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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 $a\neq 0$.

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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 $$\gcd(m,q-1)\mid\frac{q-1}r.$$
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The criterion for solvability can be written more consisely as $a^{(q-1)/\gcd(m,q-1)}=1$. –  Emil Jeřábek Jan 14 '13 at 16:40

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.

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