Timeline for How can you change the plaintext of the given secret message which is in AES-256 CBC
Current License: CC BY-SA 4.0
11 events
| when toggle format | what | by | license | comment | |
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| Jun 1, 2018 at 22:09 | comment | added | Stephen Touset | You don't need the key to do this at all. The decrypted plaintext only depends upon the IV and the decrypted block. You know all three of those components and can manipulate one of them. | |
| Jun 1, 2018 at 14:47 | vote | accept | eyesima | ||
| Jun 1, 2018 at 14:10 | comment | added | forest | It's probably looking for a key in hexadecimal. The key would have to be exactly the cipher's key size (so for AES128, 128 bits. For AES256, 256 bits. In hex this is 32 and 64 characters, respectively). | |
| Jun 1, 2018 at 14:08 | comment | added | eyesima | I entered any key (only used numbers though) and it didn't work. The key needs a specific length maybe that's why it didn't work for me? | |
| Jun 1, 2018 at 13:52 | comment | added | forest | There may be some online AES encryption and decryption tools out there where you can input things like the plaintext, key, and IV. And it doesn't matter that they require a key. Just enter a random one and pretend you don't know it. After all, the point is that you can change the plaintext without using the key to do so. You get to play Alice, Bob, and Mallory! No need to fret because the role of Mallory involves not having access to some information Alice and Bob have! | |
| Jun 1, 2018 at 13:50 | comment | added | eyesima | Problem is I don't have a key. All I have is initial vector, ciphertext, and plaintext. I have also checked for sites and they all require to enter key, here is an example: proappsoft.com/spocot/index.php#encipher You know how I can generate key? Maybe on some site because I don't have linux :( | |
| Jun 1, 2018 at 13:19 | comment | added | forest |
Do the encryption yourself! The openssl command line utility allows you to specify the key and IV directly as hex values. Any hex editor will allow you to view the ciphertext before and after. As far as I can tell, your method is correct, but it's always more enlightening to see the results with your own eyes.
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| Jun 1, 2018 at 13:16 | comment | added | eyesima | I would continue this till I worked through each character of the plaintext "Meeting tonight!". In the end I will have an initial vector which is different from the current one. But the ciphertext will be the same, right? How can I find out if I did it correctly? :) | |
| Jun 1, 2018 at 13:11 | comment | added | forest | I didn't check the math but that should be right. Try it and find out! | |
| Jun 1, 2018 at 10:02 | comment | added | eyesima | Thank you very much for answer and hint! The first character of the plaintext is M, in ASCII this is 0x4d. The first byte of the initial vector is 0x5a. So we have 0x4d XOR 0x5a which equals 0x17. Thus the first byte of our manipulated initial vector (which will change the plaintext to Meeting tonight!) is 0x17 ? | |
| May 31, 2018 at 23:53 | history | answered | forest | CC BY-SA 4.0 |