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

Question

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?

Answer 1

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.

Answer 2

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 $m.$

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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.