# in what condition, 2 different AES 128-bit Keys generate at least 6 bytes identical ciphertext on same plaintext

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.

• 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.
– fgrieu
Jun 5 at 7:52
• 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? Jun 5 at 16:19

• You could mention that for the brute-forcing one would use the birthday paradox to get a complexity of about $2^{24.5}$.