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Let $n > 1$ be an integer. Then $2^n - 1\nmid 3^n - 1$. I don't know how to prove it. Can anybody help me, please?

In general, for a fixed positive integer $a > 1$, has $a^n - 1|(a +1)^n - 1$ any integer solutions?

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    Have you tried using the standard factorization: $a^n-b^n=(a-b)(a^{n-1}+a^{n-2}b+..+a^{n-k}b^{k-1}+...+b^{k-1})$?2012-03-06
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    Maybe you could work in the case n=2k+1, since , when n is even, $2^n -1$ is divisible by 3, and $3^n-1$ is clearly not.2012-03-06
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    @ksjo3: Please tell us the source of the problem. Did you get it somewhere? Do you know there is a solution? Did you make it up yourself?2012-03-06
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    It is an exercise in a book. I don't konw how to prove it.2012-03-07
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    @ksjo3 Please tell us which book, and where in that book. This could help people know what sort of techniques you are expected to apply to the problem.2012-03-07
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    It is an exercise in the chapter on quadratic residues. The name of the book is Lecture Notes on Number Theory by Ke Chao and Sun Qi. I am not sure if you can find it.2012-03-08
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    If $2^n - 1|3^n - 1$, then we have $\frac{3^n - 1}{2^n - 1} = [(\frac{3}{2})^n] + 1$. But I don't know what to do next.2012-03-08

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