Solution to $x^q+\alpha x= \beta$

Question

Let $p$ be a prime, $q=p^a$ and $\beta\in\mathbb{F}_{q^n}$. It is known that there is solution in $\mathbb{F}_{q^n}$ to $$x^q-x=\beta$$ if and only if the trace to $\mathbb{F}_q$ of $\beta$ is $0$.

My question is, is there an easy criterion for when there is a solution in $\mathbb{F}_{q^n}$ to $$x^q + \alpha x= \beta$$ for $\alpha,\beta\in\mathbb{F}_{q^n}$?

Answer 1

Yes, there is a condition. It is a little bit more complicated though. I will denote $F=\Bbb{F}_q$, and $E=\Bbb{F}_{q^n}$

1. If $\alpha=0$ then there is the unique solution $x=\beta^{q^{n-1}}$. This is because raising to $q$th power is a bijection on $E$, and doing it $n$ times takes you back to the starting point as all the elements of $E$ are zeros of $x^{q^n}-x$.
2. If $\alpha\neq0$ then it is trickier. Consider the mapping $$L(x)=x^q+\alpha x$$ from $E$ to itself. We easily see that $L$ is linear over the smaller field $F$. Furthermore, the number of zeros of $L$ is at most $q$. We can deduce that $\operatorname{ker}(L)$ has dimension $0$ or $1$, and therefore, by rank-nullity, $\operatorname{im}(L)$ has dimension $n$ or $n-1$ respectively.
   • Assume that $t$ is a non-zero element of $\operatorname{ker}(L)$. Then $-\alpha=t^{q-1}$, and, denoting $y=x/t$, we can rewrite the equation $L(x)=\beta$, by dividing it with $t^q$, to read $$\frac{x^q}{t^q}-\frac x t=\frac{\beta}{t^q},$$ which is equivalent to the equation $$y^q-y=\beta/t^q$$ in the unknown $y=x/t$. By the fact that you cited, this has solutions if and only if $\operatorname{tr}^E_F(\beta/t^q)=0,$ when it will have exactly $q$ solutions.
   • But if $-\alpha$ is not a $(q-1)$st power in the field $E$, then $\operatorname{ker}(L)$ is trivial, and hence $L$ is surjective. Implying that the equation $L(x)=\beta$ has a unique solution $x\in E$ for all $\beta\in E$.

So the remaining task is to decide which case we have. We saw that this is equivalent to deciding whether the element $-\alpha$ is a $(q-1)$st power in $E$ or not. This may be a bit taxing. Because $E^*$ is cyclic of order $q^n-1$ and $q-1\mid q^n-1$, we can say that an element $z\in E^*$ is a $(q-1)$st power if and only if $z^{(q^n-1)/(q-1)}=1$. That is, iff its relative norm $N^E_F(z)=1$. I don't know how useful this is :-(