Diophantine: $(3n-3)(x^2+y^2+z^2)=(9n+1)(xy+yz+zx)$ For which positive integer values of $n$ does the following diophantine equation have no nontrivial solutions? :
$(3n-3)(x^2+y^2+z^2)=(9n+1)(xy+yz+zx)$
Note: I discovered this problem while trying to solve another one. I have attempted this and I know it is true for e.g. $n=3$, but it is hard to find all such $n$'s.

The original equation is equivalent to:
$$(15n-5)(x^2+y^2+z^2)=(9n+1)(x+y+z)^2,$$
so, if $n\not\equiv 1,2\pmod{5}$, both $x+y+z$ and $x^2+y^2+z^2$ must be multiples of $5$. Since the quadratic residues $\!\!\pmod{5}$ are $\{-1,0,1\}$, we can assume without loss of generality that $x\equiv 0\pmod{5}$. However, this implies $y\equiv -z\pmod{5}$ and $y^2\equiv -z^2\pmod{5}$, so $x\equiv y\equiv z\equiv 0\pmod{5}$ leads to an infinite descent.
This proves that for $n\not\equiv 1,2\pmod{5}$ there are no non-trivial solutions.
