Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $a\neq 0$.

share|cite|improve this question
up vote 3 down vote accepted

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 $$\gcd(m,q-1)\mid\frac{q-1}r.$$
share|cite|improve this answer
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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.