I am trying to solve a challenge regarding a RSA oracle which allows me to encrypt/decrypt any plaintext/ciphertext I want, but there are a few checks that I have to bypass, and my goal is to decrypt the given flag. The strategy I am using is basically trying to get the N by making the oracle encrypt some small numbers, and then just adding this N to the encrypted flag to bypass the check:

if c == flag_encrypted:

My current script works if I remove (in my local version of the oracle) the second check on the used array, but of course I cannot remove it in the remote one, which contains the flag I am trying to decrypt. Do you have any idea on how I can bypass the following check?

for no in used:
      if m % no == 0:
        print("Wait. That's illegal.")

Oracle's code:

#!/usr/bin/env python3

import signal
from binascii import hexlify
from Crypto.PublicKey import RSA
from Crypto.Util.number import *
from random import randint
from secret import FLAG
import string

TIMEOUT = 300 # 5 minutes time-out

def menu():
    print('  [0] Exit')
    print('  [1] Encrypt')
    print('  [2] Decrypt')
    return input('> ')

def encrypt(m):
    return pow(m, rsa.e, rsa.n)

def decrypt(c):
    return pow(c, rsa.d, rsa.n)

rsa = RSA.generate(1024)
flag_encrypted = pow(bytes_to_long(FLAG.encode()), rsa.e, rsa.n)
used = [bytes_to_long(FLAG.encode())]

def handle():
  print("=                      RSA Encryption & Decryption oracle                      =")
  print("=                                Find the flag!                                =")
  print("Encrypted flag:", flag_encrypted)

  while True:
    choice = menu()

    # Exit
    if choice == '0':

    # Encrypt
    elif choice == '1':
      m = int(input('\nPlaintext > ').strip())
      print('\nEncrypted: ' + str(encrypt(m)))

    # Decrypt
    elif choice == '2':
      c = int(input('\nCiphertext > ').strip())

      if c == flag_encrypted:
        print("Wait. That's illegal.")
        m = decrypt(c)

        for no in used:
          if m % no == 0:
            print("Wait. That's illegal.")
          print('\nDecrypted: ' + str(m))

    # Invalid

if __name__ == "__main__":

My current script:

from pwn import * 
from Crypto.Util.number import *
from math import gcd
import gmpy2
import sys
r = remote('oracle.challs.cyberchallenge.it', 9042)
r.recvuntil(b'Encrypted flag: ')
encrypted_flag = int(r.recvline().strip().decode())
e = 65537

# Let's first gather the ciphertext of the new num
public_exponent = 65537
numbers = [2,3,4,5,6]
numbers_bytes = [b'\x02',b'\x03',b'\x04',b'\x05',b'\x06']
ciphers = []
diffs = []
for i in range(4):
    r.recvuntil(b'Plaintext > ')
    r.recvuntil(b'Encrypted: ')
    cipher = int(r.recvline().strip().decode())
    diffs.append(gmpy2.sub(pow(numbers[i], public_exponent),cipher))

common_factor = None
for diff in diffs:
    if common_factor is None:
        common_factor = diff
        common_factor = gmpy2.gcd(common_factor, diff)
print("N: ")
encrypted_flag += int(common_factor)
r.recvuntil(b'Ciphertext > ')
r.recvuntil('Decrypted: ')
flag = int(r.recvline().decode())

1 Answer 1


Let $n$ be the derived modulus, $ct$ be the encrypted flag and $m$ the plaintext flag. You can derive $n$ using the method described here: Determine RSA modulus from encryption oracle

You might want to use some large primes, otherwise you might run into issues. If you use $3$ to derive the modulus, then there's a chance $-ct \cong 0\mod 3$. I used $97$ and $113$ to test and that worked.

The reason the second check doesn't pass is because $(m+n)^k \cong m \mod n$ for all $k\neq 0 $ and $m$ is already in the list used from the start.

Send $-ct$ to the oracle, this will be decrypted to $-m \mod n$. Since if c == flag_encrypted: are done over the integers and therefore the check will pass.

Now all you have to do is calculate $-1\cdot (-m) \cong m \mod n$.

  • 1
    $\begingroup$ normally we give hints for CTF as a comment, since it's considered off-topic $\endgroup$
    – kodlu
    Commented Aug 25, 2023 at 15:17
  • 1
    $\begingroup$ I'm sorry, I wasn't aware of this. I did however check that cyberchallenge.it was not hosting any competitions before answering. $\endgroup$ Commented Aug 25, 2023 at 16:34
  • $\begingroup$ No, cyberchallenge.it is just a portal provided to students to practice. In any case I haven't quite understood the last part. I send the negative version of the ciphertext to the oracle, and then the flag will be the number I get multiplied by -1? I got a negative number doing so. $\endgroup$
    – Shark44
    Commented Aug 25, 2023 at 16:38
  • $\begingroup$ My bad, I interpreted your answer the wrong way. Thank you, now it works! $\endgroup$
    – Shark44
    Commented Aug 25, 2023 at 16:53

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