When I am doing my study, I found that for arbitrary positive integer $a,m$. Let $\mathbf{S}=\{ a^{n} \bmod m : n \in \mathbb{Z}^{+} \}$. Obviously, $\mathbf{S}$ is a finite set. I have list some random examples, they all support that the sum of elements in a certain $\mathbf{S}$ is the multiple times of the relevant $m$.
I am a junior mathematics student, so I think my method is a little bit naive.
I attempt to prove this but I meet some difficulties. I am seeking for some help and advice.
UPDATE 1:
There seems some counter example on this statement. However, these counter example seems have similar pattern, and this statement is likely to be true for the majority case. I am wondering if it is possible to make some limitation on $a,m$.
Let's assume $\gcd(a,m)=1$. Let $r$ be the order of $a$ modulo $m$, that is, $a^r\equiv1\bmod m$, and $a^s\not\equiv1\bmod m$ for $1\le s<r$. Then $$S=\sum_0^{r-1}a^n={a^r-1\over a-1}$$ so the question is, when does $m$ divide $(a^r-1)/(a-1)$. This certainly happens if $\gcd(a-1,m)=1$, and that certainly happens if $m$ is prime and $a\ne1$. Situation appears more complicated if $\gcd(a-1,m)>1$.