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.

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.