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Oct 11, 2023 at 7:04 comment added Galois group Thank you for you answer! However you do not really convinced me. Well, what is the problem to use random $e$ and 512 bit long keys? Then it seems everything would be secure. At least I don't see any practical attack.
Jun 27, 2017 at 12:52 vote accept wlad
Jul 8, 2015 at 12:09 comment added Yehuda Lindell What do you mean by a random key? I assume that you mean that you take a random string of length say 256 and then apply the RSA function to it. In this case, the first attack works with probability 1 (up to 682 bits long).
Jul 8, 2015 at 9:24 comment added wlad If you're encrypting a random key then the probability of the first attack working is $2^{682 - 2048}$ which is extremely small. Same for the second attack. The problem I see now is when the random key is at most $682$ bits long, because $682$ bits is quite a lot for a symmetric key, in which case some padding is necessary to prevent the attack. It's important to refer back to the question I asked.
Jul 7, 2015 at 21:05 history edited Yehuda Lindell CC BY-SA 3.0
added 88 characters in body
Jul 7, 2015 at 21:01 comment added Maarten Bodewes Just to be sure; isn't the first to last paragraph describing KEM with the hash replacing a KDF? Maybe you could somehow link KEM with this paragraph in the answer?
Jul 7, 2015 at 20:59 history edited Maarten Bodewes CC BY-SA 3.0
added 6 characters in body
Jul 7, 2015 at 20:49 history edited Yehuda Lindell CC BY-SA 3.0
Removed space and added references
Jul 7, 2015 at 20:44 comment added Yehuda Lindell Thanks, I actually don't think I knew about this, so this is really important!
Jul 7, 2015 at 20:37 comment added user991 By the link I gave, both instances of "and space" should be removed. $\;$
Jul 7, 2015 at 20:25 history edited Yehuda Lindell CC BY-SA 3.0
Added attack when e=3
Jul 7, 2015 at 20:19 comment added Yehuda Lindell @RickyDemer Thanks; forgot to include that. Will edit now.
Jul 7, 2015 at 17:56 comment added user991 For small $e$, there's an even faster attack. $\:$ Even for large $e$, the amount of space used can be made trivial. $\:\:$
Jul 7, 2015 at 17:33 history answered Yehuda Lindell CC BY-SA 3.0