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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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    @GerryMyerson, how to mark this as accepted as I don't think I'll get a better answer.2012-07-24

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