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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 ...


3

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


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 ...



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