मुख्य कंटेंट तक स्किप करें

Moving Problems

I've learnt that when life gives you lemons, if you look at things the right way they taste just like pairs.

given

source.sage
import random
import hashlib
from Crypto.Cipher import AES
from sage.all import *
from Crypto.Util.Padding import pad, unpad

FLAG = b"crypto{??????????????????????????????????????}"

def gen_keypair(G, p):
n = random.randint(1, (p-1))
P = n*G
return n, P

def gen_shared_secret(P, n):
S = P*n
return S.xy()[0]

def encrypt_flag(shared_secret: int):
# Derive AES key from shared secret
sha1 = hashlib.sha1()
sha1.update(str(shared_secret).encode('ascii'))
key = sha1.digest()[:16]
# Encrypt flag
iv = os.urandom(16)
cipher = AES.new(key, AES.MODE_CBC, iv)
ciphertext = cipher.encrypt(pad(FLAG, 16))
# Prepare data to send
data = {}
data['iv'] = iv.hex()
data['encrypted_flag'] = ciphertext.hex()
return data

# Define Curve params
p = 1331169830894825846283645180581
a = -35
b = 98
E = EllipticCurve(GF(p), [a,b])
G = E.gens()[0]

# Generate keypair
n_a, P1 = gen_keypair(G, p)
n_b, P2 = gen_keypair(G, p)

# Calculate shared secret
S1 = gen_shared_secret(P1, n_b)
S2 = gen_shared_secret(P2, n_a)

# Check protocol works
assert S1 == S2

flag = encrypt_flag(S1)

print(f"Generator: {G}")
print(f"Alice Public key: {P1}")
print(f"Bob Public key: {P2}")
print(f"Encrypted flag: {flag}")
output.txt
Generator: (479691812266187139164535778017 : 568535594075310466177352868412 : 1)
Alice Public key: (1110072782478160369250829345256 : 800079550745409318906383650948 : 1)
Bob Public key: (1290982289093010194550717223760 : 762857612860564354370535420319 : 1)
Encrypted flag: {'iv': 'eac58c26203c04f68d63dc2c58d79aca', 'encrypted_flag': 'bb9ecbd3662d0671fd222ccb07e27b5500f304e3621a6f8e9c815bc8e4e6ee6ebc718ce9ca115cb4e41acb90dbcabb0d'}

Solution

This WriteUp Solution is password protected by the flag of the challenge.