5

I am interested in studying cubic equations over finite fields. For example, when does $$ ax^3 + bx^2 + cx + d = 0 $$ have a solution in $\mathbb{F}_q$ for $a,b,c,d\in \mathbb{F}_q$ (finite field of order $q$). I see this question/answer:

Solving polynomial equations over finite fields

mention the use of something called Galois theory and I know nothing about this. My question is whether there is an elementary approach to answering the question. If the general case is too difficult, then would it be possible if $q$ is a prime?

EDIT: From looking into this a bit more, I would like to know how one would go about solving a cubic equation over a finite field. What are the methods? From the formula/algorithm that works over the real numbers, I can see that this reduces to finding square roots and cube roots. I think I understand how the square roots work (maybe using something like the Legendre symbol). But (1) How can I determine if a given number in a finite field is a cube of another number and (2) how can I find this cube root? (I understand that it may not be unique). If the question is too broad or hard, I would be happy with a general outline of stuff.

John Doe
  • 3,233
  • 5
  • 43
  • 88
  • 2
    The classical formulas for solutions of cubics still do apply to cubics over finite fields not of characteristic 2 or 3. Is this relevant to your question? – paul garrett Apr 22 '21 at 19:41
  • @paulgarrett: I think so. I was thinking that all this comes down to finding square roots and the like. – John Doe Apr 22 '21 at 19:42
  • Square roots and cube roots, yes. – paul garrett Apr 22 '21 at 19:42
  • 1
    @paulgarrett: So this actually shouldn't be too hard. Just take the algorithm/formula over the real numbers. Figure out when one can square roots and cube roots, and then this should give the answer. I think I know the square root answer. What about the cube roots? – John Doe Apr 22 '21 at 19:44
  • 2
    Do you want to test whether something is or isn't a square/cube, or do you want to find a square/cube root? Guess-and-check/brute force obviously succeeds, more-or-less, but is not very explanatory... – paul garrett Apr 22 '21 at 20:31
  • 2
    You may find this old question interesting. Euclid's algorithm FTW. Formulas? I would not go beyond quadratic. Which is pretty much how I approach polynomial equations over the reals also - Cardano is not too useful due to the complex cubic roots. Cantor-Zassenhaus works over the non-prime finite fields also. – Jyrki Lahtonen Apr 23 '21 at 03:22
  • 1
    @RicardoCavalcanti Computers are, indeed, good at this. But not by brute force testing when $q$ has 100 digits or so. See the linked thread. The beauty of Cantor-Zassenhaus is that you don't care about the degree of the equation very much. Increasing the degree does slow it down soon, but see that thread for a little bit of data. – Jyrki Lahtonen Apr 23 '21 at 03:31
  • The catch is that it is an algorithm rather than a formula. I guess some may be unhappy with that. – Jyrki Lahtonen Apr 23 '21 at 03:36

0 Answers0