EVEN RSA CAN BE BROKEN???

---

picoCTF 2025 – Cryptography • Easy

The challenge provides us with the source code used to encrypt the flag.

We can observe that the code follows a standard textbook-RSA approach, but the function that is used for generating primes.

The 'get_primes' function is a custom function which codes is not provided. We assume the weakness of the system is in this step.

Connecting to the provided webshell we can observe that the number N is an even number. Our guessing was correct, the custum function is not secure.

Step 1 - Factor N

Given 'p = 2' we can easly obtain 'q' by dividing 'N' by 2 (we should also check that 'q' is a prime number).

Step 2 - Recover the flag

Now that we have the secret key, it is just a matter of standard operation to recover the plaintext.

• 1. Compute 'd'

  d = inverse(e, (p-1)*(q-1))

• 2. Find the flag

  flag = long_to_bytes(pow(cyphertext, d, N)).decode()

Lessons Learned

• When writing code, use tested libraries.