# Tag Info

4

The problem doesn't lie with curves in Weierstrass form necessarily, but with naive implementations of elliptic curve arithmetic on such curves. Basically, if you implement an ECC scheme (ECDH, ECDSA or whatever) on a smart card using a curve in Weierstrass form in the most straightforward way possible (by writing a simple double-and-add loop for ...

3

To use the proper terminology: in TLS, cipher suites which include "some Diffie-Hellman" are: Anonymous Diffie-Hellman: DH_anon Static Diffie-Hellman: DH-RSA, DH-DSS... Ephemeral Diffie-Hellman: DHE-RSA, DHE-DSS... There is no "plain DHE" cipher suite in TLS; it is called "DH_anon". As the name indicates, with DH_anon, the server is "anonymous": you ...

2

I believe that you misunderstand what DH is doing. DH-key-exchange was innovated to defence man-in-the-middle attack, because hackers can not pretend the one you want to communicate without correct share key? or hacker don't know the key generator that Alice and Bob pre-agreed? Well, no, defending against active attackers, that is, attackers who can ...

2

Unless you are absolutely sure that you don't need to and that the cost is going to be significant then I would absolutely say you should use authenticated encryption. One reason is bit-flipping attacks - flipping a few bits at the 'right' point in your encrypted message might lead well to a message that is legal (the classic example is if someone learns ...

2

A Diffie-Hellman key agreement has the following general form, presuming it is done in a group $G$ of order $q$ with generator $g$: $A$: Generate $x \in \mathbb{Z}_q$ at random. Calculate $X = g^x$ $B$: Generate $y \in \mathbb{Z}_q$ at random. Calculate $Y = g^y$ $A \to B$: $X$ $B \to A$: $Y$ $A$: Calculate $S = Y^x$ $B$: Calculate $S = X^y$ First ...

2

No, DHE is secure and allows to share a common secret between two parties over an insecure channel. But you cannot know, if the one you share the secret with is the one you want (DHE is vulnerable to man in the middle attacks). So DHE-RSA uses DHE to share a common secret and signs the communication with RSA to make sure, that both persons communicate with ...

1

I guess the problem is to find the generator $g$. Denote the factors for $p-1$ to be $p_1 =2, p_2 =2,p_3 = 37, p_4 = 709$. With $p$, and the factorization of $\phi(p)$ you can find a generator in the following way: Randomly choose an element $x$ from $Z_p$ and test whether $x^{\phi(p)/p_i}$ mod $p \ne 1$ for every $i =1,2,3,4$. If this is the case, $x$ is ...

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