Let F be a finite field containing $q$ elements. Let $a \in F$ be nonzero. If n divides $q -1$ prove that $x^n - a = 0$ has either n solutions in F or no solutions in F.
I am unsure of how to begin this problem. I know that since $q -1$ is the order of the cyclic group of the nonzero elements of F under multiplication, n divides the order of this group, but I am having a hard time seeing how it is useful. Is there a better way to approach the problem?
Suppose that the equation $x^n-a$ has a root $c\in F$.
Let $k=\frac{q-1}{n}$, and let $\beta$ be any generator for the multiplicative group of nonzero elements of $F$.
I claim that $\;c,\; c\cdot \beta^k, \;c \cdot \beta^{2k} , \;c \cdot \beta^{3k}, ..., \;c \cdot \beta^{(n-1)k}$ are all roots of the equation $x^n-a$, and that these roots are all distinct.
Indeed, whenever $0 \leq i \leq n-1$, we have that $$\big(c \cdot \beta^{ik}\big)^n = c^n \cdot \beta^{i(q-1)} =c^n \cdot 1 = a$$ where the second equality follows because $\beta$ has order $q-1$.
The reason that all of the elements $\{c \cdot \beta^{ik}\}_{0\leq i\leq n-1}$ are distinct is that if $0\leq i\leq j\leq n-1$, then $c \cdot \beta^{ik} = c \cdot \beta^{jk}$ implies that $\beta^{(j-i)k}=1$, which in turn implies that $i=j\;$ (because $\beta^k$ has order $n$ in $F^{\times}$).
Hence we have proved that if a solution to the equation "$x^n-a=0$" exists, then $n$ distinct solutions exist, which is exactly what we wanted.