GryphonCTF_2016: The Keymaker

Category: Crypto Points: 30 Description:

You've intercepted an encrypted flag that was being transmitted between two GCTF admins, along with a 256-bit RSA public key that was used in the encryption of the flag... Creator - Shawn Pang (@Optixal)

Write-up

DISCLAIMER: TOOLS WERE USED, CREDITS TO @Ganapati ON GITHUB

Okay, this around, we are given a ciphertext and a public key. It appears that the public key happens to be a short, 256-bit RSA key and thus we can search up ways to break RSA keys. As most people reading this should know, current standards of RSA keys are currently at 2048-4096 bits and bitsizes of below 2048 should never be used.

Since we have a 256-bit key, we can take advantage of a well-known exploit on the RSA algorithm known as the Wiener's attack. As of this writing, I'm no expert in cryptography but what I do have are skills in Google. Googling, to be exact, led me to a certain git repository. Simple enough, downloading it and running,

./RsaCtfTool.py --publickey ~/Downloads/pub_key.txt --uncipher ~/Downloads/flag_ciphertext 
Clear text : GCTF{1f_0nly_17_w45_1n_1337}

Voila!~ Therefore, the flag is GCTF{1f_0nly_17_w45_1n_1337}.

P.S On hindsight, I probably could've tried to make my own script knowing the algorithms, too lazy.

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