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Math Help - Divisibility problem

  1. #1
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    Divisibility problem

    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.
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  2. #2
    Senior Member I-Think's Avatar
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    Re: Divisibility problem

    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
    Last edited by I-Think; December 8th 2011 at 07:42 PM. Reason: typos
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  3. #3
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    Re: Divisibility problem

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