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17 September 2018

Tokyo Westerns CTF 2018 - Revolutional Secure Angou Writeup

by mr96

We’re given a short Ruby code used to encrypt the flag, the public key and the encrypted flag.

require 'openssl'

e = 65537
while true
  p = OpenSSL::BN.generate_prime(1024, false)
  q = OpenSSL::BN.new(e).mod_inverse(p)
  next unless q.prime?
  key = OpenSSL::PKey::RSA.new
  key.set_key(p.to_i * q.to_i, e, nil)
  File.write('publickey.pem', key.to_pem)
  File.binwrite('flag.encrypted', key.public_encrypt(File.binread('flag')))
  break
end

Code Analysis

The code is a simple textbook RSA implementation, the only difference is that we have , and this is the vulnerability.

Attack

From we can write for some and then . So we have to find such ; rewriting we can see that . Because we have and running

openssl rsa -pubin -in publickey.pem -text -noout

we can see bits, so because bits also bits and . Now we only have to check the values of from to and then decrypt the flag (in less than 3 seconds with the following script).

import gmpy2
from Crypto.PublicKey import RSA

e = 65537L
n = 16809924442712290290403972268146404729136337398387543585587922385691232205208904952456166894756423463681417301476531768597525526095592145907599331332888256802856883222089636138597763209373618772218321592840374842334044137335907260797472710869521753591357268215122104298868917562185292900513866206744431640042086483729385911318269030906569639399362889194207326479627835332258695805485714124959985930862377523511276514446771151440627624648692470758438999548140726103882523526460632932758848850419784646449190855119546581907152400013892131830430363417922752725911748860326944837167427691071306540321213837143845664837111L
p = 0

for k in range(int(e/2),e):
    delta = 1+4*k*n*e
    if gmpy2.is_square(delta):
        y = gmpy2.isqrt(delta)
        if (y-1)%(2*k) == 0:
            p1 = (y-1)/(2*k)
            if n%p1 == 0:
                print("Found!")
                p = p1
                break

q = n/p
phi = (p-1)*(q-1)
d = gmpy2.invert(e,phi)
key = RSA.construct((n,e,long(d)))
data = open('revolutional-secure-angou/flag.encrypted', 'r').read()
print(key.decrypt(data))

and we can find the flag TWCTF{9c10a83c122a9adfe6586f498655016d3267f195}

tags: crypto  rsa