Following up on my previous post, I thought I might get a more concrete answer if I gave a more concrete question.

I require 128-bit security so I choose a 3072-bit RSA modulus ($\ell_n=3072$). Specifically I choose $n=pq=(2p'+1)(2q'+1)$ that is a product of safe primes $p$ and $q$.

Now, I want to choose $\ell_\Lambda$ such that finding DL with $\ell_\Lambda$-bit exponents is hard in $QR_n$, for an adversary who does not know the factorization of $n$, with 128 bits of security.

Note that the order of $QR_n=\phi(n)/4=p'q'$, so each element has large order.

Is choosing $\ell_\Lambda=256$ sufficient as suggested in the answer to my previous post?

  • $\begingroup$ Is the factorization of $n$ secret? With $p$ and $q$ known, we can reduce a DLP modulo $n$ to two DLP modulo $p$ and $q$, and with each 1536-bit these would not be 128-bit secure (more like 100-bit, give or take a lot). $\endgroup$
    – fgrieu
    Feb 9, 2023 at 17:42
  • 2
    $\begingroup$ @fgrieu: as explained in his previous post, the factorization of $n$ is secret $\endgroup$
    – poncho
    Feb 9, 2023 at 18:32
  • 1
    $\begingroup$ @fgrieu edited question to clarify $p$ and $q$ are unknown $\endgroup$ Feb 9, 2023 at 19:00

1 Answer 1


I second that with the factorization of $n$ remaining secret, as assumed in the paper linked in that previous question (in particular by making the strong RSA assumption), the Discrete Logarithm Problem modulo $n$ is believed no easier that if $n$ was prime. And therefore, the best methods to solve that DLP have expected cost the lowest of:

  1. a few times $2^{\ell_\Lambda/2}$ multiplications modulo $n$, for methods based on collision search like Pollard's rho/kangaroo and distributed variants.
  2. about the cost of (G)NFS factorization of $n$; see this paper for about the state of the art, implying that the cost of factorization of $n$ and DLP for prime $n$ have roughly similar cost.

With $\ell_\Lambda=256$ and $\ell_n=3072$, (1.) is believed to be the least infeasible, and this parametrization is believed to give 128-bit security, disregarding hypothetical CRQC.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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