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This is a follow up question to this:

In general, how does one solve a quartic equation over a finite field?

The Question:

What is the formula for $x$ in $\Bbb F_q$ when $$f(x)=ax^4+bx^3+cx^2+dx+e=0,$$ where $f\in \Bbb F_q[x]$?

Bonus points go to whoever provides a reference too.

Thoughts:

I guess there'll be different formulae for whether $\operatorname{char}(\Bbb F_q)=p$ is odd or even.

I reckon this is close to cryptographic stuff, so it could be that there's no solution publicly available. (I don't know how to justify this feeling.)

Context:

In my research, I have a quartic equation over a finite field, that is pretty much as general as it gets.

I want:

  • to see whether it always has a solution in $\Bbb F_q$,
  • and if it doesn't, to understand why and for what values of $q$ and the coefficients it fails.

I cannot share much more than that!

Shaun
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    The general solution (in an algebraic closure) using Galois theory works over every field of high enough characteristic. – Martin Brandenburg Sep 24 '24 at 11:43
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    The general solution does work, but the necessary roots often reside in an extension fields, even if the solutions themselves are in the field of definitions. In other words, you have casus irreducibilis on steroids. – Jyrki Lahtonen Sep 24 '24 at 13:58
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    Characteristic two is particularly tricky. For starters, it's impossible to depress the quartic because a linear substitution does not change the cubic term at all ($(x+r)^4=x^4+r^4$). Instead, you can make the linear term vanish by such a process (assuming $b\neq0$), and after that you can go reciprocal, when you at long last get a polynomial without a cubic term. Then you can use the fact that (still even $q$) the polynomial $L(x):=Ax^4+Cx^2+Dx$ is an additive homomorphism (Frobenius FTW). Existence of solutions is then basically a problem in linear algebra. – Jyrki Lahtonen Sep 24 '24 at 14:04
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    (cont'd) Alternatively it may be possible to write $L(x)$ as a composition of two quadratic additive homomorphisms (you may need to go to an extension field for that to work out): $L(x)=A(f_1(f_2(x)))$, where $f_i(x)=x^2+\alpha_i x$, $i=1,2$. This turns the problem into one of solving a quadratic. When I get home, I will hopefully remember to check whether Lidl & Niederreiter describes the above process. – Jyrki Lahtonen Sep 24 '24 at 14:09
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    Couldn't find it anywhere. I do suspect somebody to have worked out the characteristic two case. – Jyrki Lahtonen Sep 24 '24 at 17:07
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    (still about even $q$) And the existence of a decomposition like the one I described depends on the solvability of a cubic. Actually, when that cubic has no solutions in $\Bbb{F}_q$, the quartic will always have a unique solution in $\Bbb{F}_q$, settling some of your questions – Jyrki Lahtonen Sep 24 '24 at 17:17

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