The Riemann Zeta function $\zeta(s)$ for complex numbers whose real part is greater than $1$ is given by $$ \zeta(s) = \sum_{k=0}^{\infty} \frac{1}{k^s}$$ I came across an interesting problem : Show that $$ {\sum_{m=1}^{\infty} \sum_{n=1}^{\infty}}_{gcd(m,n)=1}\frac{1}{m^2n^2} = \frac{\zeta^2(2)}{\zeta(4)}$$
I know that if $a(1)=0$ and $a(n)=k$ when the prime decomposition of $n$ consists $k$ distict prime factors then $$ \sum_{n=0}^{\infty}\frac{2^{a(n)}}{n^s} = \frac{\zeta^2(s)}{\zeta(2s)} $$ But many attempts of mine were futile to conquer the above problem. Is there any way to do that?