Regarding a number theory proof Before I begin I should mention that I have only recently started studying number theory.
It is required to prove that there are no positive integers  $a,b,n>1$ such that
$$a^n - b^n\mid a^n+b^n$$ 
that is $k$ can never be an integer if 
$$\frac{a^n+b^n}{a^n-b^n} = k$$
Now if $b|a$ then we can write 
$$\frac{(a/b)^n+1}{(a/b)^n-1} = k$$
which is clearly not true for all $a/b$. However I have no idea how to prove the general result. Help will be appreciated.
 A: Set $\gcd(a,b)=d$, $a=da_1$ and $b=db_1$, then $\gcd(a_1,b_1)=1$. We proceed by proof by contradiction. If $a^n-b^n\mid a^n+b^n$, then $a_1^n-b_1^n\mid a_1^n+b_1^n$, so $a_1^n-b_1^n\mid2a_1^n$ and $a_1^n-b_1^n\mid2b_1^n$, leading to $a_1^n-b_1^n\mid\gcd(2a_1^n,2b_1^n)=2$. Hence $a_1^n-b_1^n\mid2$.
Without loss of generality we assume $a>b$, and so $a_1>b_1$. Since $a_1^n-b_1^n\mid2$ we have $a_1^n-b_1^n=1$ or $2$, but $$a_1^n-b_1^n=(a_1-b_1)(a_1^{n-1}b_1+a_1^{n-2}b_1^2+\cdots+a_1b_1^{n-1})>n\geq2,$$ which is a contradiction!
A: Notice if $a^n - b^n|a^n + b^n$ then $a^n - b^n|a^n + b^n + a^n - b^n = 2a^n$ and $a^n - b^n| a^n + b^n -(a^n- b^n) = 2b^n$.
So if $p|a^n - b^n$ and $p$ is prime either $p = 2$ or $p|a$ and $p|b$.
Let $d = \gcd (a,b)$ and let $a = a'd; b = b'd$ then $\frac{a^n - b^n}{a^n + b^n}= \frac {a'^n - b'^n}{a'^n + b'^n}$ and $a^n - b^n| a^n + b^n \iff a'^n - b'^n|a'^n + b'^n$.
So $a'^n - b'^n|2a'^n$ and $a'^n- b'^n|2b'^n$ and for any prime $p \ne 2$ if $p|a'^n - b'^n$ then $p|a'$ and $p|b'$ so $p|\gcd(a'b') =1$ which is impossible.  So no prime other than $2$ divides $a'^n - b'^n$.  
So $a'^n - b'^n = 2^m$.  But only $2$ can't divide both $a'$ and $b'$ so $a'^n - b'^n = 2$ or $a'^n - b'^n = 1$.  
So $(a'^n - b'^n) = (a'-b')(a'^{n-1}+ a'^{n-2}b'+ ... + a'b'^{n-2} + b'^{n-1})$.
Clearly $a'-b' < (a'^{n-1}+ a'^{n-2}b'+ ... + a'b'^{n-2} + b'^{n-1})$ and $(a'^{n-1}+ a'^{n-2}b'+ ... + a'b'^{n-2} + b'^{n-1}) > 0$ so 
$a -b = 1$ and $(a'^{n-1}+ a'^{n-2}b'+ ... + a'b'^{n-2} + b'^{n-1})= 2$
$(a'^{n-1}+ a'^{n-2}b'+ ... + a'b'^{n-2} + b'^{n-1}) = 2$ is only possible if $n= 1$.  And indeed if $a-b =1$ then $1|a+b$ but... $n$ must be bigger than $1$ and that is impossible.
