# Solution to 35C3 Junior CTF challenge pretty linear pretty_linear

crypto

500 (variable)

#### Writeup

We were given a PCAP file full of recorded network traffic, and the source code for the server.

The server code that generated the network traffic looks like this:

``````    print(' '.join(map(str, challenge)))
response = int(input())
if response != sum(x*y%p for x, y in zip(challenge, key)):
exit(1)

print('ACCESS GRANTED')
``````

`challenge` and `key` are both arrays consisting of 40 integers that are 16 bytes in length and the response is another 16 byte integer. `challenge` is sent in plain text from the server to the client, and therefor known to us. `key` is a shared secret between the client and the server and never sent over the network. We need to calculate that key in order to crack the challenge. `response` is also sent over the network, from the client to the server.

Line three has the algorithm that is used to ensure that the client and the server both know the same `key`. We can expand that to a more readable form like this:

``````response = c * k % p + c * k % p + ... + c * k % p;
``````

Both variables `response` and `c` are known and `p` is hard coded to 340282366920938463463374607431768211297 in the source code.

Looking into the PCAP file we see that we have 40 of these interactions. Combining those gives us an equation system with 40 equations, each of them containing 40 variables. This is a problem that can be solved by a linear equation solver, but before we can do that we need to get the data out of the PCAP file in a structured form.

We wrote the following program to get the data out of the PCAP file, and also generate a sage program that solves the equation system in Z over 340282366920938463463374607431768211297.

``````from collections import defaultdict
import pyshark
import string
cap = pyshark.FileCapture('c5c0d261333729feb801834d5168ba4c-surveillance.pcap')

streams = defaultdict(list)

for p in cap:
t = p['tcp']
if p:
data = "".join([chr(int(c, 16))
streams[t.stream].append(data)

print("R = IntegerModRing(340282366920938463463374607431768211297)")

challenges = []
responses = []
for k, v in streams.items():
challenge, response, out = v
print(out)
challenges.append(list(map(int, challenge.split())))
responses.append(int(response))
print("M = Matrix(R, [")
first = True
for c in challenges:
if not first:
print(",")
first = False
print(c, end="")
print("])")
print("b = vector(R,[")
first = True
for r in responses:
if not first:
print(",", end="")
first = False
print(r)
print("])")
print("print(M.solve_right(b))")
``````

Running the generated sage program gives us the key, and lets us attack the next part of the server code:

``````    cipher = AES.new(
sha256(' '.join(map(str, key)).encode('utf-8')).digest(),
AES.MODE_CFB,
b'\0'*16)
print(binascii.hexlify(cipher.encrypt(flag)).decode('utf-8'))
``````

Like this:

``````import os
import math
import sys
import binascii
from hashlib import sha256
from Crypto.Cipher import AES

key = (151166356399959194245460055888166966126,
23349654305343746371904146512921179610,
303231127335861985008837572586617401477,
52564325979162295713031020943288299431,
318561098467762156502271721157519784045,
263049694618319332492436935081367988962,
151925705582116739255625584197651639678,
46319333286788790879399387215584902926,
144250191566113115015826218788418570765,
95097625879948609497612754022619234195,
40890527924981050968775993543458295905,
73015657936779070795829412187806965634,
17764129701686300306686689106838999642,
325835500394544926294581718484613749556,
71443020776832402486826429105359001130,
328905290970722092344104084599942510400,
246319993494260311894585740502008352891,
339251916682414225894494357646852524504,
270753355547506496805860877660621175158,
266604583518913012106937436764867155955,
132952188910249324219774647464400732439,
229485064954594431373138165566214808548,
273124499649767430591820642695664426994,
161206428662237066098654588615704724656,
191676246712534509807283243359699775780,
110791878778380133926865862999743362183,
121869512181659437298676494294916884080,
81324902884339942138294016318959955113,
219404824444265280645688236691554688702,
169041597038940530794876375975659802012,
131851490945732599957487956170326572223,
337190018815691236060142455413012785269,
215436829468576180414177636304832181536,
174614268507338543165725749934608091983,
316523955444804263394840392424504742312,
215434679427738924369625297037020081680,
103769840624100781721896803697739863413,
302813910848119681638497129402557822574,
104414047167186149419822776294661649936,
124689157029586638342169541932443340723)

chipher = "923fa1835d8dbdcd9f9b0e6658b24fce54512fbba71d7a0012c37d2c9dd094a6278593d8d9f7a4aa9fecb66042"

cipher = AES.new(
sha256(' '.join(map(str, key)).encode('utf-8')).digest(),
AES.MODE_CFB,
b'\0'*16)
print(cipher.decrypt(binascii.unhexlify(chipher)).decode('utf-8'))
``````

The flag was: 35C3_G4uss_w0uld_b3_so_pr0ud_of_y0u_r1ght_n0w