# Tag Info

5

No, since finding $a$ allows offline checking of passwords. $\:$ No, although I can't back this part up.

4

When using a Discrete Logarithm based scheme, such as SRP, the rule of thumb is to always use private exponents with a bit length twice the desired security strength. Hence, a 128 bit exponent $a$ will at most give you 64 bits of security. If you want 128 bit security, you need (at least) a 256 bit exponent. This is because the algebraic structure of the ...

3

Solving a 256-bit discrete log is absolutely doable, and quite quickly, these days; there are public tools that can do it, though they may require some expertise to use. On that note, even a 1024-bit modulus is not particularly conservative: it is generally agreed that well-funded organizations today could break logs of that size as well, but at a very ...

2

It seems like a better solution would be to have the server that is providing the Javascript file, also provide a random seed. The Javascript can then use that random seed (and anything other maybe-random bits it can scrounge up, such as the output from Math.random()) to see a cryptographic PRNG, and then use the output of that crypto-PRNG for generating s, ...

2

The purpose is to prevent a two-for-one guessing attack, where an active adversary, impersonating the server, can test two password guesses per attempt. The attack and why the multiplier prevents it is described in Section 2 of the SRP-6 paper (ps). (According to MacKenzie, it was discovered by Bleichenbacher.) In brief, the attack goes like this: Instead ...

2

Being able to solve the discrete logarithm in SRP-6 allows an eavesdropping attacker to dictionary attack the password. It will not directly reveal a strong password or its hash. It requires the attacker to observe a successful authentication, $B$ alone does not suffice. The attacker eavesdrops $s$, $A = g^a$, $B$ and $M_1$. The attacker solves $a$ from ...

2

"Would it be possible for an attacker to launch an offline dictionary/brute-force attack on the B public key: ..." That is possible if and only if the attacker can distinguish b's distribution from the uniform distribution on {0,1,2,3,...,N-3,N-2}. $\:$ If so, an attacker could compute verifiers v for candidate passwords, subtract kv from B mod N, and ...

1

SRP protocol is quite abstract so to provide matching implementation for version 6a you need to know following: N, g - group parameters H - hash function, there can be different hash functions used for different values how is private key x calculated how is shared session key K calculated how are evidence messages (M1, M2) calculated In addition you need ...

1

The multiplier parameter $k$ is different between SRP 6 and 6a. You can see that RFC 5054 calculates it using a hash of the domain parameters (modulus $N$ and generator $g$), so it is using SRP 6a, as opposed to SRP 6 where $k$ is constant. Likewise, in section 6.2.1 of IEC 11770-4 – the October 2005 draft at least – the equivalent value $c$ is defined as a ...

1

There is an explicit RCF 5054 which uses SRP to negotiate a shared key for a TLS connection. There are also hooks for OpenSSL to be able to use SRP to setup an SSL connection without using certificates using the SRP generated shared session key.

1

Look at where $a$ is used in the protocol: The user calculates the public $A = g^a$ using it. The user computes the session key as $K = H(S)$, with $S = (B - kg^x) ^ {a + ux}$. An attacker should never find out $S$, because even if the session key $K$ leaks due to e.g. a flawed encryption algorithm, she would only know the hashed value. So knowing or ...

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