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Show that $(n-1)^2\vert n^k-1$ if and only if $n-1\vert k$, for naturals $n,k$.

I couldn't go much further, the only thing that i could conclude is that $n-1\vert n^k-1$, since it is true that $a-b\vert a^k-b^k$ for all natural $k$. Now i can't proceed from here.

Any hints on this question?

user2345678
  • 3,015

1 Answers1

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$$\frac{n^k-1}{n-1}=n^{k-1}+n^{k-2}+\cdots+n+1\equiv k\pmod{n-1}$$

Angina Seng
  • 161,540