
Divisibility problem
Let n $\displaystyle \geq$ 2 and k be any positive integer. Prove that $\displaystyle (n1)^{2} \mid n^{k}1$ if and only if $\displaystyle (n1) \mid k$. I've tried considering $\displaystyle (n1) \mid (n^{k}1)/(n1)$ 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.

Re: Divisibility problem
Use the factorization $\displaystyle n^k1=(n1()n^{k1}+n+n^{k2}+...+n+1)$
So it reduces to $\displaystyle (n1)(n^{k1}+n^{k2}+...+n+1)$
then use the fact that $\displaystyle n\equiv{1}$ $\displaystyle (mod$ $\displaystyle n1)$
P.S. Remember the multiplication of congruences

Re: Divisibility problem
you can factor $\displaystyle (n^k1)$ into $\displaystyle (n1)(1+n+n^2+\ldots +n^{k1})$