Re: finding primes modulo which x^m mod f(x) has a prescribed result
Max Alekseyev <[email protected]>
| Newsgroups | gmane.comp.mathematics.pari.user |
|---|---|
| Message-ID | <CAJkPp5O_w8aGgP70W8_+nha_KVTOqhGv-5VxYvJiWmCBqEqsTA@mail.gmail.com> |
Hi Bill, Unfortunately, not. Number m = 10^10 was just randomly picked as an example to show its magnitude. Regards, Max On Fri, Jun 6, 2025 at 3:32 PM Bill Allombert < [email protected]> wrote: > On Wed, Jun 04, 2025 at 03:44:11PM -0400, Max Alekseyev wrote: > > Hi Bill, > > > > First off, I should have said that I'd like to have O(poly(log(m))) > > not O(log(m)) algorithm, but even with my honest typo, I don't quite > follow > > how your exercise implies the absence of the algorithm with O(log(m)) > space. > > >From m = p-1 (mod p^2-1), it follows that p <= m+1 and so it fits the > > O(log(m)) space (if it was the size of p you referred to). > > > > Anyway, I'd like to have the algorithm be bounded in both time and memory > > resources by a polynomial in log(m), not in m itself. > > If you only need m = 10^k then it might possible to prove the content > is always 4 for k>=2. > > Cheers, > Bill. >