOV        byte ptr [RBP + local_2d],'k'
        001016fd c6 45 eb 74     MOV        byte ptr [RBP + local_1d],'t'
        00101701 c6 45 ed 6f     MOV        byte ptr [RBP + local_1b],'o'
        00101705 c6 45 d6 65     MOV        byte ptr [RBP + local_32],'e'
        00101709 c6 45 d7 63     MOV        byte ptr [RBP + local_31],'c'
        0010170d c6 45 d5 73     MOV        byte ptr [RBP + local_33],'s'
        00101711 c6 45 e8 72     MOV        byte ptr [RBP + local_20],'r'
        00101715 c6 45 dd 79     MOV        byte ptr [RBP + local_2b],'y'
        00101719 c6 45 de 75     MOV        byte ptr [RBP + local_2a],'u'
        0010171d c6 45 df 73     MOV        byte ptr [RBP + local_29],'s'
        00101721 c6 45 ec 69     MOV        byte ptr [RBP + local_1c],'i'
        00101725 c6 45 e0 65     MOV        byte ptr [RBP + local_28],'e'
        00101729 c6 45 e1 64     MOV        byte ptr [RBP + local_27],'d'
        0010172d c6 45 e5 65     MOV        byte ptr [RBP + local_23],'e'
        00101731 c6 45 e9 79     MOV        byte ptr [RBP + local_1f],'y'
        00101735 c6 45 d8 72     MOV        byte ptr [RBP + local_30],'r'
        00101739 c6 45 d4 72     MOV        byte ptr [RBP + local_34],'r'
        0010173d c6 45 ea 70     MOV        byte ptr [RBP + local_1e],'p'
        00101741 c6 45 ef 21     MOV        byte ptr [RBP + local_19],'!'
        00101745 c6 45 ee 6e     MOV        byte ptr [RBP + local_1a],'n'
        00101749 c6 45 dc 65     MOV        byte ptr [RBP + local_2c],'e'
        0010174d c6 45 d3 65     MOV        byte ptr [RBP + local_35],'e'
        00101751 c6 45 e6 6e     MOV        byte ptr [RBP + local_22],'n'

Rewriting it into the right order, it gives us the key :

key="supersecretkeyusedforencryption!"

Finally, now that we know the inner workings of the program. We can decrypt 32bytes of data in packet no. 5 :

from Crypto.Cipher import AES
from Crypto.Util.Padding import unpad

#key = b"setktiupncoecsryrforneyusedeerp!"
#key = b"seupforctktoecsryusiedeyrrp!neen"
key = b"supersecretkeyusedforencryption!"
iv = b"someinitialvalue"
ct = bytes.fromhex("5f558867993dccc99879f7ca39c5e406972f84a3a9dd5d48972421ff375cb18c")
cipher = AES.new(key, AES.MODE_CBC, iv)
pt = cipher.decrypt(ct)
print(pt)

output : b'HTB{vryS3CuR3_F1L3_TR4nsf3r}\x04\x04\x04\x04'