Smooth Criminal

The title of the challenge hints at a clue - the modulo used in the challenge is a smooth prime. There are no obvious weakness in the encryption scheme given in the source code. Again, revising the old lesson - Pohlig-Hellman algorithm is insane at solving DLP with primes of smooth order.

Otherwise, the challenge is simply a matter of figuring out the proper Sage syntax to solve the challenge. discrete_log of Sage is powerful when it comes to solving DLP, as the underlying implementation of Sage involves both Pohlig-Hellman and BSGS.

Sage Implementation:

xxxxxxxxxx
from Crypto.Hash import SHA1
from Crypto.Cipher import AES
from Crypto.Util.Padding import pad, unpad
import hashlib


def is_pkcs7_padded(message):
    padding = message[-message[-1]:]
    return all(padding[i] == len(padding) for i in range(0, len(padding)))


def decrypt_flag(shared_secret: int, iv: str, ciphertext: str):
    # Derive AES key from shared secret
    sha1 = hashlib.sha1()
    sha1.update(str(shared_secret).encode('ascii'))
    key = sha1.digest()[:16]
    # Decrypt flag
    ciphertext = bytes.fromhex(ciphertext)
    iv = bytes.fromhex(iv)
    cipher = AES.new(key, AES.MODE_CBC, iv)
    plaintext = cipher.decrypt(ciphertext)

    if is_pkcs7_padded(plaintext):
        return unpad(plaintext, 16).decode('ascii')
    else:
        return plaintext.decode('ascii')

p = 310717010502520989590157367261876774703
F = GF(p)
E = EllipticCurve(F, [2, 3])

g_x = 179210853392303317793440285562762725654
g_y = 105268671499942631758568591033409611165
G = E(g_x, g_y)

b_x = 272640099140026426377756188075937988094
b_y = 51062462309521034358726608268084433317
B = E(b_x, b_y)

public = E(280810182131414898730378982766101210916, 291506490768054478159835604632710368904)

iv = '07e2628b590095a5e332d397b8a59aa7'
encrypted_flag = '8220b7c47b36777a737f5ef9caa2814cf20c1c1ef496ec21a9b4833da24a008d0870d3ac3a6ad80065c138a2ed6136af'

n = discrete_log(public, G, operation='+')
# n = 47836431801801373761601790722388100620
shared_secret = (B * n)[0]

print(decrypt_flag(shared_secret, iv, encrypted_flag))