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Your conclusion is valid, and for this we only need to assume that $\varphi(3n)$ is divisible by $3$. Indeed, this weaker condition is equivalent to the existence of a prime divisor $p\mid n$ that is either $3$ or congruent to $1$ modulo $3$. It is known that $p$ is the norm of an Eulerian integer $z=x+y\frac{1+i\sqrt{3}}{2}$. Multiplying $z$ with a sixth root of unity, we can achieve that $0<\arg(z)<\pi/3$. Then, $x$ and $y$ are positive integers satisfying $p=x^2+xy+y^2$, and so $n$ is three times the Heronian mean of $(n/p)x^2$ and $(n/p)y^2$.

Your conclusion is valid, and for this we only need to assume that $\varphi(3n)$ is divisible by $3$. Indeed, this weaker condition is equivalent to the existence of a prime divisor $p\mid n$ that is either $3$ or congruent to $1$ modulo $3$. It is known that $p$ is the norm of some Eulerian integer $z=x+y\frac{1+i\sqrt{3}}{2}$. Multiplying $z$ with a sixth root of unity, we can achieve that $0<\arg(z)<\pi/3$. Then, $x$ and $y$ are positive integers satisfying $p=x^2+xy+y^2$, and hence $n$ is three times the Heronian mean of $(n/p)x^2$ and $(n/p)y^2$.

Your conclusion is valid, and for this one only needs to assume that $\varphi(3n)$ is divisible by $3$. Indeed, this condition is equivalent to $n$ having a prime divisor $p$ that is either $3$ or congruent to $1$ modulo $3$. It is known that $p$ is the norm of an Eulerian integer $x+y\frac{1+\sqrt{-3}}{2}$, and multiplying this Eulerian integer with a unit, we can achieve that it has argument lying in $(0,\pi/3)$. Then, $p=x^2+xy+y^2$, where $x$ and $y$ are positive integers, and hence $n$ is three times the Heronian mean of $(n/p)x^2$ and $(n/p)y^2$. Done.

Your conclusion is valid, and for this we only need to assume that $\varphi(3n)$ is divisible by $3$. Indeed, this weaker condition is equivalent to the existence of a prime divisor $p\mid n$ that is either $3$ or congruent to $1$ modulo $3$. It is known that $p$ is the norm of an Eulerian integer $z=x+y\frac{1+i\sqrt{3}}{2}$. Multiplying $z$ with a sixth root of unity, we can achieve that $0<\arg(z)<\pi/3$. Then, $x$ and $y$ are positive integers satisfying $p=x^2+xy+y^2$, and so $n$ is three times the Heronian mean of $(n/p)x^2$ and $(n/p)y^2$.

Your conjecture is true, and for this one only needs to know that $\varphi(3n)$ is divisible by $3$. Indeed, this condition is equivalent to $n$ having a prime divisor $p$ that is either $3$ or congruent to $1$ modulo $3$. It is known that $p$ is the norm of an Eulerian integer $x+y\frac{1+\sqrt{-3}}{2}$. Without loss of generality, this Eulerian integer has argument lying in $(0,\pi/3)$. Then, $p=x^2+xy+y^2$, where $x$ and $y$ are positive integers, and hence $n$ is three times the Heronian mean of $(n/p)x^2$ and $(n/p)y^2$. Done.

Your conclusion is valid, and for this one only needs to assume that $\varphi(3n)$ is divisible by $3$. Indeed, this condition is equivalent to $n$ having a prime divisor $p$ that is either $3$ or congruent to $1$ modulo $3$. It is known that $p$ is the norm of an Eulerian integer $x+y\frac{1+\sqrt{-3}}{2}$, and multiplying this Eulerian integer with a unit, we can achieve that it has argument lying in $(0,\pi/3)$. Then, $p=x^2+xy+y^2$, where $x$ and $y$ are positive integers, and hence $n$ is three times the Heronian mean of $(n/p)x^2$ and $(n/p)y^2$. Done.