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this post was submitted on 19 Nov 2023
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I'm not a cryptographer (so maybe this is wrong), but my understanding is that although it's possible to modify the cipher text, how those changes modify the plaintext are very difficult (or impossible) to predict. That can still be an attack vector if the attacker knows the structure of the plaintext (or just want to break something), but since the checksum is also encrypted, the chances that both the original file and checksum could be kept consistent after cipher text modification is basically zero.
A checksum and a digital signature aren't the same thing. If you have a data block and a checksum of the data block, the data block can be modified and a new checksum can be computed to reflect the modifications. Instead of a checksum would be a digital signature using an asymmetric key. The data block would be modified but the signature of that block can't be recomputed without the key used to sign it, which is not part of the transfer.
Isn't that also true of an encrypted checksum, though? For some plaintext block q there is a checksum r, but the attacker can only see and modify the encrypted q (Q) and encrypted r (R). How any change to Q would modify q (and R to r) can't be known without knowing the encryption key, but the attacker would need to know that in order to keep q and r consistent.
Possibly the source of any confusion here is when the encryption and when the compression takes place? Maybe some more details about how you are using xz and encryption would help.
As far as I can tell, xz doesn't do anything with signatures or encryption, but it does perform checksums like you stated, which is very cool and I'm glad you shared this.
Edit: I am re-reading your post above. You are compressing with xz, then encrypting, got it. So yes, if any part of the payload is tampered with, then it would be detected by the decryption, depending on the algorithm, or by the decompression because of the checksums like you said. Sorry for the confusion! You've got it all straight lol.