Timeline for Is there a maximum number of encryptions I should make with the same RSA key?
Current License: CC BY-SA 4.0
7 events
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| Apr 30, 2021 at 0:20 | comment | added | LWChris | Yes, that's what I meant with "running out of options". It's the pigeon hole problem. There is only a finite number of different possible padding sequences, therefore there's only a finite number of different ciphertexts that encrypting the same plaintext over and over again could yield. Eventually, you'll by chance hit the same padding value, hence the same ciphertext, and the adversary will know it must have been the same plaintext. | |
| Apr 14, 2020 at 16:00 | comment | added | fgrieu♦ | Yes. In "running out of options" it is not kept track of the options previously used; it is relied on chance not to use twice the same option. Hence collision occurs per the birthday bound. As noted in the answer, this happens mostly when close to the maximum capacity of RSAES-PKCS1-v1_5, but this is precisely what happens when one enciphers sizable plaintext in some APIs that attempt to give the same interface for symmetric and asymmetric crypto. Java/BouncyCastle is an example of that dubious design philosophy, which is a common cause of security and reliability issues. | |
| Apr 14, 2020 at 14:52 | comment | added | LWChris | So if I send the same plaintext over and over, usually the padding algorithm would ensure that each ciphertext looks totally different, but with the old algorithm and a short padding it's literally "running out of options" and since a bitwise identical ciphertext can only be decrypted to the bitwise identical plaintext, this is some information an adversary would gain (not what it is, but that it was sent at least twice). | |
| Apr 6, 2020 at 16:57 | history | edited | fgrieu♦ | CC BY-SA 4.0 |
Oups
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| Apr 6, 2020 at 16:51 | history | edited | fgrieu♦ | CC BY-SA 4.0 |
Polish
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| Apr 6, 2020 at 16:45 | history | edited | fgrieu♦ | CC BY-SA 4.0 |
Polish
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| Apr 6, 2020 at 16:39 | history | answered | fgrieu♦ | CC BY-SA 4.0 |