Pwn2win2020 Androids Encryption

Challenge Info

​ The challenge is a crypto challenge from the pwn2win event , it’s focused on the symmetric cryptography and especially the aes block cipher.So we are given remote connection nc encryption.pwn2.win 1337 and the python script that is running in the remote server.py .

Writeup Summary

• gain general information

• deep look into encrypt your secret

• Solution

gain general information

​ By the first look at the server.py script we notice there are 3 main functions first one is def encrypt(txt, key, iv) where you pass the plaintext the key and iv this function will check that the plaintext length is multiple of Block Size which is 128 and then it will encrypt the plaintext with custom implementation of AES , the second one is enc_plaintextthis function that we will interact with it will take our plaintext and decode it as a base64 and pass it to encrypt with key1 and iv1 which are secrets . The last function is enc_flag() it will encrypt the flag with key2 and iv2 which their difinition is :

1
2

	iv2 = AES.new(key1, AES.MODE_ECB).decrypt(iv1)
	key2 = xor(to_blocks(flag))

by looking at the xor function we noticed that if two parametres a,b are passed then it will calculate a xor b else if one parametre a is passed it will return a[0] xor \x00 *len(a[0]) so it will return the first block xored with 00 and that meen it will return the first block .

deep look into encrypt your secret

​ After looking in the function i noticed it doing 2 things interesting :

• first thing is that it is returning to us the iv passed in parametre + the cipher so for example if we passed iv2 to the function than we get as a result iv2+ cipher

  1	base64.b64encode(iv+ctxt)

  PS: of course all inputs and outputs are encoded with base64

• the second interesting thing is that it is overriding the iv2 and key2 that are used in encrypting the flag :

  1 2	iv2 = AES.new(key2, AES.MODE_ECB).decrypt(iv2) key2 = xor(to_blocks(ctxt))

  so the new iv2 is the decryption of the previous iv2 with key2 and the new key2 is the xor of the cipher calculated and as we have discussed xor function when we pass one parametre it will return the first block of the passed object so key2 = cipher[0] so from the result iv+ctx we can get iv1 and key2 from the cipher and to get the next value of iv2 we need the value of key2 before change .so the idea of the challenge is to try to guess the iv2 and key2 that will be used next time we encrypt the flag .

Solution

​ after trying in a paper the different combination of commandes that will allow us to get the key and initial vector i finnaly found the solution it is bit tricky so what we will do is :

1. we will send a random payload with 16 bytes to the oracle that will return us iv1 + cipher and from the cipher we can get the new value key2 because as we said key2 = a[0] the new key is the first block of cipher

2. next we will send the encrypt flag command that will return to us iv2 + flag_cipher and here we can use the result of the first step which is key2 and get the new value of iv2 because it will be changed by iv2 = AES.new(key2, AES.MODE_ECB).decrypt(iv2) and also we update the value of key2 now we have the key2 and iv2 values and we can use it to decipher the flag next time because this value will be used next time

3. we will send the encrypt flag command and decrypt the cipher :

   1 2 3 4 5 6 7 8

   aes = AES.new(key2, AES.MODE_ECB) curr = iv2 bs = len(key2) text=b"" for block in blocks: text +=xor(curr, aes.decrypt(block)) curr = xor(text[-bs:], block) print(text)

   the challenge used a modified version of aes ecb it work like this cipher1 = aes(text1 xor iv) then cipher2 = aes(text2 xor (text1 xor cipher1)) and it do this for each block . so for the decryption part we can do this

   text1 = iv xor aes.decrypt(cipher1) and for other blocks textI= (cipherJ xor textJ) xor aes.decrypt(cipherI) where J = I-1.

and finnaly we got the flag CTF-BR{kn3W_7h4T_7hEr3_4r3_Pc8C_r3pe471ti0ns?!?} . awesome challenge had so much fun solving it .

​