1
$\begingroup$

So I'm trying to figure out if it's possible to extract the private key if you have the ciphertext and the IV corresponding to that ciphertext. It's for some kind of a challenge, and can't really seem to figure it out.

$\endgroup$
1
  • 1
    $\begingroup$ It's usually called a secret key instead of a private key for symmetric block ciphers such as AES. If anybody takes up this challenge for a well randomized AES key then I would consider them challenged indeed. $\endgroup$ – Maarten Bodewes Mar 4 '15 at 18:26
3
$\begingroup$

No, that's not really possible without blatant flaws of the implementation. Modern modes of operation of ciphers are resistent to attacks even if you know many pairs of plaintext and ciphertext - and the IV is public knowledge. Knowing it is the normal case. You also didn't mention what operation mode was used.

Well, of course you could brute force the key, but that would take more time and computational power than you can ever muster.

Oh, and a small note: AES is a symmetric cipher, so there's just one key, which is normally not called private key. It may be true, but this term is generally reserved for asymmetric encryption.

$\endgroup$
0
$\begingroup$

We know this is theoretically possible, but your search space is 2^126 right now and no sign of this improving. Good luck building enough supercomputer (all the computers on the planet combined aren't even close right now).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.