Let $F$ be a finite field of characteristic $2$. Let $m \geq 2 $ be a positive integer.
Seems unknown
if the trinomial
$$
T(t,x) = x^{2m+1}+x^2+s(t) \in F[t][x]
$$
(more explicitly, the `constant` coefficient $s(t)$ is a polynomial in $t$,
i.e., $s(t) \in F[t]$)


has monic factors $D(t,x)$ in $F[t][x]$ with degree $3$ relative to $x$.


I.e., 
$$
 D(t,x) = x^3+ a\_2(t)x^2+a\_1(t)x + a\_0(t)
$$
with $a\_2(t), a\_1(t),a\_0(t) \in F[t]$ and $T(t,x) = D(t,x)K(t,x)$ for some $K(t,x) \in F[t][x].$


We will then say that $T(t,x)$ have $D(t,x)$ as a factor or that $D(t,x)$ divides $T(t,x)$.


Question: Assume that $D(t,x)$ as above divides the trinomial $T(t,x).$


Do we have
$$
\deg(a\_1(t)) = 2 \deg(a\_2(t)),\;\;\deg(a\_0(t))=3\deg(a\_2(t)).
$$
???


I am aware of the work of Schinzel on trinomials. The trinomials $T(t,x)$ do not seem
to be worked out in these papers. 



Reducible trinomials $x^{odd}+x^2+s(t)$ in characteristic $2$
We show the following:



> 
> **Theorem.** Let $F$ be a field of characteristic $2$, and $s(t)\in F[t]$ a non-constant polynomial. Then $f(t,X)=X^{2m+1}+X^2+s(t)$ (where $m\in\mathbb N$) is either irreducible, or a product of irreducible factors of degrees $1$ and $2m$.
> 
> 
> 


In particular, this answers the question about the non-existence of cubic factors for $m\ge2$.


We use Galois theory, group theory and a little bit of valuation theory: Set $y=s(t)$, and let $x$ be a root of an irreducible factor of degree $k$ of $f(t,X)$. Note that $x^{2m+1}+x^2=s(t)=y$. We obtain the following field degrees: $[F(x):F(y)]=2m+1$ and $[F(x,t):F(t)]=k$.


It is easy to see that we may assume that $s(t)$ is not a polynomial in $t^2$. (Take the derivative with respect to $t$ of a factorization). So $x$ and $t$ are separable over $F(y)$. Let $L$ be the smallest Galois extension of $F(y)$ which contains $x$ and $t$, and set $G=\text{Gal}(L/K(y))$.


The group $G$ need not act faithfully on the conjugates of $x$. Let $N$ be the kernel of this action. So $G/N$ acts faithfully on the conjugates of $x$. Note that $G/N$ is the Galois group of $X^{2m+1}+X^2+y$ over $F(y)$.


We first claim that $G/N$ is the symmetric group $S\_{2m+1}$: The polynomial $X^{2m+1}+X^2$ is functionally indecomposable. To see this, take the derivative of $X^{2m+1}+X^2=A(B(X))$. A short calculation shows that either $A$ or $B$ is linear. So, by Luroth's Theorem, $G/N$ is primitive. Furthermore, $X^{2m+1}+X^2=(X^{2m-1}-1)X^2$, with the first factor being separable. Valuation theory shows that the inertia generator of a place of $L$ lying above the place $y\mapsto 0$ is a transposition on the conjugates of $x$. A well-known theorem of group theory then shows that $G/N$ acts as the symmetric group $S\_{2m+1}$ on the conjugates of $x$. Let $G\_x$ and $G\_t$ be the stabilizers of $x$ and $t$ in $G$. As $x$ has degree $k$ over $F(t)$, we get that $G\_t$ has an orbit of length $k$ on the conjugates of $x$. But $NG\_t$ has the same orbits as $G\_t$ on this set. Thus, we may (and do) replace $G\_t$ with $NG\_t$, which amounts to replacing $s(t)$ with a polynomial $s\_1(t)$ such that $s(t)=s\_1(s\_2(t))$ for another polynomial $s\_2(t)$, and $X^{2m+1}+X^2+s\_1(t)$ has the same factorization pattern as $f(t,X)$.


So we have $N\le G\_t$, that is $N$ fixes $t$. But $N$ is normal in $G$, so $N$ fixes all conjugates of $t$ (and those of $x$ by definition), hence $N=1$. 


So $G=S\_{2m+1}$ acts faithfully on the conjugates of $x$.


Now let $T$ be the inertia group of a place of $L$ lying above the place $p=(y\mapsto\infty)$. As $p$ is totally ramified in $F(x)$, and $F(x)/F(y)$ is tame, we get that $T$ is a cyclic group permuting regularly the conjugates of $x$. In particular, $T$ has order $2m+1$. However, $p$ is totally ramified in $K(t)$ too, so $T$ permutes the conjugates of $t$ transitively as well. Therefore $[G:G\_t]\le 2m+1$.


Now suppose that $2\le k\le 2m-1$, so $G\_t$ has an orbit of length $k$ on the conjugates of $x$. This yields $\lvert G\_t\rvert\le k!(2m+1-k)!$, hence $[G:G\_t]\ge\binom{2m+1}{k}>2m+1$, contrary to the inequality above.


If $k=1$, then $G\_x$ and $G\_t$ are conjugate in $G$. In particular $G\_t$ has orbit lengths $1$ and $2m$ on the conjugates of $x$, and the claim follows again.


