AngstromCTF2016 - Smartest Encryption (re, 70)

We have  a PNG encrypted image :  https://angstromctf.com/static/problems/re/image_encryptor/flag.encrypted .
Softawre used to crypt the image : https://angstromctf.com/static/problems/re/image_encryptor/encryptor.apk

Hint : PNG image

first step decompile encryptor.apk :

The application ask for a password , create the hash of this password and than XOR the PNG image using key=md5(password) .
Routine to encrypt is  :
private byte[] encryptData(byte[] paramArrayOfByte1, byte[] paramArrayOfByte2)
  {
    int i = (byte)paramArrayOfByte2.length;
    byte[] arrayOfByte = new byte[paramArrayOfByte1.length];
    for (int j = 0; j < paramArrayOfByte1.length; j++) {
      arrayOfByte[j] = ((byte)(paramArrayOfByte1[j] ^ paramArrayOfByte2[(j % i)]));
    }
    return arrayOfByte;
  }









 Extract 32 Byte from flag.ecrypted :




 crypted=553430f1536d0beb5537521568637d36


png_header= 89504e470d0a1a0a0000000d49484452


key = crypted XOR png_header

def xor_strings(xs, ys):
    return "".join(chr(ord(x) ^ ord(y)) for x, y in zip(xs, ys))
crypted='553430f1536d0beb5537521568637d36'
png_header='89504e470d0a1a0a0000000d49484452'
binary_a = crypted.decode("hex")
binary_b = png_header.decode("hex")
key= xor_strings(binary_a, binary_b).encode("hex")
print key

dc647eb65e6711e155375218212b3964

Online md5 :















Now xor the   flag.encrypted using key = dc647eb65e6711e155375218212b3964  using xortool https://github.com/hellman/xortool




The flag is flag{all_encryption_is_equal_but_some_are_More_equal_than_others}