Backtalk

Task

"We caught two WPI students sending illegal secrets on our network... can you find out what they said?"

Also we are given pcapng file.

Solution

There are five packets with any data in this pcapng:

  • STARTING KEY EXCHANGE PROTOCOL
  • g: 10| mod: 667247790729629168801868651264033309682519458519152541737787866083418205742829005514311734375000001| pub: 490305926196059599023102212147002263206141591048662034088541405656196275339142044588265256800520651
  • pub: 395845978879527614395747015710220346649266534299630917403355837833474281961610537213685243037337675| enc_key: 212772881309582973820829804871086196222372777546441069077968628884317585160442388385222503910529273
  • 32 bytes of binary data
  • v=h_D3VFfhvs4

So as it says "Key exchange" in the first packet I instantly thought about Diffie-Hellman key exchange. It makes sense cause then we have generator g, modulus mod and two public values.
In D-H pub = pow(g, random_number, mod), so if we wanna break it we gotta find discrete logarithm.

My first attempt was to use sagemath. Assuming I got all variables set the code is just.

K = GF(mod)
discrete_log(K(pub1), K(g), mod)

It returns 2100 and as we can check in python/sage pow(10, 2100, mod) - pub1 == 0. As for discrete_log(K(pub1), K(g), mod) it doesn't want to end counting.

But stop.
I added third argument because that's how it is on the sagemath page, but that's actually really strange. The third argument should be ord - integer (multiple of order of base, or None) and the order of the base is actually mod - 1 but that doesn't work. Maybe I don't get the documentation or sth. BUT if we drop the third argument (cause it's optional) both discrete_log(K(pub1), K(g)) and discrete_log(K(pub1), K(g)) count without a problem. My only guess is that setting ord which is actually moultiple of order makes something worse?
But that's me sidetracking, I didn't realize this until writing this writeup.

Also given the logarithm is so small if one would just try bruteforcing his way e.g. counting powers of g until 1e10 he would also find it with no problem.

So we know discrete logarithm of one of the secrets and that allows us to count shared secret which is secret = pow(pub2, discrete_log(pub1), mod).
Actuall number is 470025532326429509257190794699658985614265142446015884571648783710068051626363138484599386732557854.
So then we gotta decode the enc_key (the guess is that's key to symmetric block cipher).

My first idea was that enc_key = secret * key % mod, but to not go in blind I checked on Wikipedia.
I looked for use in encryption and here we find in paragraph 5.1 link to ElGamal Encryption and it says the exact thing I just proposed.

At this point I was sure that I was right so into decrypting we go.
We just have to get enc_key / secret and as we are in Zmod it's key = enc_key * pow(secret, mod - 2, mod) % mod.
The result is key = 128009196690239019135781346654390395054.
That's a pretty small number compared to mod so it must be a hit.
Then I used Crypto.Util.number.long_to_bytes() to convert it to byte-like object. It's exactly 16 bytes so it's perfect.

So I copied data from fourth packet as hexstream and did:

>>> cipher = '8daa192c19dc4037b58def2935623704856779cefe83ff9042677b9b62661c59'
>>> cipher = int(cipher, 16)
>>> cipher = long_to_bytes(cipher)

Now we have 16-byte key and 32-byte data to decrypt. The obvious guess is AES as the most popular symmetric key algorithm AFAIK.
There are multiple modes but the simplest one is ECB it also gives us full 32-byte of data (e.g. CBC has 16-byte IV).
So the last thing to do is:

>>> from Crypto.Cipher import AES
>>> aes = AES.new(key, mode = AES.MODE_ECB)
>>> aes.decrypt(cipher)
b'WPI{sTRuk_byA_$m0otH_cR!mIn@1}\x00\x00'

Other stuff

So after the competition ended we were chatting with the task author.
He explained what was the last packet about (notice I didn't use it).
Notice that YT links are in format https://www.youtube.com/watch?v=<video id>.
That gives us the video.
That's a hints a fact that mod - 1 is extremely smooth integer (highly divisible):
factor(mod - 1) == 2^6 * 3^13 * 5^12 * 7^14 * 11^2 * 13^13 * 17^10 * 19^4 * 23^14 * 29^12'
Prime numbers that are highly divisible integers plus one are actually really unsafe cryptographically, that's offen modulus = 2 * q + 1 where q is prime. Those primes are called "safe primes".
Given mod - 1 is highly divible one can efficiently count discrite logarithm using this algorithm.
I didn't notice that, but my guess sagemath uses this or similar algorithm since it can count discrete_log(pub2, g).