lean_workbook_plus_63550

For $\boxed{\Rightarrow}$ let $b=ka+r$ , with $k \in \mathbb{Z}_{\ge 0}$ and $0 \le r < b.$ Then, as we proved before, $x^a -1 \mid x^{ka}-1,$ so $x^a -1 \mid (x^{ka+r}-1)-x^r(x^{ka}-1),$ i.e. $x^a - 1 \mid x^r-1.$ However, since $\deg(x^r-1)<\deg(x^a-1),$ we must have $a=0,$ so $a \mid b.$