# Tagged Questions

The Diffie–Hellman key agreement is an anonymous, non-authenticated key-agreement protocol.

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Let $p$ be a prime number, and let $g_{1},g_{2},...,g_{n}$ be $n$ generator of $Z^{*}_{p}$. We have a list $y_{1},y_{2},\dotsc,y_{n}$ of elements in $Z^{*}_{p}$ such that for every $i\in ... 1answer 452 views ### In the STS Authentication Protocol, why are the signatures encrypted? From Wikipedia: (1) Alice → Bob : g^x (2) Alice ← Bob : g^y, E_K(S_B(g^y, g^x)) (3) Alice → Bob : E_K(S_A(g^x, g^y)) I know there should be something I'm ... 1answer 334 views ### Is there proof to the relation between the gap Diffie-Hellman problem and the the Cha-Cheon signature scheme? I am trying to prove that: "If the gap Diffie-Hellman problem is easy, then the Cha-Cheon signature scheme will be broken." Can you help me to prove it? Is there any proof to the relation between ... 1answer 81 views ### One-time pad using RSA and Diffie-Hellman functions Suppose that Alice and Bob use an encryption algorithm based on a one-way function$F$(Eve knows it), and$r$, a secret key shared between Alice and Bob (Eve doesn't know it). Key is computed as:$K ...
If you have the cyclic group of integers modulo $p$, where $p$ is not a safe prime, as well as a generator $g$ with which for all factors $q$ of $(p-1)$, $g^{(p-1)/q} \ne 1$, This answer says that ...
Must Oscar change only the Key $K_{abc}$ (Alice, Bob and Charlie) or does he have one key for everyone of them ?