Exercise 2.8.6 (No two of $a, a^2,a^3,\ldots, a^h$ are congruent modulo $m$)

Proof. Assume for contradiction that $a^i\equiv a^j(\mathit{mod}m)$ for some exponents $i,j$ such that $1\le i<j\le h$. Then $m\mid a^j-a^i=a^i(a^{j-i}-1)$. By definition 23.6, $a\wedge m=1$, thus $a^{j-i}\equiv 1(\mathit{mod}m)$. But $j-i\le h-1$. Since the order of $a$ is $h$, $j-i=0$, so $i=j$. This is a contradiction. This proves that no two of $a,a^2,a^3,\dots ,a^h$ are congruent modulo $m$. □