# Why does $a^n - b^n$ never divide $a^n + b^n$?

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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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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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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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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How does this produce a contradiction? – Sky Jul 20 at 17:40