$x^{\ell}-1$ divides $x^{m}-1$ if and only if $\ell$ divides $m$

\(\implies\)

• Write \(m = \ell q + r\) with \(0\leq r < \ell\).
• Write \[ p(x) = {x^m-1 \over x^\ell - 1} = {x^{lq+r} -1 \over x^\ell - 1} = x^r{x^{lq} - 1 \over x^\ell - 1} + {x^r - 1 \over x^\ell - 1} = q(x) + {x^r-1 \over x^\ell - 1} ,\] where \(p,q\) are polynomial by divisibility.
• So the remaining ratio must be polynomial, but since \(r<\ell\) is strict this forces \(r=0\). Thus \(\ell \divides m\).

\(\impliedby\):

• Write \(m = \ell q + r\), then \(r=0\) by divisibility.
• Then \(x^m-1 = x^{\ell q} - 1 \da z^q-1\) where \(z\da x^\ell\).
• Use that \(z-1 \divides z^q - 1\), so \(x^{\ell}-1 \divides x^{\ell q} -1 = x^m-1\).