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as mentioned in the title, and read some past discusstion on AES collison here.

it shows it's possible that, AES(k1, m1)=AES(k2, m1)

just wondering what condition those 2 keys need to meet. Thanks.

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    $\begingroup$ If the question's "AES(k1, m1)=AES(k2, m1)" is as intended, then the title's "at least 6 bytes" should be "16 bytes". Please fix the discrepancy one way or the other. $\endgroup$
    – fgrieu
    Jun 5 at 7:52
  • $\begingroup$ thanks Fgrieu, yes, at first i dont even know whether even 6 bytes are identical is possible or not. after did some readup, i found it was discussed before on AES(k1, m1)=AES(k2, m1). a continuous 6 bytes identical is what i want to achieve, and find out a possible set of keys? is there pattern for those keys? $\endgroup$
    – hunter lic
    Jun 5 at 16:19
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AES is (as far as is known) a very good approximation of a Pseudo-Random Permutation. There's no discernible pattern to how keys transform plaintext into ciphertext blocks. If there were, that would be usable in a Known-Plaintext attack to recover the key given some number of plaintext-ciphertext pairs. That would be a massive security break for AES.

AES used in a Mode of Operation, however, can be different. There extra inputs aside from the key and plaintext block are allowed. This could make it easier to find a collision, or it might make it harder (eg Merkle-Damgaard mode turns a block cipher into a collision-resistant hash function).

With only 6 colliding bytes needed it's likely fast enough to simply brute-force two keys.

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  • $\begingroup$ Thx SAI. so i believe there is no pattern to follow between the 2 keys which can generate 6 bytes identical ciphertext. i think i will just randam pick a 128 bit key, and use brute-force mechanism to find another key? in my applicaiton, the IV is the same, and requires the first 6 bytes of ciphertext are same. basically this application is to allow info encrypted by different parties (using different keys) recognize each other by comparing the first 6 bytes of the ciphertext. $\endgroup$
    – hunter lic
    Jun 6 at 1:10
  • $\begingroup$ just to add that my plaintext is only 14-16 bytes, i think only 1 cipher block to encrypt, was using AES-CTR operation mode. $\endgroup$
    – hunter lic
    Jun 6 at 1:29
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    $\begingroup$ You could mention that for the brute-forcing one would use the birthday paradox to get a complexity of about $2^{24.5}$. $\endgroup$
    – j.p.
    Jun 6 at 8:02

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