TGHack - Transmission Control

A TGHack writeup by Edward Pwnden

Posted by Nikolai Magnussen on April 19, 2017

This is the first in a series of blog posts that will explain challenges from hacking competition held at The Gathering 2017, called TGHack. TGHack is organized by The Gathering, created and held by a number of students at UiO, and sponsored by NSM.

I implore you to read the blog and writeups of Martin Ingesen, another member of the Edward Pwnden CTF team. You should also read the writeup of this challenge by it’s creator here

Transmission Control

Category: Crypto

Points: 275

Author: Chabz

We’ve intercepted some messages, but we’ve had no luck deciphering it. Can you help us?

The transmission

After downloading the transmission:

B: UmVxdWVzdGluZyBzZWN1cmUgY29ubmVjdGlvbg==
B: VYOMsCq1yqv0J+8UMe0pFaCtUi5EFlrhqKh+Y4bmR1MvoGzgSxAI6ZD8QJV0VFl9
A: atU5UvHp5bRbUMeIiJLVxkdLaGF8tivfSKlGpOJQvLRE/74LKZ9DD42xTE6mwbmk2P7Po2Y2Ryh80aYAJY8atA==
B: grl0/CNUkhes9mdZPsLieUC5/kWx3G0n/yEuQNbE+Ao=
B: IvDZ++ZH9jx3LGCJzkS0TbZYz1N6siy2OD/A0xQbuLk=
A: wXEf37wwVGHuYaoD/nULOoUN7JBA9Kq/rH6Rduj/ZW7oLbUfQcPW/0Zto5rqI0T+

we see lines ending with one or more equal signs. This is often an indicator of the data being base64-encoded.

The transmission file can be decoded as follows:

from base64 import b64decode

for line in open("transmission.txt", "r"):
    sender, message = line.split(" ")
    print("{} {}".format(sender, b64decode(message)))

Yielding the actual transmission data:

B: b'Requesting secure connection'
A: b'Initiating protocol: AES(SHA256(ECDH), CBC, IV)'
A: b'Message format: IV + AES(message)'
A: b'420173635064, 4558608748405, 15517495227187'
A: b'13847371191292, 7178231866375'
B: b'12740127109992, 2657998351758'
A: b'12108314104269, 7751833675327'
B: b"U\x83\x8c\xb0*\xb5\xca\xab\xf4'\xef\x141\xed)\x15\xa0\xadR.D\x16Z\xe1\xa8\xa8~c\x86\xe6GS/\xa0l\xe0K\x10\x08\xe9\x90\xfc@\x95tTY}"
A: b'j\xd59R\xf1\xe9\xe5\xb4[P\xc7\x88\x88\x92\xd5\xc6GKha|\xb6+\xdfH\xa9F\xa4\xe2P\xbc\xb4D\xff\xbe\x0b)\x9fC\x0f\x8d\xb1LN\xa6\xc1\xb9\xa4\xd8\xfe\xcf\xa3f6G(|\xd1\xa6\x00%\x8f\x1a\xb4'
B: b"\x82\xb9t\xfc#T\x92\x17\xac\xf6gY>\xc2\xe2y@\xb9\xfeE\xb1\xdcm'\xff!.@\xd6\xc4\xf8\n"
B: b'"\xf0\xd9\xfb\xe6G\xf6<w,`\x89\xceD\xb4M\xb6X\xcfSz\xb2,\xb68?\xc0\xd3\x14\x1b\xb8\xb9'
A: b'\xc1q\x1f\xdf\xbc0Ta\xeea\xaa\x03\xfeu\x0b:\x85\r\xec\x90@\xf4\xaa\xbf\xac~\x91v\xe8\xffen\xe8-\xb5\x1fA\xc3\xd6\xffFm\xa3\x9a\xea#D\xfe'

Initially, we see the initialization of the protocol and the message format before we see some strange numbers, and finally data. In this scenario, A plays the role of the server, while B is the client.

The server initially specify the protocol to be used, namely AES(SHA256(ECDH), CBC, IV). It might not be obvious, but this is the protocol specification for the AES usage in the message, which has the format IV + AES(message).

Meaning each message contain it’s very own IV to be used for decryption of that particular message. The protocol specify the AES key to be the SHA256 digest of the shared ECDH key

Breaking ECDH

After reading about ECDH key exchange, we see that the transmission is as follows:

Where , and are the curve parameters, and must be prime, used to generate the elliptic curve by the means of the following function:

Next up, the generator base point is provided in the pair

Now, the client generate it’s secret key as a random integer before computing and sending Finally; the server does the same, only we call this key and the transfered data is

Because both client and server have their own secret key, and the communication partner’s secret key multiplied with the generator they can calculate the shared key, which is calculated using the following formula:

But we do not have any knowledge of either nor meaning we can’t calulate the shared key. Perhaps this is a poor implementation due to bad parameters? We can easily check that the prime is only 44 bits, which result in the possibility of breaking the discrete logarithm problem that Diffie-Hellman relies on.

Let’s try to do just that using sagemath:

# First, the three curve parameters
sage: a = 420173635064
sage: b = 4558608748405
sage: p = 15517495227187

# Create curve function
sage: E = EllipticCurve(GF(p), [a, b])

# Get the initial point
sage: G  = E(13847371191292, 7178231866375)

# Calculate the 'public key' for each
sage: cG = E(12740127109992, 2657998351758)
sage: dG = E(12108314104269, 7751833675327)

# Try to break the 'public key' to expose the secret ones.
sage: c = G.discrete_log(cG)
sage: d = G.discrete_log(dG)
sage: c
sage: d

# Check that they result in the same shared key
sage: cG*d
(27356762530 : 7671609982327 : 1)
sage: dG*c
(27356762530 : 7671609982327 : 1)

# Because they are equal, this is the shared key, and we succeeded in breaking poor ECDH
sage: shared = cG*d

Because elliptic curves have both an X and a Y coordinates, we should try both, but X is generally the one used, so we say for the rest of the writeup.

Decrypting messages

Putting all of the above information together; we must decrypt each message by creating a new AES crypter, where the key is SHA256(ECDH), the mode of operation is CBC, and IV is the first 16 bytes of the message.

These steps are combined in the following code snippet:

from base64 import b64decode
from Crypto.Cipher import AES
from Crypto.Hash import SHA256

messages = [

secret = "27356762530"
hasher =
key = hasher.digest()

def decrypt(message):
    iv = message[:16]
    encrypted = message[16:]

    decrypter =, AES.MODE_CBC, iv)
    return decrypter.decrypt(encrypted)

for m in messages:

Executing this yield the following conversation:

Send me flag plz
Here you go: TG17{large_primes_are_nice}
End connection
Connection closed

And the flag is TG17{large_primes_are_nice}.

My appreciations to the team behind TGHack for creating such an awesome CTF!