If $a,m,n \in \mathbb{Z}$ with $m>0, n >0$, Prove that [$a^m]=[a^n]$ in $\mathbb{Z_2},$

Question

Hello I need help with this proof

If $a,m,n \in \mathbb{Z}$ with $m>0, n >0$, Prove that [$a^m]=[a^n]$ in $\mathbb{Z_2},$

Answer 1

Hint : $[a^m] = [a]^m$, and consider the cases $a\equiv 1 \mod(2)$ and $a\equiv 0\mod(2)$ separately.

Answer 2

Can you see this amounts to proving that if $a$ is even (odd), so is $a^m$ for $m>0$?