HTB Business CTF - Vitrium Stash
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# HTB Business CTF - Vitrium Stash

We have to forge a DSA signature for the admin being able to ask the server a signature for an arbitrary username. We exploit the fact that the message is not hashed in the signature, and hence find two messages that are equal mod q giving us a valid signature.

I solved this challenge together with @robin.

The full solution scripts is composed by:

## Challenge Description

The challenge setting is quite simple. It consists in a server implementing DSA signature, with fixed (secure) p, q and g. We are given two options. First of all, we can create a new user, providing a username. The server returns us the message json.dumps({"username": username, "admin": False}), together with the signature (r,s) of the message. We can also send a message and a signature to the server. If the signature is valid and the message contains "admin":true, the server gives us the flag.

## Solution

The vulnerable part of the challenge is the sign function:

1 2 3 4 5 6 def sign(message): m = bytes_to_long(message) k = randbelow(p) r = pow(g, k, p) % q s = (inverse(k, q) * (m + x*r)) % q return r, s 

The difference between this and the actual DSA protocol is that s here is computed using the message m, and not a hash of it as suggested. The consequence is that given a message m, its signature (r, s) will be valid for each message m' = m mod q, since

$((k^{-1}\bmod q)(m + nq + rx)) \equiv ((k^{-1}\bmod q)(m + rx)) \bmod q$

We can ask the server to sign messages of the form {"username": username, "admin": False}, where username is an arbitrary string we can provide. We win if we manage to find a value for username such that the resulting string is congruent mod q to another string containing "admin":true.
Recall that a string here is just a large number: each letter is in fact a 8-bit integer, so we can see ab = 97*2^8 + 98 (a is 97 in ASCII and b is 98). We can easily generalize this idea including variables instead of characters: the string $a_2a_1a_0$ correspond with the number $a_0 + 2^8 a_1 + 2^{16}a_2$, and so on. In our problem, some of this variables will be fixed by the structure of the message. The holes instead will have a strictly bounded value: we want them to be printable, so they can be integers between 32 and 126 (theoretically we should be more careful here, since not every printable character can go inside a json string, but fine). We can also impose more strict requirements, for example if we want only printable digits we may ask numbers between 48 and 57. To sum up, we have a bunch of integer inequalities, which can be solved quite effectively as a CVP instance using Babai’s algorithm (see here for the implementation).
First of all, we need a template for our strings. The target string is fixed, we can put holes only on the username. For the forged string we decided to keep it as simple as possible, setting only the admin parameter.

1 2 s1 = b'{"admin": 1################################}' s2 = b'{"username": "????????????????????????????????", "admin": false}' 

Here ? represents a printable ASCII, and # a printable digit.
The implementation is conveniently based on versesrev’s wrapper: we only have to set up the variables and the constraints, and it does the rest of the job with the lattice. The constraints are quite simple: we want left_string - right_string + zq = 0, where z is a variable and left_string and right_string are computed as above. Specifically, every time we find a regular character we add its value (mutiplied by the right power of $2$) to the sum, while every time we encounter ? or # we add a new variable to the sum and then add a constraint to ensure that the variable is a printable ASCII or digit respectively. The function match_strings(s1, s2, p) accepts two such strings and return two valid strings satisfying the requirements (if possible). In this case, we got very quickly

1 2 s1 = b'{"admin": 144454544455544565622743625683666}' s2 = b'{"username": "UWdaNYLKAR/F[>1C_VQETCQSX@HBYu;", "admin": false}' 

Sending this username to the server, and using the signature for s1` finally gave us the flag.