# In $\Bbb{Z}/m\Bbb{Z}$, show that $([a]_m)^{qd+r} = (([a]_m)^d)^q([a]_m)^r$

In $$\Bbb{Z}/m\Bbb{Z}$$, show that $$([a]_m)^{qd+r} = (([a]_m)^d)^q([a]_m)^r$$.

My first attempt at this question was to use simple arithmetic properties to prove this true, however, this is incorrect.

What is the best way to prove this?

You could use $$([a]_m)^{qd+r}=\underbrace{[a]_m[a]_m\cdots[a]_m}_{qd+r\textrm{ times}}=_*[a^{qd+r}]_m=[(a^d)^qa^r]_m$$ where $$(*)$$ happens by definition.
• I was thinking of doing this, but then wouldn't it entail proving $([a^d])([a^q]) = [a^{qd}]$? – flutterbug98 Sep 27 '19 at 2:51
Start by showing the following is well-defined: $$[x]_m \cdot [y]_m = [x\cdot y]_m$$
That is, if you take representatives $$x_1, x_2, y_1, y_2 \in \mathbb{Z}$$ such that $$[x_1]_m = [x_2]_m$$ and $$[y_1]_m = [y_2]_m,$$ then $$x_1 y_1 - x_2 y_2 = (x_1-x_2)y_1 + x_2(y_1-y_2)$$ is divisible by $$m.$$
This property allows us to write $$([a]_m)^n = [a^n]_m$$ and these two statements together allow you to use those simple arithmetic properties to prove your statement.