I was trying to see that if the sums like the following is calculable
$$\sum_{(x, y)=1} f(x,y)$$
where $x>y$ are positive integers with different parity. How can we prove that the sum is divergent or convergent? Like the following
$$\sum_{(x, y)=1} \frac{x^2-y^2}{x^2+y^2} $$
Actually, what I'm looking for is to see that if the sum is convergent, how can we calculate it? Like
$$\sum \frac{x^2}{x^4+y^4}$$
Hint:
Since $\gcd(n,1)=1$ for any positive integer $n$, the sum contains $\sum\limits_{k=1}^\infty\dfrac{k^2-1}{k^2+1}$
Also, we have $\lim\limits_{k\to\infty}\dfrac{k^2-1}{k^2+1}=\lim\limits_{k\to\infty}\dfrac{1-1/k^2}{1+1/k^2}=1\neq 0$
One could also argue that since we're summing over all coprime tuples of $\Bbb Z^2$, for $\gcd(x,y)=1$, the term $\dfrac{x^2-y^2}{x^2+y^2}$ is cancelled out by the term $\dfrac{y^2-x^2}{x^2+y^2}$ which is from the tuple $(y,x)$ since $\gcd(x,y)=\gcd(y,x)$, so the overall sum is $0$.
Hence, my original answer assumes (wlog?) that $x\gt y$ (i.e., the tuple $(x,y)$ is assumed to be the same as $(y,x)$, so we only take the tuple $(x,y)$ in the sum) to show divergence.