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I need this for a proof of Wedderburn's theorem: $$q^m - 1 | q^n - 1 \quad \Rightarrow \quad m|n$$ with $q>1 \in \mathbf{N}$ and $m,n \in \mathbf{N}$.

I'd also like to know if it works the other way around: $$m|n \quad \Rightarrow \quad q^m - 1 | q^n - 1.$$

Thanks.

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    The other direction follows easily from the identity $(x^N - 1) = (X - 1)(X^{N - 1} + X^{N - 2} + \cdots + 1)$.2012-06-10

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