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 | <CAJkPp5PV1yCoEDU-PrmHeaQqjq-tRkAq2n-15w0DzYND3TD2Lg@mail.gmail.com> |
Hi Bill, This would be my approach for small m, but it's impractical for m about 10^10. Another approach is to try testing small primes, that is, for a candidate p to compute the power x^m in GF(p)[x]/<f(x)>. It'll be fast for any chosen p, but it's impossible to guess/find a large prime solution p when one exists. All in all, I'm hoping for an algorithm that works in O(log(m)) time and space. I do not expect many primes to be produced this way, but my goal is to systematically search for non-trivial cases when such primes do exist and may be large. Regards, Max On Wed, Jun 4, 2025 at 6:00 AM Bill Allombert < [email protected]> wrote: > On Tue, Jun 03, 2025 at 11:55:39AM -0400, Max Alekseyev wrote: > > Hello, > > > > Suppose I have a large number m, a quadratic polynomial f(x) and linear > > polynomial g(x). > > Is there a fast way to find all primes p such that the remainder of > > division of (x^m - g(x)) by f(x) vanishes modulo p ? > > To give a specific example, let m = 10^10, f(x) = x^2 - 3*x - 3, and > g(x) = > > x - 4. > > For m=10^5, you can do this: > > factor(content(Mod(x,x^2 - 3*x - 3)^m-(x - 4))) > %22 = Mat([2,2]) > > Cheers, > Bill > >