Identity involving product of the $\zeta$ function for different values

Question

I would like to prove the identity $$\sum_{\substack{b,d>0 \\ (b,d)=1}}\frac{1}{b^n}\frac{1}{d^m}=\frac{\zeta(n)\zeta(m)}{\zeta(m+n)},$$ where $\zeta$ is the Riemann zeta function and $n,m\ge 2$. Any help would be welcome.

Answer 1

Every pair $(r,s)$ of positive integers has the form $(tb,td)$ where $t=\gcd(r,s)$ and $\gcd(b,d)=1$. Therefore $$\zeta(m)\zeta(n)=\sum_{r,s>0}\frac1{r^n}\frac1{s^m} =\sum_{b,d>0\atop\gcd(b,d)=1}\sum_{t=1}^\infty\frac1{(tb)^n}\frac1{(td)^m} =\zeta(m+n)\sum_{b,d>0\atop\gcd(b,d)=1}\frac1{b^n}\frac1{d^m}.$$