Copperbox

We are given the following Python source code that encrypts the flag:

import secrets

p = 0x31337313373133731337313373133731337313373133731337313373133732ad
a = 0xdeadbeefdeadbeefdeadbeefdeadbeefdeadbeefdeadbeefdeadbeefdeadbeef
b = 0xdeadc0dedeadc0dedeadc0dedeadc0dedeadc0dedeadc0dedeadc0dedeadc0de

def lcg(x, a, b):
    while True:
        yield (x := a*x + b)

flag = open('flag.txt', 'rb').read()
x = int.from_bytes(flag + secrets.token_bytes(30-len(flag)), 'big')
gen = lcg(x, a, b)

h1 = next(gen) * pow(next(gen), -1, p) % p
h2 = next(gen) * pow(next(gen), -1, p) % p

with open('output.txt', 'w') as o:
    trunc = 48
    # oops, i forgot the last part
    o.write(f'hint1 = {h1 >> trunc}\n')
    o.write(f'hint2 = {h2 >> trunc}\n')

And we also have the output.txt file:

hint1 = 77759147870011250959067600299812670660963056658309113392093130
hint2 = 50608194198883881938583003429122755064581079722494357415324546

Source code analysis

Luckily for us, the code is short. It starts by defining three integer numbers , and , which are used to define a generator lcg:

p = 0x31337313373133731337313373133731337313373133731337313373133732ad
a = 0xdeadbeefdeadbeefdeadbeefdeadbeefdeadbeefdeadbeefdeadbeefdeadbeef
b = 0xdeadc0dedeadc0dedeadc0dedeadc0dedeadc0dedeadc0dedeadc0dedeadc0de

def lcg(x, a, b):
    while True:
        yield (x := a*x + b)

The above generator implemente a Linear Congruential Generator (LCG), which is normaly used as a pseudo-random number generator (PRNG). For instance, Glibc and Java use LCG-based PRNGs.

The seed of the LCG is the flag as an integer value with some random bytes at the end:

flag = open('flag.txt', 'rb').read()
x = int.from_bytes(flag + secrets.token_bytes(30-len(flag)), 'big')
gen = lcg(x, a, b)

Then, the program computes two values h1 and h2 that depend on the LCG steps (4 steps in total):

h1 = next(gen) * pow(next(gen), -1, p) % p
h2 = next(gen) * pow(next(gen), -1, p) % p

Finally, we are given truncated values of these h1 and h2. Particularly, the program removes the 48 least significant bits:

with open('output.txt', 'w') as o:
    trunc = 48
    # oops, i forgot the last part
    o.write(f'hint1 = {h1 >> trunc}\n')
    o.write(f'hint2 = {h2 >> trunc}\n')

Linear Congruential Generator

Let’s define the LCG in math terms as follows:

Notice that we are missing the “Congruential” part of the LCG because there is no modulus.

The LCG steps are usually defined in a recursive way, but we can transform the expression:

With this, we have expressed the LCG steps in terms of the initial seed :

Notice that LCGs are not secure, because leaking a single will allow to recover previous states and to predict future steps. For this reason, the steps of the LCGs are usually truncated, in order to prevent adversaries from cracking the PRNG (or at least, making it more difficult to crack).

Analyzing the output

Let’s express in math terms:

If we expand the values and , we get

We can do the same with , which depends on and :

But we are not given nor … That would be too easy. Instead, we have:

Solution

Whenever you see truncated outputs in a cryptography CTF challenge, it is very likely that the solution is related to Coppersmith method or some lattice-reduction technique. Indeed, the challenge name is Copperbox, which suggests the use of Coppersmith method.

For this, we need to define one or more polynomials modulo an integer such that there is a small root modulo a divisor of the modulo. This time, the modulo is a prime number, so we don’t have to worry about it.

This small root will be found by Coppersmith method. Technique employs lattice reduction to find another polynomial that shares the same root with the provided modular polynomial. However, the new polynomial is defined in the integers, so computing solutions is efficient.

So, let’s see what values we are missing. Basically,

• Initial seed (the flag)
• Truncated value from (48 bits). Let it be
• Truncated value from (48 bits). Let it be

Hence, we can express:

At this point, we can substitute the above values in the previous expressions. To improve readability, let , and also , , so

So far, so good. But let’s get rid of the modular inverse by multyplying in both sides:

Now, we can group variables:

We are left with three variables and two equations. If we didn’t have more information about this system, it would not be solvable. However, this time we know that , so they are short values compared to , or . These are good candidates to look for small roots. Notice that can be large, since it is a flag (usually around 30 characters, so around ).

One trick is to eliminate the variable in the equations by combining the two of them. Since the expressions are a bit long, let

So, we start with

Then, we put the same coefficient on variable :

Now we subtract both equations:

Notice that we can simplify a bit the expression and remove the :

So we have found a single equation on variables and . Now we can easily define the polynomial:

Notice that , so it is a root, and indeed small because they are bounded by . As a result, we can use a bivariate Coppersmith method to find both values. Once we have them, it is trivial to find :

• First we find :

• Next, we solve for :

And with this, we find the flag .

Implementation

In SageMath, we can define the variables and in , define the auxiliary values and then define the polynomial :

#!/usr/bin/env sage

p = 0x31337313373133731337313373133731337313373133731337313373133732ad
a = 0xdeadbeefdeadbeefdeadbeefdeadbeefdeadbeefdeadbeefdeadbeefdeadbeef
b = 0xdeadc0dedeadc0dedeadc0dedeadc0dedeadc0dedeadc0dedeadc0dedeadc0de

trunc = 48

hint1 = 77759147870011250959067600299812670660963056658309113392093130
hint2 = 50608194198883881938583003429122755064581079722494357415324546


Y, Z = PolynomialRing(Zmod(p), 'Y, Z').gens()

A = [sum(a ** m for m in range(i + 1)) for i in range(4)]

H1 = hint1 << trunc
H2 = hint2 << trunc

HY = a * H1 + a * Y - 1
HZ = a * H2 + a * Z - 1

P = a ** 2 * HZ * ((H1 + Y) * A[1] - A[0]) - HY * ((H2 + Z) * A[3] - A[2])

Now, we use Coppersmith method to compute the small roots and . I will be using defund’s implementation:

load('coppersmith.sage')
y, z = small_roots(P, (2 ** trunc, 2 ** trunc))[0]
print(f'{y = }; {z = }')

Finally, we compute the value to get the flag:

h1 = H1 + y
h2 = H2 + z

x = b * (A[0] - A[1] * h1) * pow(a ** 2 * h1 - a, -1, p) % p
assert x == b * (A[2] - A[3] * h2) * pow(a ** 4 * h2 - a ** 3, -1, p) % p

print(bytes.fromhex(hex(x)[2:]))

Flag

If we run the script, we will get the flag (a nice easter egg is that the random padding L\xc6 looks like LCG):

$ sage solve.sage
y = 53006259096585; z = 248699398699637
b'HTB{sm1th1ng_mY_c0pp3r_fl4G}L\xc6'

The full script can be found in here: solve.sage.