Awesome Mix1

Challenge

• The challenege was to break the Crypto Algorithm being used to get access.
• Source code service.py is:

from Crypto.PublicKey import RSA
from Crypto.Signature import PKCS1_v1_5 as RSAsign
from Crypto.Hash import SHA
from pad import PKCS1_pad as pad
from SECRET import flag
import sys

def verify(s,m,n,e):
    if pow(s,e,n) == pad(m):
        return True
    else:
        return False

key = RSA.generate(1024)

message = "super important information for admin only"

h = SHA.new(message)
signer = RSAsign.new(key)
signature = signer.sign(h)
s = int(signature.encode("hex"),16)

print "Welcome to admin's music portal.\nTo verify that you are the owner of this service\nsend the public key which will verify the following signature :\n"
print "Message   ->", message
print "Signature ->", sig
print
sys.stdout.flush()
n = long(raw_input("Enter n:"))
print "n: " , n
e = long(raw_input("Enter e:"))
print "e : " , e
sys.stdout.flush()

input_key = RSA.construct((n,e))

print "input key : ", input_key
print

if verify(s,h.hexdigest(),n,e):
    print flag
else:
    print "Music is only for admin's eyes."

• source code for pad.py :

def PKCS1_pad(data):
    asn1 = "003021300906052b0e03021a05000414"
    ans = asn1+data
    n=len(ans)
    padding = '0001'+'f'*(1024/4-n-4)
    return int((padding + ans),16)
    

Analysis

• the if statement in verify checks the input against pad(m) which is always h.hexdigest() and thus constant
• so now we need the correct values of s, n, and e
• pow(s,e,n) =: s^e % n
• if e=1 , s%n = pad(m)
• thus n = pad(m) - s

Own Code:

from Crypto.PublicKey import RSA
from Crypto.Signature import PKCS1_v1_5 as RSAsign
from Crypto.Hash import SHA
from pad import PKCS1_pad as pad
import sys

message = "super important information for admin only"

h = SHA.new(message)

padding = pad(h.hexdigest())

sig = raw_input('enter signature > ').strip()

s  = int(sig,16)

print "s : " , s
print 
print "n = ", s - padding
print
print "e = 1"

The values of n and e when input to the service gives the flag!