Are there infinitely many natural numbers that can be represented as $\frac{a^2+b^2}{c^2+d^2}$ with coprime $a$, $b$, $c$, $d$? 
Are there infinitely many natural numbers that can be represented as the ratio
$$
\frac{a^2 + b^2}{c^2+d^2}
$$
where $a$, $b$, $c$, $d$ are coprime integers?

Is there a way to characterize all such numbers?

Suppose the number $N > 1$ is such that it's prime decomposition has no term of the form $p^k$ with $k$ odd and $p \equiv 3 (4)$. Then it can be written as $\frac{a^2 + b^2}{1^2 + 1^2}$. This doesn't mean $a$ and $b$ are coprime though.
Now suppose the prime factorization of $N = p_1^{k_1} \cdots p_r^{k_r}$ with some being congruent to 3 modulo 4 and having odd powers. Then if we were to write $N = \frac{M}{O}$ then $M = p_1^{k_1 + \ell_1} \cdots p_r^{k_r + \ell_r} p_{r+1}^{\ell_{r+1}} \cdots p_s^{\ell_s}$ and $O = p_1^{\ell_1} \cdots p_r^{\ell_r} p_{r+1}^{\ell_{r+1}} \cdots p_s^{\ell_s}$ where we are free to choose the $\ell$'s among natural numbers as long as they are not all $0$. Then $M$ and $O$ are integers greater than $1$ that we want to write as sum of two squares each.
Suppose $p_j$ was such an offending prime. Then we know that $k_j$ was odd. But now we have it in $k_j + \ell_j$ and $\ell_j$ as it's powers in $M$ and $O$. If we use $\ell_j$ as odd, then $O$ can't be written as a sum of two squares. If we use $\ell_j$ as even, then $M$ can't be written as a sum of two squares because of the factor $p_j^{k_j+\ell_j}$ has an odd power.
So even by allowing the seemingly more general form of $\frac{a^2 + b^2}{c^2 + d^2}$ you have not gained any more natural numbers than can be written besides those that were already of the form $a^2 + b^2$.
