I've set up an open cryptography challenge for everyone. The goal is to recover the original plaintext from this ciphertext using only the provided public key. Here's what you need:
- **Ciphertext:** `44d31351849553eda9fde3261ae7a22cbe837fc5b5f4d5d9a7bc4813631eec5d194977bb372ba1555c4221f92cca45aa108123190de4c7025248136f323bc6bc60854e5cf8e0d5c959824f2dcac288e6`
- **Public Key:** {"pubkey": "4733008843232521885267611839866408727362590782235583946042001273391542507556911039580558489377236615745119239138992070467620135888095023838956048317962928", "p": "6819873056954028096669600527154144091583685931523071039913161651935675643166368485469593800563313709741457680625762752271150575880140441725750462091516463", "hashN": "b6fea046ce49db4258a7f72fe6b228c60df2541c667e9768e3a8447bb469af2c"}
You can use any programming language or approach you like, including brute-force. If you're successful, please share your code and the plaintext below!
2 Answers
It sounds like you're asking people to break AES encryption. Just so you know, AES is known for being extremely secure, and decrypting it without the right keys and parameters is nearly impossible.
I don't think anyone can actually decrypt this, including your intended recipient. Your paper mentions needing the "IV" and "nonce" for decryption, but those aren't provided. Without them, the decryption seems impossible. If you share the IV and nonce, it would make this whole thing trivial; they shouldn't be accessible to an attacker!
Actually, in DIAC ∞, the IV and nonce are not random; they represent a secret window that is never shared. Only the recipient has access to this info, which is what keeps it secure. If someone can decrypt it with just the public key and ciphertext, it would point out a serious vulnerability.
True, but it's not necessarily about breaking AES itself. The challenge is to derive the symmetric key while missing some hidden parameters. That's what makes it interesting!