When can $a^m-b^m$ divide $a^m+b^m$. where $a$, $b$, $m$ are natural numbers, $a\gt b$

When can $(a^m-b^m)$ divide $(a^m+b^m)$, where $a$, $b$, $m$ are natural numbers, $a \gt b$.

I approached this way:

Let $(a,b)=d$ and $\frac{a}{A}=\frac{b}{B}=d$, so $(A,B)=1$ and $A>B$.

$(a^m-b^m)$ will divide $(a^m+b^m)$

iff $(A^m-B^m)$ divides $(A^m+B^m)$.

If $D|(A^m-B^m)$ and $D|(A^m+B^m)$ where $D$ is natural number greater than $1$, then $D| (A^m-B^m) \pm (A^m+B^m)$, \begin{align*} &\implies D|2A^m\text{ and }D|2B^m\\ &\implies D|(2A^m, 2B^m)\\ &\implies D|2(A^m, B^m)\\ &\implies D|2(A, B)^m\\ &\implies D|2\\ &\implies D=2\text{ as }D>1. \end{align*}

So in that case, the only divisor>1, that $(A^m-B^m)$ admits is 2 => $(A^m-B^m)= 2$

Now, we observe $2^2-1=3>2$

So, $(A^m-B^m)>2$ if m>1.

So, $(A^m-B^m) | (A^m+B^m)$ iff $A=B+2$ and $m=1$.

Is it ok?

Any other approach is also welcome.

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this is trivial because if $A=B+2$ then it means that $A+B=2*n$ for some natural $n$ and $A-B=2$ –  dato datuashvili Jul 22 '12 at 9:27
@dato, what about the non-trivial cases, have I missed anything? –  lab bhattacharjee Jul 22 '12 at 14:42
It looks good, except you've missed $A=B+1$, $m=1$. –  Gerry Myerson Jul 22 '12 at 23:43
@GerryMyerson, how to mark this as accepted as I don't think I'll get a better answer. –  lab bhattacharjee Jul 24 '12 at 5:14