The roots in a finite field Let $\mathbb F$ be a finite field of order $q=p^k$, where $p$ is an odd prime number. For an element $a\in\mathbb F$ how can we count the m-th roots of $a$? That is, the number of solutions of the equation
$$x^m=a$$
Suppose that $a\neq 0$.
 A: The multiplicative group of $\mathbb F$ is cyclic of order $q-1$, and as already pointed out by DonAntonio, all solutions of $x^m=a$ can be obtained from any fixed solution by multiplying it by all $m$th roots of unity in $\mathbb F$. This implies:


*

*If there is at least one solution, the total number of solutions is the number of $m$th roots of unity in $\mathbb F$, which is
$$\gcd(m,q-1).$$

*In order to determine whether the equation has a solution, let $r$ be the multiplicative order of $a$ (i.e., the least $r$ such that $a^r=1$). Then $x^m=a$ is solvable iff
$$\gcd(m,q-1)\mid\frac{q-1}r.$$

A: As in any other field, if $\,\alpha\,$ is a root, i.e. $\,\alpha^m=a\,$ ,then all the roots are $\,\alpha\, w^k\,,\,k=0,1,...,m-1\,$ , where $\,w\,$ is a primitive $\,m-$th root of unity: $\,w^m=1\,\,\,,\,\,w^t\neq 1\,\,,\,\forall\,0\leq t<m\,$ .
I don't think something in general can be said: it'll depend on $\,a\,\,,\,\operatorname{char}\Bbb F$\, and on $\,m\,$ , though it can be divided in cases.
