Solutions to an equation of the form $x^n=x$ in a finite field.

Question

I'm trying to count the number of solutions to an equation of the form $x^n=x$ in the finite field $F_{q^n}.$ In case of $n=q$, it's clear that every element of $F_q$, namely the base field, is a solution to this equation. I also know that $F_{q^n}-\{0\}$ is a cyclic multiplicative group of size $q^n-1$, and hence in order for $x^n=x$ to have a nontrivial answer, $(n-1,q^n-1) \neq 1$. Any thoughts would be appreciated.

Answer 1

Clearly $x=0$ is always a solution. Next we assume $x \neq 0$ and the equation simplifies to $x^{n-1}=1$. Considering the fact that the multiplicative group $F_{q^n} - \{0\}$ is cyclic of size $q^n-1$, we have $x^{n-1}=1$ precisely when $x^{gcd(n-1,q^n-1)}=1$. Let $t=gcd(n-1,q^n-1)$; Assume $y$ is a generator for the cyclic group $F_{q^n} - \{0\}$, which clearly means $F_{q^n} - \{0\} = \{y^k | k=0,1,...,q^n-2\}$. This means we have to find elements $y^k$ such that $(y^k)^t=y^{kt}=1$. This happens if and only if $q^{n}-1$ divides $kt$. So let $A=\{k: 0\leq k \leq q^n-2, q^{n}-1 | kt\}=\{k: 0\leq k \leq q^n-2, \frac{q^{n}-1}{t} | k\}$. Considering $0$ is always a solution, it follows the number of solutions to $x^n=x$ equals $|A|+1=t+1$. (My special thanks to @Jyrki Lahtonen for his very helpful remarks.)