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
$\textbf{Hint:}$Assume such positive integers exist.Then,
There exists $n$ such that $a^2+b^2+c^2=3n(ab+bc+ca)$.
Now,add $2(ab+bc+ca)$ on both sides.That would give $(a+b+c)^2=(3n+2)(ab+bc+ca)$
Now,$3n+2$ has a prime divisor say $p$ appearing odd number of times of the form $3k+2$ for some $k$.From here deduce that
$p \mid a+b+c$ and $p \mid ab+bc+ca$.
Deduce from here that, $p \mid a^2+ab+b^2$
After that use quadratic reciprocity to derive some contradiction.
Given integers $B > A > 0,$ with $\gcd(A,B) = 1,$ then there is a solution in integers $x,y,z,$ not all $0,$ to $$ A(x^2 + y^2 + z^2) - B (yz + zx + xy) = 0, $$ if and only if both $B+2A$ and $B-A$ are integrally represented by the binary form $u^2 + 3 v^2.$
In your case $A = 1$ and $B = 3k$ for some integer $k$
My $B+2A = 3k+2$ and cannot be represented by $u^2 + 3 v^2,$ so that is it since your vriables are restricted to positive.
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.
