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

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

from decimal import Decimal
getcontext().prec = 3000 # To get all digits

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

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

alpha, beta = quadratic(-alpha_plus_beta, alpha_times_beta)

Now that we have \alpha and \beta, we can try GCD'ing them against pq to get p and q:

from math import gcd

p = gcd(pq, int(alpha))
q = gcd(pq, int(beta))
assert p * q == pq # Success!

Alternative method

@sahuang used the sympy library to do this part instead, resulting in much less manual math. It's based on this proof from Math StackExchange that $p \cdot (p^{-1} \mod q) + q \cdot (q^{-1} \mod p) = pq + 1$.

from sympy import *
p,q = symbols("p q")
eq1 = Eq(ip * p + iq * q - pq - 1, 0)
eq2 = Eq(p * q, pq)
sol = solve((eq1, eq2), (p, q))

Decrypting the ciphertexts

Now that we know p and q, it's time to plug them back into the cryptosystem and get our plaintexts. c_2 is actually easier than c_1, because with $e_2 = 5$ we can just find the modular inverse:

phi = (p - 1) * (q - 1) * (r - 1)
d2 = pow(e2, -1, phi)
m2 = pow(c2, d2, n)
print(long_to_bytes(m2))
# ... The last part of the flag is: 8ut_num83r_sy5t3m_15_3v3n_m0r3_1nt3r35t1n6} ...

This trick won't work with c_1 however:

d1 = pow(e1, -1, phi)
# ValueError: base is not invertible for the given modulus

Because \phi is even (it's the product of one less than 3 primes), there can't possibly be a d_1 such that 2 \cdot d_1 \equiv 1 \mod \phi. According to Wikipedia, the decryption for a standard two-prime n takes 3 steps:

1. Compute the square root of c \mod p and c \mod q:
   • m_p = c^{\frac{1}{4}(p + 1)} \mod p
   • m_q = c^{\frac{1}{4}(q + 1)} \mod q
2. Use the extended Euclidean algorithm to find y_p and y_q such that $y_p \cdot p + y_q \cdot q = 1$.
3. Use the Chinese remainder theorem to find the roots of c modulo n:
   • r_1 = (y_p \cdot p \cdot m_q + y_q \cdot q \cdot m_p) \mod n
   • r_2 = n - r_1
   • r_3 = (y_p \cdot p \cdot m_q - y_q \cdot q \cdot m_p) \mod n
   • r_4 = n - r_3
4. The real message could be any r_i, but we don't know which.

Converting this to work with n = pqr, it looks like:

1. Compute the square root of c \mod p, c \mod q, and c \mod r:
   • m_p = c^{\frac{1}{4}(p + 1)} \mod p
   • m_q = c^{\frac{1}{4}(q + 1)} \mod q
   • m_r = c^{\frac{1}{4}(r + 1)} \mod r
2. Using the variable names from AoPS's definition of CRT:
   • For k \in \\{ p, q, r \\}, b_k = \frac{n}{k}.
   • For k \in \\{ p, q, r \\}, a_k \cdot b_k \equiv 1 \mod k.
3. Let r = \displaystyle\sum_k^{\\{ p, q, r \\}} \pm (a_k \cdot b_k \cdot m_k) \mod n.
4. The real message could be any r, but we don't know which.

In code this looks like:

# Step 1
mp = pow(c1, (p + 1) // 4, p)
mq = pow(c1, (q + 1) // 4, q)
mr = pow(c1, (r + 1) // 4, r)

# Step 2
bp = n // p
bq = n // q
br = n // r
ap = pow(bp, -1, p)
aq = pow(bq, -1, q)
ar = pow(br, -1, r)

# Step 3
from itertools import product
for sp, sq, sr in product((-1, 1), repeat=3):
  m = (sp * ap * bp * mp + sq * aq * bq * mq + sr * ar * br * mr) % n
  m = long_to_bytes(m)

  # Step 4
  # We know that the real flag starts with `cybergrabs{`...
  if b"cybergrabs" in m: print(m)

# Congratulations, You found the first part of flag cybergrabs{r481n_cryp70sy5t3m_15_1nt3r35t1n6_ ...

The final flag, then, is:

cybergrabs{r481n_cryp70sy5t3m_15_1nt3r35t1n6_8ut_num83r_sy5t3m_15_3v3n_m0r3_1nt3r35t1n6}

🎉

Big thanks to @10, @sahuang, and @thebishop in the Project Sekai discord for doing a lot of the heavy-lifting to solve this challenge.