67th Czech and Slovak Mathematical Olympiad

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 $$a^2+b^2-c^2=\frac{1}{4}((y+z{)}^2+(x+z{)}^2-(x+y{)}^2)=\frac{1}{2}(z(z+x+y)-xy),$$ hence $z\mid xy$ and likewise $y\mid xz$ and $x\mid yz$.

For a prime $p$, let $i_p$ be the largest exponent such that $p^{i_p}\mid xyz$. It suffices to show that for all odd primes $p$ the corresponding $i_p$ is even. If $i_2$ is also even then $xyz$ 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 $x\mid yz$ we infer $\alpha \le \beta$. Likewise, from $y\mid xz$ we infer $\beta \le \alpha$. Hence $\beta =\alpha$ and $i_p=\alpha +\beta +\gamma =2\alpha$ is an even number as desired.