1
$\begingroup$

Can I use Diffie-Hellman over, say, $GF(2^{128}) \bmod$ irreducible poly in $GF(2^{128})$ instead of $GF(p)$? If not, why?

Or increase it to $GF(2^{2^{\text{whatever}}})$.

$\endgroup$

1 Answer 1

6
$\begingroup$

Always use a randomly chosen prime field.

First of all, $GF(2^{2^{whatever}})$ is nonsensical. The field is still $GF(2^k)$ for some $k.$

The security of Diffie Hellman rests on the difficulty of discrete logarithms.

The discrete logarithm problem for fields of the form $GF(2^k)$ is much easier, having quasi-polynomial algorithms developed by Joux and others for some values of $k$. See also the answer to this question where the complexity for general $k$ of the form you are asking is stated as:

As for the case of $q=2^k$, the best asymptotic complexity is provided by Barbulescu et al.'s version of the function field sieve (FFS), which in that case runs in heuristic quasipolynomial time $2^{O((\log k)^2)}$.

Even for prime fields $GF(p)$, if the prime is known ahead of time there are attacks state actors can mount, but $p$ is randomly chosen for proper implementation of Diffie Hellman, thwarting those attacks.

$\endgroup$
2
  • $\begingroup$ Given the weakdh.org attacks I would say if you go to something like 3k bits it's probably acceptable to use a fixed prime, right? $\endgroup$
    – Elias
    Aug 18 at 10:19
  • $\begingroup$ @Elias, I believe so, but my expertise in this area is not that deep. You might want to ask a separate question. $\endgroup$
    – kodlu
    Aug 18 at 20:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.