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I'm working through the problems in Niven's number theory book, and problem 46 in section 1.2 (page 19) has me stumped.

Prove that there are no positive integers $a, b, n > 1$ such that $(a^n - b^n) | (a^n + b^n)$.

I've tried playing around a bit (e.g. noticing that $a^n - b^n$ must divide $2a^n$ and $2b^n$), but in general I've just been going around in circles. Can anyone please provide a (small) hint in the right direction?

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As you said, it would imply that there is an integer $k$ such that $ka^n - kb^n = 2b^n$. Then, $ka^n = (2+k)b^n$. If $a$ and $b$ have a common factor, we can divide both sides by the nth power of it to get the same equality with $a$ and $b$ relatively prime.

The same way, if $k$ is even, we can divide both sides by $2$, to get $ka^n = (k+1)b^n$. In this case, $k$ and $k+1$ are relatively prime, so $a^n = k+1$ and $b^n = k$. But this cannot happen because consecutive numbers are not nth power of some integer.

If $k$ is not even, then $k$ and $k+2$ are relatively prime. Again, $ka^n = (k+2)b^n$, then implies that $a^n = k+2$ and $b^n = k$. But this cannot happen either.

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  • $\begingroup$ In this case, $a^n-b^n=1$ and $a^n-b^n=2$ cannot happen because $a,b,n>1$. Notice $1^{2k+1}-(-1)^{2k+1}=2$, $k\in\mathbb Z$, $\left|1^n-0^n\right|=\left|0^n-(-1)^n\right|=1$, $n\in\mathbb Z$. $\endgroup$ – user236182 Aug 24 '17 at 12:57
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  1. Prove that it is enough to search for $a$, $b$ relatively prime.

  2. Use the useful fact that you have already noticed.

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We can assume $a,b$ is coprime. Suppose we do have the relationship, then we have $(m-1)a^{n}=(m+1)b^{n}$. This would imply $\frac{m+1}{m-1}=(\frac{a}{b})^{n}$.

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  • $\begingroup$ How does this produce a contradiction? $\endgroup$ – mysatellite Jul 20 '16 at 17:40
  • $\begingroup$ There no 2 such numbers who have difference 2 between them when powered greter than 2 or equal to. $\endgroup$ – Lalla95 Feb 14 '19 at 16:11

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