Timeline for Is direct RSA encryption of AES keys secure?
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
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| Jan 7, 2020 at 11:44 | comment | added | alberto123 | thank you very much for answering me @kelalaka. you've both complimented each other and i learnt a lot from both answers. i feel that the answer who gave me the most tools to understand the matter was SEJPM's one, so i marked him. however, i learnt a lot from your answer to apply the correct measures and how to fix the problem. thank you very much | |
| Jan 6, 2020 at 19:34 | history | edited | kelalaka | CC BY-SA 4.0 |
polish thanks to SEJPM
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| Jan 6, 2020 at 17:04 | history | edited | kelalaka | CC BY-SA 4.0 |
polish
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| Jan 6, 2020 at 13:46 | comment | added | kelalaka | @fgrieu thanks. Added as 2. note with some clarification. | |
| Jan 6, 2020 at 13:38 | history | edited | kelalaka | CC BY-SA 4.0 |
polish
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| Jan 6, 2020 at 12:53 | comment | added | kelalaka | @alberto123 Since the exponentiation with 3 did not pass the modulus, there is no modulus operation and the value of the ciphertext can be found by cube-root algorithms. Cube calculation is easy (cube-root attack). To mitigate this a good padding scheme is necessary. | |
| Jan 6, 2020 at 12:44 | comment | added | fgrieu♦ | Note: using RSAES-OAEP makes it safe to send the same $k$ encrypted to multiple recipients, which in turn is handy to have the bulk of the ciphertext common to all recipients (as PGP/GPG does). RSA-KEM does not, for it would fall to Håstad's broadcast attack. | |
| Jan 6, 2020 at 11:52 | history | edited | kelalaka | CC BY-SA 4.0 |
polish
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| Jan 6, 2020 at 11:37 | history | answered | kelalaka | CC BY-SA 4.0 |