blog/content/posts/2022-02-07-cybergrabs-ctf-unbreakable/index.md

+++ title = "The Cyber Grabs CTF: Unbr34k4bl3 (942)" draft = true date = 2022-02-02 tags = ["ctf", "crypto"] languages = ["python"] layout = "single" math = true +++

Crypto challenge Unbr34k4bl3 from the Cyber Grabs CTF.

No one can break my rsa encryption, prove me wrong !!

Flag Format: cybergrabs{}

Author: Mritunjya

output.txt source.py

Looking at the source code, this challenge looks like a typical RSA challenge at first, but there are some important differences to note:

• N = pqr (line 34). This is a twist but RSA strategies can easily be extended to 3 prime components.
• p, q \equiv 3 \mod 4 (line 19). This suggests that the cryptosystem is actually a Rabin cryptosystem.
• We're not given the public keys e_1 and e_2, but they are related through x.

Finding e_1 and e_2

We know that e_1 and e_2 are related through x, which is some even number greater than 2, but we're not given any of their real values. We're also given through an oddly-named functor function that:

 1 + e_1 + e_1^2 + \cdots + e_1^x = 1 + e_2 + e_2^2 

Taking the entire equation \mod e_1 gives us:

$$\begin{aligned} 1 &\equiv 1 + e_2 + e_2^2 \mod e_1 \\ 0 &\equiv e_2 + e_2^2 \\ 0 &\equiv e_2(1 + e_2) \end{aligned}

This means there are two possibilities: either e_1 = e_2 or e_1 is even (since we know e_2 is a prime). The first case isn't possible, because with $x > 2$, the geometric series equation would not be satisfied. So it must be true that \boxed{e_1 = 2}, the only even prime.

Applying geometric series expansion, 1 + e_2 + e_2^2 = 2^{x + 1} - 1. We can rearrange this via the quadratic equation to $e_2 = \frac{-1 \pm \sqrt{1 - 4 (2 - 2^{x + 1})}}{2}$. Trying out a few values we see that only $\boxed{x = 4}$ and \boxed{e_2 = 5} gives us a value that make e_2 prime.

Finding p and q

We're not actually given p or q, but we are given ip = p^{-1} \mod q and iq = q^{-1} \mod p. In order words:

$$\begin{aligned} p \times ip &\equiv 1 \mod q \\ q \times iq &\equiv 1 \mod p \end{aligned}

We can rewrite these equations without the mod by introducing variables $k_1$ and k_2 to be arbitrary constants that we solve for later:

$$\begin{aligned} p \times ip &= 1 + k_1q \\ q \times iq &= 1 + k_2p \end{aligned}

We'll be trying to use these formulas to create a quadratic that we can use to eliminate k_1 and k_2. Multiplying these together gives:

$$\begin{aligned} (p \times ip)(q \times iq) &= (1 + k_1q)(1 + k_2p) \\ pq \times ip \times iq &= 1 + k_1q + k_2p + k_1k_2pq \end{aligned}

I grouped p and q together here because it's important to note that since we have x, we know r and thus pq = \frac{N}{r}. This means that for purposes of solving the equation, pq is a constant to us. This actually introduces an interesting structure on the right hand side, we can create 2 new variables:

$$\begin{aligned} \alpha &= k_1q \\ \beta &= k_2p \end{aligned}

Substituting this into our equation above we get:

$$\begin{aligned} pq \times ip \times iq &= 1 + \alpha + \beta + \alpha\beta \end{aligned}

Recall from whatever algebra class you last took that $(x - x_0)(x - x_1) = x^2 - (x_0 + x_1)x + x_0x_1$. Since we have both \alpha\beta and $(\alpha + \beta)$ in our equation, we can try to look for a way to isolate them in order to create our goal.

$$\begin{aligned} pq \times ip \times iq &= 1 + k_1q + k_2p + k_1k_2pq \\ k_1k_2pq &= pq \times ip \times iq - 1 - k_1q - k_2p \\ k_1k_2 &= ip \times iq - \frac{1}{pq} - \frac{k_1}{p} - \frac{k_2}{q} \end{aligned}

\frac{1}{pq} is basically 0, and since k_1 and k_2 are both smaller than p or q, then we'll approximate this using k_1k_2 = ip \times iq - 1. Now that k_1k_2 has become a constant, we can create the coefficients we need:

$$\begin{aligned} \alpha + \beta &= pq \times ip \times iq - 1 - k_1k_2pq \\ \alpha\beta &= k_1k_2pq \end{aligned}

$$\begin{aligned} (x - \alpha)(x - \beta) &= 0 \\ x^2 - (\alpha + \beta)x + \alpha\beta &= 0 \\ x &= \frac{(\alpha+\beta) \pm \sqrt{(\alpha+\beta)^2 - 4\alpha\beta}}{2} \end{aligned}

Putting this into Python, looks like:

>>> k1k2 = ip * iq - 1
>>> alpha_times_beta = k1k2 * pq
>>> alpha_plus_beta = pq * ip * iq - 1 - k1k2 * pq

>>> def quadratic(b, c):
>>>   disc = b ** 2 - 4 * c
>>>   return (-b + sqrt(disc)) / 2, (-b - sqrt(disc)) / 2

I'd like to thank @10, @sahuang, and @thebishop in the Project Sekai discord for doing a lot of the heavy-lifting to solve this challenge.