Exercise 1.2.32 ($n- 1 \mid n^k - 1 \iff n - 1 \mid k.$)

Question

Let $n\geq 2$ and $k$ be any positive integers. Prove that $(n-1)^2 \mid (n^k - 1)$ if and only if $(n-1) \mid k$.

Richard Ganaye · Accepted Answer

\begin{proof} Since $n \equiv 1 \pmod {n-1}$,
\begin{align*}
(n- 1)^2 \mid n^k - 1 & \iff n- 1 \mid \frac{n^k - 1}{n-1} = \sum_{j=0}^{k-1} n^j\
&\iff \sum_{j=0}^{k-1} n^j\equiv 0 \pmod {n-1}\
&\iff k \equiv  0 \pmod {n-1}\
&\iff n - 1 \mid k.
\end{align*}
So
$$(n- 1)^2 \mid n^k - 1 \iff  n - 1 \mid k.$$
(If we want to write this proof without congruences, we replace $n^j$ bt $(1 + (n-1))^j$ in the sum.)
\end{proof}

Exercise 1.2.32 ($n- 1 \mid n^k - 1 \iff n - 1 \mid k.$)

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Exercise 1.2.32 ($n- 1 \mid n^k - 1 \iff n - 1 \mid k.$)

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