Problem 1493

Let $z = a + b - c$, $x = b + c - a$, $y = c + a - b$ be the (positive) denominators. Then $a = (y + z)/2$, $b = (x + z)/2$, $c = (x + y)/2$ and
$$a^2 + b^2 - c^2 = \frac{1}{4}((y + z)^2 + (x + z)^2 - (x + y)^2) = \frac{1}{2}(z(z + x + y) - x y),$$
hence $z \mid x y$ and likewise $y \mid x z$ and $x \mid y z$.

For a prime $p$, let $i_p$ be the largest exponent such that $p^{i_p} \mid x y z$. It suffices to show that for all odd primes $p$ the corresponding $i_p$ is even. If $i_2$ is also even then $x y z$ is a square. Otherwise, it is twice a square.
Fix odd prime $p$ and consider the largest exponents $\alpha, \beta, \gamma$ such that $p^\alpha \mid x$, $p^\beta \mid y$, $p^\gamma \mid z$. Without loss of generality, assume $\min\{\alpha, \beta, \gamma\} = \gamma$. If $\gamma > 0$ then $p$ divides each of $x, y, z$ and thus it divides each of $a, b, c$ ($p$ is odd), contradicting $\text{GCD}(a, b, c) = 1$. Therefore $\gamma = 0$.
From $x \mid y z$ we infer $\alpha \le \beta$. Likewise, from $y \mid x z$ we infer $\beta \le \alpha$. Hence $\beta = \alpha$ and $i_p = \alpha + \beta + \gamma = 2\alpha$ is an even number as desired.