Divisibility problem

**wsldam** · December 8th 2011, 04:23 PM

Let n $\geq$ 2 and k be any positive integer. Prove that $(n-1)^{2} \mid n^{k}-1$ if and only if $(n-1) \mid k$ . I've tried considering $(n-1) \mid (n^{k}-1)/(n-1)$ but it still got me no where.

I'm looking more for a hint than an actual answer as I do want to figure this out mostly on my own.

**I-Think** · December 8th 2011, 06:20 PM

Use the factorization $n^k-1=(n-1()n^{k-1}+n+n^{k-2}+...+n+1)$
So it reduces to $(n-1)|(n^{k-1}+n^{k-2}+...+n+1)$

then use the fact that $n\equiv{1}$ $(mod$ $n-1)$
P.S. Remember the multiplication of congruences

**MonroeYoder** · December 8th 2011, 06:24 PM

you can factor $(n^k-1)$ into $(n-1)(1+n+n^2+\ldots +n^{k-1})$

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