Prove that $a^n - b^n$ does not divide $a^n + b^n$ [duplicate]

Solution 1:

Actually the answers in Why does $a^n - b^n$ never divide $a^n + b^n$? say it much more clearly than I do.

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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.)