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Wouldn't encrypting a message with AES, then encrypting the (randomly generated) AES key and IV with the EC public key suffice? Yes it would suffice and is what is usually done. However for this to work you'd have to have a way to reliably convert a random integer to a curve point and back which isn't trivially possible. And even if you could reliably ...


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I guess the answer is no, as long as you are using ECIES then this protocol does not work - you cannot trust the public key of Bob, which is required for ECIES. You could however use ephemeral-static Diffie-Hellman, using ECDH as cryptographic algorithm. Alice would supply the static part as her public key is trusted, Bob may use any key pair. That means ...


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It depends on exactly what protocol you're asking about. ECIES as design gives no assurance to Bob that the message really came from Alice. This is, with standard ECIES, Alice does not use her private key -- instead, everything that Alice does (encrypt using Bob's public key) could have been done equally well by someone else - hence, Bob has no ...


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ECIES may seem complex, but if you try another approach, you would end up with something very much like it. If you only encrypt with AES, then you are not authenticating, which is most cases you also need to do. If you then encrypt and authenticate by yourself, you pretty much reinvented ECIES. But yes, ECIES is in higher layer of abstraction compared to ...


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Can $A$ assume that only $B$ can read the message? Only someone with the private key corresponding to $P_B$ can read the message. If $A$ can assume that only $B$ has it, that works. As $B$ does not know anything about $A$ what is the cryptographic purpose of the MAC? So that someone else cannot modify the encrypted message. What we want to prevent ...


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I'm answering based on the protocol in your question, which seems to be (EC) Diffie-Hellman. My question is; if Bob is confident that Alice's public key really belongs to Alice, then can he be confident that it was Alice who encrypted the message? If Bob is confident no one else has access to either secret key or the shared secret, then yes, ...



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