# Tag Info

1

The reason for both is that the generated values are trivial to detect / exploit and should be avoided and your RNG is deeply flawed if you actually get those values because the chance for this lies around $2^{-2000}$ if you use an appropriate parameter set. Now for the math: You need to choose your secret exponent $x$ such that $1\leq x\leq p-2$, with ...

0

Because $a^{p-1} \equiv 1 \mod p$ and $a^{p-2} \equiv a^{-1} \mod p$. In both cases it is easy to solve for the exponent.

5

Short key fingerprints are indeed vulnerable. But those are different from the short-authentication-string (SAS) used by ZRTP. A simple SAS based protocol using one-time keys could look like this: Alice sends a (collision resistant) hash of her public key to Bob. Bob sends his public key to Alice Alice sends her public key to Bob The short ...

1

This can be derived from two simple facts about the $mod$ operation: $a \bmod b = a + bi$ for some integer $i$ (for any $a, b$) $a \bmod c = b \bmod c$ if $a - b = ci$ for some integer $i$ With these two facts, we can look at $(g^a \bmod p)^b$; that can be simplified to $(g^a + pi)^b$ (for some integer $i$), and by the binomial expansion, this is $g^{ab} ... 3 I see you use the same generator on both sides (this need not work for any$n_1, n_2$of course...). But even if this holds (trying a small example): Let 1 use$n_1 = 11, M=2$, and 2 uses$n_2 = 13, M= 2$. Check that$2$is a generator for both of the multiplicative groups. If$d_1 = 7, d_2 = 9$, then$A = 2^7 \bmod 11 = 7$, while$B = 2^9 \bmod 13 = 5\$. ...

1

Elliptic curve security relies on the hardness of discrete logarithm on that curve. (Well, that's a simplification, but this will do for this answer.) When the curve contains N points, it takes an effort of roughly sqrt(N) "elementary operations" to break discrete logarithm. A prime p of "k bits" means that p is less than 2k, but greater than 2k-1. The ...

0

The [very] simplified answer is that one of the parameters for an elliptic curve is a prime p, the addition and multiplication in your elliptic curve is done (mod p), so the larger your p, the larger the set of integers you're working with, the more guesses an attacker has to make to break your key. For example, if your prime was 3, then the only possible ...

0

It has to do with the Diffie–Hellman assumption. The DH key exchange is secure in groups where the computational DH assumption holds. One of the simplest such groups is the multiplicative group modulo a large prime. However, that is not necessarily required. At least some composite integers with unknown factors would make a secure Diffie–Hellman modulus, ...

1

I assume the recommended approach is to use a KDF function like HKDF, but what is the security implication of taking an SHA-256 hash and using it directly for AES-256 or truncating it for AES-128 (Alice and Bob are using Java which doesn't have a native implementation of HKDF and I don't think it is a good idea to try and write your own). HKDF(-Expand) ...

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