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how does $A$ know $g^y$ to include in the formula above? Actually, what they mean here is, in fact, $g^y \bmod p$; that is, the value that $A$ received. It wouldn't work to insert the literal values $g^x$ and $g^y$; apart from the fact that $A$ doesn't know the second one, there's also the practical difficulty that since $x$ and $y$ are (perhaps) 256 ...

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ECDH is included in the ciphersuites, so the only answer is: yes, this should be possible. For your further research, it might help to know that Crypto.SE features a lot of Q&As related to “OpenSSL ECDH”. Also see the related documentation at the OpenSSL wiki for practical code examples showing how to use ECDH in OpenSSL, how to use the low-level APIs ...

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I am trying to better understand authentication. Lets say I have posted my 128-bit AES symmetric key on some forum, encrypted asymmetrically to my friend using 256-bit ECC (25519). The forum isn't controlled by us so this key message could potentially be tampered with. Well first of all, note that encrypting with curve25519 isn't as trivial as ...

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If the key is send beforehand it is not required but highly recommended to sign it. Otherwise anybody could post an AES key, encrypted using the public encryption key of your friend. In that case your friend may only find out after receiving the right messages. Furthermore, your friend may not be able to distinguish between an invalid key and an invalid ...

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Here is such a protocol. Alice chooses a random AES key $K_a$, encrypts it using Bob's public key to get $C_a = \mathrm{Enc}_{pk_b}(K_a)$, and sends $C_a$ to Bob. Likewise, Bob sends $C_b = \mathrm{Enc}_{pk_a}(K_b)$ for a random AES key $K_b$. Upon receiving $C_a$, Bob recovers $K_a = \mathrm{Dec}_{sk_b}(C_a)$ and computes the shared AES key \$K_{ab} = ...

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So, I does anybody know how to design a module for remainder operation in >verilog that computes the remainder in a single clock cycle? Any links to >literature discussing the algorithm would suffice. A nieve implementation would be something like. parameter bitwidth; input [bitwidth-1:0] a; input [bitwidth-1:0] b; output reg [bitwidth-1:0] result; reg ...

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