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

Question

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?

Answer 1

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 $a^2 + b^2$.

Answer 2

Excuse my late reaction, but I think the answer is already contained in the so called "Fermat Xmas theorem", which states that an odd prime $p$ is the sum of two (integral) squares iff $p \equiv 1$ mod $4$ (*). A consequence (due to Euler ?) is that a strictly positive integer is a sum of two squares iff in its prime decomposition, every factor $p\equiv 3$ intervenes with an even exponent. Consider then an integer $N$ equal to a fraction whose numerator and denominator are both sums of two squares. The integrality of $N$ implies that each prime divisor of the denominator divides also the numerator, and with a lesser exponent. So that finally $N$ is already a sum of two squares.

(*) There exist quite a few different proofs of this Xmas theorem, but the record of brievity belongs to Don Zagier:« A one-sentence proof that every prime $p\equiv 1$(mod 4) is a sum of two squares », The American Mathematical Monthly, vol. 92, no 2,‎ 1990, p. 144.

Answer 3

Observation: if n is a prime of the form of the form 4n + 1, not only is it the sum of two relatively prime squares, so is 2n. So make your denominator 1 + 1 and you're done. The problem is more fun if you exclude this trivial case.