If $\frac{a^n-1}{b^n-1} $ is a natural number for every $n$ then $a = b^m$

Asked Sep 08 '16 at 18:44

Active Sep 08 '16 at 19:26

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Consider two natural numbers $a,b$. Prove that $a = b^m$ if $ \frac{a^n-1}{b^n-1} $ is natural for all $n \in N$.

I tried to assume that there is exist some $k $, such : $a = b^m + k$, so i get $$\displaystyle \frac{\sum_{0}^{n}b^{n-i+m}k^i -1}{b^n-1}$$ but I guess I just waste my time.

Need some good idea.

edited Sep 08 '16 at 19:26

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asked Sep 08 '16 at 18:44

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@AlexM. my bad , sorry – openspace Sep 08 '16 at 18:55
I found one solution: a=8, b=2, n=1, and m=3. – Jeffrey Young Sep 08 '16 at 19:03
Ok. So we have to prove (or disprove) that a has to be a power of b when one less than a power of a is divisible by one less than the same power of b. – Jeffrey Young Sep 08 '16 at 19:18
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@JeffreyYoung Yes, it's very well know that $\frac{c^n - 1}{c-1} = (c^{n-1} +.....+ 1)$ so if replace $c$ with $b^n$ and $a$ with $b^m$ this result will always be true for all $n$. The question is is this the only time it is true for all $n$. – fleablood Sep 08 '16 at 20:03
See first answer which is answer to this question. – simonzack Sep 08 '16 at 20:49

If $\frac{a^n-1}{b^n-1} $ is a natural number for every $n$ then $a = b^m$

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