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What if I will use 96-bit integer counter starting from 0 as IV for AES256-GCM cipher. Is it safe? Is there any other ways to ensure uniqueness of IV for GCM cipher?

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  • $\begingroup$ Yes GCM was designed to work with simple counters. $\endgroup$ – SEJPM Aug 6 '18 at 14:50
  • $\begingroup$ Just be careful to NEVER, EVER reuse the counter - the IV is a nonce. Safely implementing a counter isn't as easy as it might seem at first. $\endgroup$ – Swashbuckler Aug 20 '18 at 15:04
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There is no difference in terms of security as long as no AES input block (the counter/IV not the plaintext) is repeated. (And you're not doing something extremely silly like making the input $AES_k^{-1}(ctr)$.)

You probably don't want/need to use a 96 bit IV if you can use a counter. Using a larger IV is good when you need to use random IVs to prevent repeats. But you're not using random IVs and it only leaves 32 bits for the other counter. You probably won't encrypt $2^{96}$ messages but you could want to encrypt something larger than 64 GB. (Or $2^{32} * (128 / 8)$ bytes.)

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  • $\begingroup$ Thank you for the answer. Unfortunately I don't understand what "making the input $AES_k^{-1}(ctr)$" means. $\endgroup$ – igor.sol Aug 6 '18 at 16:28
  • $\begingroup$ No I don't think I will encrypt something bigger than 64GB $\endgroup$ – igor.sol Aug 6 '18 at 16:29
  • $\begingroup$ @igor.sol The inverse of AES block encryption (decryption) using key $k$ and some value $ctr$ short for counter. But $ctr$ could be from any non-repeating pattern. $AES_k(AES_k^{-1}(x)) = x$ just as $AES_k^{-1}(AES_k(y)) = y$. It's unrealistically silly, but it was a necessary warning if we want to be pedantic. $\endgroup$ – Future Security Aug 6 '18 at 16:36
  • $\begingroup$ Well, I probably won't encrypt $2^{96}$ messages. On the other side there are some difficulties with maintaining last used counter value between sessions. I need very fast generation of IV values so I cannot store counter value into durable memory after each increment. So I will reserve counter range blocks of some size (maybe something like $2^{15}$). This is why I will need at least 64 bits for counter. Anyway I will need to estimate these numbers $\endgroup$ – igor.sol Aug 6 '18 at 16:39
  • $\begingroup$ Ah, ok now I see what $AES_k^{-1}(ctr)$ means. This can happen randomly in a very rare case. $\endgroup$ – igor.sol Aug 6 '18 at 16:43
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See page 20 of the NIST recommendations (28 in the pdf, 20 on the page) - basically this is the deterministic construction (assuming you're doing less than $2^{64}$ messages), so yes this should be fine - assuming (!) that you don't reuse IV (even across restarts / crashes).

Alternatively, you can use the RBG construction, and use a secure random source to generate the IV.

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