Prove that $a^n - b^n$ does not divide $a^n + b^n$ 
Prove that $$a^n - b^n \text{ does not divide } a^n + b^n \text { and } a,b,n \in \mathbb{Z}^+. n > 1$$

I have tried to prove this but have had no success. My efforts till now were concerned with showing that $a-b \mid a^n-b^n$ while it does not divide $a^n+b^n$.
I would prefer a number theoretic answer. Thanks.
Source: Introduction to Theory of Numbers
 A: Actually the answers in Why does $a^n - b^n$ never divide $a^n + b^n$? say it much more clearly than I do.
====
Wolog we can assume $a$ and $b$ are relatively prime.  If $a = ka'$, $b= kb'$ and $gcd(a',b') = 1$ then $a^n - b^n|a^n+b^n \iff k^n(a'^n - b^n)|k^n(a'^n + b'^n) \iff a'^n - b'^n|a'^n + b'^n$.
Wolog we can assume $a > b$.  If $a = b$ we get $0|a^n + b^n$ which only is possible if $a^n + b^n = 0$.
So if $a^n - b^n | a^n + b^n$ then $a^n - b^n|(a^n + b^n)\pm(a^n - b^n)$ and $a^n- b^n|2a^n$ and $a^n-b^n|2b^n$.
So $a^n - b^n|\gcd(2a^n, 2b^n) =2$.  So $a^n - b^n = 1 or 2$.
Let $k = a - b > 0$.  Then $a^n - b^n = (b + k)^n - b^n= \sum_{i=0}^n {n \choose i}b^ik^{n-k} - b^n = \sum_{i=0}^{n-1} {n \choose i}b^ik^{n-k}$.  If $n \ge 2$ and $b \ge 1$ then $\sum_{i=0}^{n-1} {n \choose i}b^ik^{n-k} \ge  {n \choose 1}bk^{n-1} + k^n \ge nb + 1 \ge 3$.
So $a^n - b^n \ge 3$ and $a^n - b^n =1$ or $2$ so we have a contradiction.  This is impossible under the conditions $a,b$ are positive and $n > 1$.
(Several trivial solutions for $n=1$ and $a=2$, $b=1$ or $a$ or $b$ equal zero or $a = -b$ etc. do exist.)
