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My Galois theory is a little rusty, so this might be a quick question for you to solve. There doesn't seem to be anything about this on MSE (based on the search "[galois-theory] [finite-fields] quartic").

I'm sorry if it's obvious.

The Question:

In general, how does one solve a quartic equation $$f(x)=ax^4+bx^3+cx^2+dx+e=0$$ for roots in $\Bbb F_q$, where $f\in \Bbb F_q[x]$ with $q$ a power of a prime?

Thoughts:

I think it's matter of modifying the formula for a quartic equation over the reals.

Shaun
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    Trying out all field elements is a valid option. I'm not aware of any other method that solves a generic quartic over finite fields. – schiepy Sep 19 '24 at 19:08
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    It's worth noting that when the field is isomorphic to the integers mod $p$ the usual quartic formula will work (although taking 4th, 3rd and 2nd roots mod $p$ might be tricky for LARGE $p$). Tinkering with just the quadratic formula suggests when these formulas are not solvable (ex: $\sqrt{2} \mod 3$) then there is no solution to the original polynomial). I think these ideas extend to arbitrary rings $Z_n$ so perhaps this problem is better posed over commutative rings, of which a subset can also afford to be finite fields. – Sidharth Ghoshal Sep 20 '24 at 02:05
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    FWIW I vaguely having worked out the use of Cardano's formula (so a cubic) for a problem (with all solutions in a finite prime field) posted at Math.SE. The cube roots take some (re)interpretation. I don't recall enjoying the exercise. Can't find it, so it is possible that I chose to junk it in the end :-) – Jyrki Lahtonen Sep 21 '24 at 17:21

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Instead of using a "quartic formula", I think it will be easier to use that if a polynomial of degree $n$ over $\Bbb F_q$ has a root, then you just need to compute $\gcd(f(x), x^q - x)$ using the Euclidean algorithm. Indeed, the roots of $x^q - x$ are precisely the elements of $\mathbb{F}_q$. Essentially this is the idea of Berlekamp's algorithm, yielding a factorization into irreducible polynomials over $\Bbb F_q$. For example, the Berlekamp algorithm yields that $$ x^4+x^3+6x^2+3=0 $$ has the solutions $x=1$ and $x=3$ over $\Bbb F_{11}$. In fact, $$x^4+x^3+6x^2+3=(x-1)(x-3)(x^2+5x+1).$$ Note that this quartic polynomial has no real root at all.

Dietrich Burde
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