I wonder if it's possible to do public key authentication (preferably not vulnerable to replay attacks) before modern computers became a thing (e.g. in a medieval setting).


  1. Authentication can be done in a reasonable time by hand.
  2. Brute force by hand requires unreasonable time. (But it may be possible with modern computers)

As I understand it, this would require that the disparity between the time complexities of normal authentication and brute-force is very large even for small n, where n is something like the length of the key. Is there anything that can achieve this?


1 Answer 1


I could conceive of this in theory. If we look back to when the word "computer" referred to a job done by a human being as early as Renaissance times, computers were capable of lengthy and sophsticated calculation.

In a near miss for the discovery of public key cryptography in 1874 William Stanley Jevons created what could be thought of as the first RSA-like challenge in his book Principles of Science. Jevons thoughts were not far from cryptography at this point and it could certainly be argued that he understood that his "RSA key generation" was simpler than factoring by trial division. Notably though Charles Busk used Fermat's method to factor Jevons number five years later with a moderate amount of effort. Both Jevons and Busk likely had access to mechanical devices or human computers and its possible that if Jevons had devoted longer to generating his key, it would have taken a much longer (infeasible?) amount of time for Busk to factor. Arguably, Fermat's method is more sophisticated than brute forace and so falls outside the parameters of the question.

  • $\begingroup$ Good point that Fermat's primality test is feasible by hand to sizes making factorization by hand much harder. The strong pseudoprime test thus also is feasible (since it's even less costly). Using the CRT, RSA signature of short messages (with crude padding adding redundancy) is of cost comparable to such primality test for both factors, thus doable too, with comparably significant but feasible effort. I wonder for ECC. The parts I see somewhat problematic by hand are finding the group order, and making the equivalent of a hash. $\endgroup$
    – fgrieu
    Jan 13, 2023 at 11:44

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.