Roots of Artin-Schreier equation

Let $a \in \mathbb F_q, q=p^f$. Is it true that $x^p-x-a$ has a root in $\mathbb F_q$ iff $tr_{\mathbb F_q/\mathbb F_p}a=0$?

By the so called freshman's dream, $(x+y)^p=x^p+y^p$ for all $x,y\in\Bbb{F}_q$, the function $$ g:\Bbb{F}_q\to \Bbb{F}_q, g(x)=x^p-x $$ is linear over the prime field $\Bbb{F}_p$. Little Fermat tells us that $\ker g= \Bbb{F}_p$. Thus by rank-nullity theorem the image of $g$ is of dimension $f-1$ over $\Bbb{F}_p$. Therefore the equation $g(x)=a$ has a solution $x\in \Bbb{F}_q$ for exactly $p^{f-1}$ choices $a\in\Bbb{F}_q$.

Consider the trace map $$ tr(x)=x+x^p+x^{p^2}+\cdots+x^{p^{f-1}}. $$ It is linear, and it has the property that $tr(x^p)=tr(x)$ for all $x\in\Bbb{F}_q$. Therefore $$ tr(x^p-x)=tr(x^p)-tr(x)=tr(x)-tr(x)=0 $$ for all $x\in\Bbb{F}_q$. Thus $tr(a)=0$ whenever $a\in\mathrm{Im}\, g$, so $\mathrm{Im}\, g\subseteq\ker tr$. On the other hand the trace is a polynomial of degree $p^{f-1}$, so $\ker tr$ has at most $p^{f-1}$ elements. Therefore we must have equality $$ \ker tr=\mathrm{Im}\, g. $$ This is exactly the claim.

This result is occasionally called additive Hilbert 90, as it makes a similar statement about the action of the Galois group on the additive group of a finite field as the (multiplicative version of) Hilbert Satz 90 makes about the multipliactive group of a cyclic extension of fields.