<s>[INST] Olympiads Number Theory Problem: Show that there are no positive $a,b,c\in\Bbb Z$ such that $\frac{a^2+b^2+c^2}{3(ab+bc+ca)}\in\Bbb Z$ I have been struggling to solve this problem, could someone give me an idea/solution? please!

Show that there are no positive $a,b,c\in\Bbb Z$ such that $$\frac{a^2+b^2+c^2}{3(ab+bc+ca)}\in\Bbb Z$$

Edit source image: https://i.stack.imgur.com/mk9cl.png
 [/INST] I thought that I had a solution using Vieta Jumping but this was not a complete solution so we can use quadratic residues to solve it...
We can choose WLOG $\text{gcd} (a,b,c)=1$.
After letting $\frac{a^2+b^2+c^2}{ab+bc+ca}=3k$, we have $(a+b+c)^2=(3k+2)(ab+bc+ca)$.
Choose a prime $p \equiv 2 \pmod{3}$ with $v_p(3k+2) \equiv 1 \pmod{2}$
(clearly there exists one since $3k+2$ is not a square).
Then $p|a+b+c$, and we have $p|ab+bc+ca$ using our condition on $v_p$.
$\implies c \equiv -a-b \pmod{p}$, so $p|ab+(a+b)(-a-b)$, so $p|a^2+ab+b^2$.
If $p=2$, we check that $a$ and $b$ must be even.
Otherwise, we have $p \equiv 5 \pmod{6}$, and $\left( \frac{-3}{p} \right)=-1$.
This gives us $p|a,b,c$, contradiction.
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