# WRITE UP - CSTV CTF 2023

## Crypto:

### RSA:
> Solve by i4mh3nry (aka tuankiet.trinh0r)

Here, the exam show you **ciphertext** và **mykey.pem** and after read python file you'll know that the datas were stored in **ciphertext**. Now, opening the **mykey.pem** i saw the `PUBLIC KEY`


Researched for few minutes, i found this [WEB](https://travistidwell.com/blog/2013/09/06/an-online-rsa-public-and-private-key-generator/)


And i know that `PUBLIC KEY` has been made from `SECRET KEY` now you just find the `SECRET KEY`

The program create `SECRET KEY` from `PUBLIC KEY`

from Crypto.PublicKey import RSA
import sympy

with open('mykey.pem', 'rb') as f:
    key = RSA.import_key(f.read())

n = key.n

p, q = sympy.factorint(n)

phi = (p - 1) * (q - 1)
d = pow(key.e, -1, phi)


private_key = RSA.construct((n, key.e, d, p, q))

# Export
with open('private_key.pem', 'wb') as f:
    f.write(private_key.export_key('PEM'))

Done ! Now, just decrypt RSA

from Crypto.Cipher import PKCS1_v1_5
from Crypto.PublicKey import RSA

with open('private_key.pem', 'rb') as key_file:
    private_key = RSA.import_key(key_file.read())

with open('ciphertext', 'rb') as cipher_file:
    ciphertext = cipher_file.read()

cipher = PKCS1_v1_5.new(private_key)
decrypted_message = cipher.decrypt(ciphertext, None)

print(decrypted_message.decode('utf-8'))

You had the **FLAG** lol~


> Flag: CSTV_2023_{408f8e7210d60d78caac8ff6b204459276411da1}