TITLE: Solving cubic equations in characteristic 2
QUESTION [9 upvotes]: Consider the cubic polynomial
$$
f = x^3+px+q,
$$
where $p,q$ are elements of a fixed algebraic closure $\overline{\mathbb{F}}_2$ of $\mathbb{F}_2$. 

Is there an elegant criterion for deciding whether $f$ has $0$, $1$ or $3$ roots in the field $\mathbb{F}_2(p,q)$?

For example, I would consider as "elegant" any criterion stipulating that certain polynomial expressions in $p,q$ be "of a certain form," e.g., are themselves values of certain polynomials (with coefficients in $\mathbb{F}_2$).

REPLY [2 votes]: Let $K$, the field generated by the coefficients of $f$, have $t=2^n$ elements.
If $p=q=0$, $f$ has a triple root. If $q$ is not $0$, then $f$ is separable. Assume $q$ not $0$.
If $p=0$ and n odd, then $f$ has exactly one root in $K$. If $p=0$ and $n$ even then $f$ has $0$ or $3$ roots in $K$.
Assume now that $f$ has at least 1 root in $K$. Let $tr$ be the trace of $K$ over the prime field. Put $B=1+\frac{p^3}{q^2}$. Then $f$ has $3$ roots iff $tr(B)=0$.
The question if $f$ has no roots (i.e. $f$ irreducible) can be done by Berlekamp's algorithm.