PicoCTF - It's Not My Fault 1

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PicoCTF - It’s Not My Fault 1

tags: PicoCTF CTF Crypto

Source code

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#!/usr/bin/python3 -u
import random
import string
import hashlib
import time

from Crypto.Util.number import inverse, getPrime, bytes_to_long, GCD
from sympy.ntheory.modular import solve_congruence

FLAG = open('flag.txt', 'r').read()

def CRT(a, m, b, n):
	val, mod = solve_congruence((a, m), (b, n))
	return val

def gen_key():
	while True:
		p = getPrime(512)
		q = getPrime(512)
		if GCD(p-1, q-1) == 2:
			return p, q

def get_clue(p, q, BITS):
	while True:
		d_p = random.randint(1, 1 << BITS)
		d_q = random.randint(1, q - 1)
		if d_p % 2 == d_q % 2:
			d = CRT(d_p, p - 1, d_q, q - 1)
			e = inverse(d, (p - 1) * (q - 1))
			print("Clue : ", e)
			return

def get_flag(p, q):
	start = time.time()
	ans = int(input())
	if (time.time() - start) > (15 * 60):
		print("Too long!")
		exit()
	else:
		if ans == p + q:
			print(FLAG)
		else:
			print("oops...")


#PoW

vals1 = "".join([random.choice(string.digits) for _ in range(5)])
vals2 = "".join([random.choice(string.hexdigits.lower()) for _ in range(6)])
user_input = input("Enter a string that starts with \"{}\" (no quotes) which creates an md5 hash that ends in these six hex digits: {}\n".format(vals1, vals2))
user_hash = hashlib.md5(user_input.encode()).hexdigest()

if user_input[:5] == vals1 and user_hash[-6:] == vals2:
	p, q = gen_key()
	n = p * q
	print("Public Modulus : ", n)
	get_clue(p, q, 20)
	get_flag(p, q)

:::

  • 一開始先設立PoW的Challenge，之後才進到RSA的部分
  • 可以看到在54行先產出大質數$p$, $q$，並把$n$ release出來
  • 隨後進到get_clue function，可以看到他先create一個$d_q$和$d_p$，且$d_p$的range被限制在$(1,\ 2^{20}=1048576)$
  • 在28行的地方進入CRT function，這一段是在找一個$d$使其符合中國餘式定理，也就是:
\[d\ \%\ m=a\\ d\ \%\ n=b\]

帶入參數就是

\[d\ \%\ (p-1)=d_p\\ d\ \%\ (q-1)=d_q\]

Recon

這一題超難，看了三篇Write Up還是看不懂其中的原理

Exploit

  • Prove of Work 
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      from pwn import *
  from tqdm import trange
  import hashlib

  context.arch = 'amd64'
  # r = process(['python', 'not_my_fault.py'])
  r = remote('mercury.picoctf.net', 41175)

  r.recvuntil(b'Enter a string that starts with "')
  tmp = r.recvline().strip().decode()

  value1 = tmp.split('"')[0]
  value2 = tmp.split(": ")[-1]

  log.info("Prefix = {}, Postfix = {}".format(value1, value2))
  for i in trange(20000000000):
      guess_collision = hashlib.md5((value1 + str(i)).encode()).hexdigest()
      if guess_collision[-6:] == value2:
          r.sendline((value1 + str(i)).encode())
          print("Collision Found: {}".format(value1 + str(i)))
          break

  r.interactive()

Reference

maple3142 - It’s Not My Fault 1 It’s Not My Fault 1 - Write Up Attack on CRT-RSA picoCTF 2021 It’s Not My Fault 1 - Video