Seeking explanation for a curious reciprocity-like identity.

Question

I just stumbled, by pure accident, on this: $$ \gcd \left(\frac{(m+1)^n-1}{m},m\right)=\gcd \left(\frac{(n+1)^m-1}{n},n\right) $$ It probably has a straightforward proof but more than a proof I want a conceptual background for it. Is it related to any known reciprocity laws or something?

Answer 1

Note that $$(m+1)^n=1+nm+\frac{n(n-1)}2m^2+\ldots $$ where everything hidden in the dots is also a multiple of $m^2$. Hence $\frac{(m+1)^n-1}{m}$ is just $n$ plus a multiple o $m$, which can be ignored when taking the $\gcd$ with $m$. In other words, the left hand side is just$\gcd(n,m)$ (and the right hand side $\gcd(m,n)$).