Let $G$ be a finite cyclic group. is it true that for each positive integer $m$ dividing the order of $G$, the number of solutions $x$ of the equation $x^m=e$ in $G$ is exactly $m$?
using Frobenius theorem statement and the summation of euler-phi formula, along with the fact that number of elements of order $n$ dividing order of cyclic group $G$ is $Phi(n)$. using these I proved the problem statement is true. did I miss anything?
Yes, it is true.
Let $g\in G$ be a generator, and say $|G| = n$ (so we have $m\mid n$). Then any power of $g^{n/m}$ is a solution, and there are $m$ of those, since $g^{n/m}, g^{2n/m}, \ldots, g^{mn/m} = e$ are all distinct.
Conversely, if $g^k$ is a solution for some integer $k$, that means that $(g^k)^m = g^{km} = e$, so $km$ must be an integer multiple of $n$: $$ km = rn\\ k = r\frac nm $$ so $k$ is an integer multiple of $\frac nm$.
In general, if we don't require $m\mid n$, there are $\gcd(m, n)$ solutions. The proof is more or less the same.