<s>[INST] Number theory: $a\in A\iff \frac{1}{2}-a\in A$. Let $A$ be the set of all $a\in \mathbb{Q}$ for which there exist $x,y,z\in \mathbb{Z}$ not all $=0$ such that $$a=\frac{xy+yz+zx}{x^2+y^2+z^2}.$$ Prove that $$a\in A\iff \frac{1}{2}-a\in A.$$
Note: This is a special case of a more general result I discovered while investigating Diophantines of the form $a(x^2+y^2+z^2)=b(xy+yz+zx)$ and already have a solution using methods I invented for that purpose, but this is indirect/unmotivated. I am looking for a more direct solution.
It might help to know the "more general result:"
Proposition. If $b\in A$ then $a\in A\iff \frac{b-a}{1+a-2ab}\in A.$ 
However, none of these look as nice as the case where $b=\frac{1}{2}=\frac{1\cdot 1+1\cdot 4+4\cdot 1}{1^2+1^2+4^2}$, which gives rise to the above problem: $a\in A\iff \frac{\frac{1}{2}-a}{1+a-a}=\frac{1}{2}-a\in A.$
 [/INST] Let
$$
a = \frac{xy+yz+zx}{x^2+y^2+z^2}
$$
as above, note that this is equivalent to 
$$ 
1+2a = \frac{(x+y+z)^2}{x^2+y^2+z^2}
$$ and let $a'=\frac12-a$.
If $x=y=z$ then $a=1$ and
$$
a' = -\frac12 = \frac{-1\cdot 2 -1\cdot 2 + 1}{(-1)^2+(-1)^2+2^2} \in A
$$
Otherwise, if $x,y,z$ are not all equal then
$$
1+2a' = 2-2a = \frac{u+v+w}{x^2+y^2+z^2} = \frac{(u+v+w)^2}{u^2+v^2+w^2}
$$
where
$$
u = x^2+y^2-z(x+y) \\
v = y^2+z^2-x(y+z) \\
w = z^2+x^2-y(z+x) \\
u+v+w = (x-y)^2+(y-z)^2+(z-x)^2>0
$$
clearly $u,v,w$ are not all zero and hence
$$
a' = \frac{uv+vw+wu}{u^2+v^2+w^2} \in A
$$
Thus $a\in A \implies \frac12-a \in A$, and the other direction follows immediately because $a'\in A \implies \frac12-a' = a \in A$.
</s>