Tags: python rsa-crypto cryptography-rsa crypto rsa
# Santa's List 2.0
Reading the source provided in Santa's List 2.0, the only difference is that there is a limit of 5 requests to the server. Fortunately, our previous solution only requires 3 requests: 2 for encrypting '!' and '#', and 1 for decrypting the modified ciphertext. This is the same solution.
Logging into the server, we see the following prompt:
Ho, ho, ho and welcome back!
Your list for this year:
Sarah - Nice
Bob - Nice
Eve - Naughty
Galf - 25898466163a90f42702135d867da28e83bdfb8e65cd8e63b2a55912dcfc7cf27dc90ea3b7704e322c71a0ca94dbb082c4ed60d8f047cd97799b199371dff074f7cfb65fe720a091d89b0a517e42f2698d069a7187f5eb5b852078ca88ba1a0f1c7fa66435df4eeadce529799e95e258ad45813c778633c3d1f9f3375f8097d8
Alice - Nice
Johnny - Naughty
Exploring the options, we have the ability to:
1. Encrypt an arbitrary string. This gives us a large integer.
2. Decrypt an integer. This also gives us a large integer.
3. Exit. Not very interesting.
Clearly, "Galf" has an interesting Naughty/Nice value. It appears to be a hex encoded value. Unfortunately it does not decode to a string, and it unpacks to a very large integer. Decrypting this integer gives the result `Ho, ho ho, no...`. Note that opening the server again will give Galf a different value.
* Generates a random RSA key
* Opens the file `flag.txt`, encrypts it, converts it to hex, and displays it in the opening message
* Listens for user input
* When encrypting, it simply converts the text to hex, converts that to a long, and encrypts it. However, it also adds the numberical plaintext to the list `used`.
* When decrypting, it first checks that the value isn't the encrypted flag (which is why we were unable to decrypt it previously). After decrypting, it also checks that `message % previous != 0` for every previous in the list `used`. This will be important later.
So we have access to an encrypted flag (C = M<sup>e</sup> (mod N)), an RSA encrypt function (X<sup>e</sup> (mod N)), and an RSA decrypt function that doesn't allow us to decrypt `C` directly. There is an attack called [Blinding](https://en.wikipedia.org/wiki/Blinding_(cryptography)) iwhich allows us to modify the ciphertext, decrypt that modified value, then un-modify it. Basically, we need an arbitrary message `S` which we will encrypt (S<sup>e</sup> (mod N)), then multiply it by the ciphertext (CS<sup>e</sup> = M<sup>e</sup>S<sup>e</sup> = MS<sup>e</sup> (mod N)), decrypt that ((MS<sup>e</sup>)<sup>d</sup> ≅ MS (mod N)), then divide by `S` to get the original `M`. Note that we can't necessarily use normal integer division because of modular arithmetic. Instead, we must calculate the modular inverse of `S (mod N)` and multiply by that.
However, the modulo check in the decrypt step blocks this plan. If we encrypt `S` using the Encrypt function, it will be recorded in `used`. When we try to multiply the ciphertext by the encrypted `S` and decrypt that, it will notice that the result is divisible by `S` and block our decryption attempt. We need a way to avoid this check. If we can find a way to encrypt `S` without using the Encrypt function, we will be able to use Decrypt.
The RSA encrypt function is simply M<sup>e</sup> (mod N). By running `list.py` and adding some debug statements, we can determine that `e` is 65537, a very common value. Unfortunately, it appears that N is randomized every time `list.py` is run, so we will have to find a way to determine it using only what we can get out of the prompt. Let's do some number theory:
x<sup>e</sup> ≅ C<sub>x</sub> (mod N)
x<sup>e</sup> = C<sub>x</sub> + kN (for some value of k)
x<sup>e</sup> - C<sub>x</sub> = kN
y<sup>e</sup> - C<sub>y</sub> = jN (same equation, but different value for x)
Note that the Greatest Common Denominator (GCD) of kN and jN is N. For reasonably small values of `x` and `y`, we can use [Euclid's Algorithm](https://en.wikipedia.org/wiki/Greatest_common_divisor#Using_Euclid's_algorithm) to determine N in a reasonable amount of time.
Practically speaking, the smallest values of `x` and `y` I could easily find were 33 and 35, which correspond to characters '!' and '#'. We can Encrypt these strings using the prompt, then calculate `N` using the following code:
cx = Encrypt('!') # 33
cy = Encrypt('#') # 35
kn = (33 ** 65537) - cx
jn = (35 ** 65537) - cy
N = gcd(kn, jn)
Now that we know `N`, we can encrypt anything we want without using the server. We will now pick an arbitrary `S` (I used `N/2`), then use the above blinding attack to decrypt the encrypted ciphertext. See https://github.com/dchiquit/xmasctf-2018/blob/master/santaslist/santaslist_solve.py for my solution.