$M_3=M$, and therefore $$M_1 = \lambda^{2(n+1)} M_2 = \lambda^{2(n+1)-2} M_3 = \lambda^{2n} M.$$ Hence $M$ is either 0 or homogeneous of homogeneity degree $2n$. To see that $M \ne 0$, it is easy to see that, and it follows from Lemma 3.3 of [@impurity], the determinant obtained from $M$ by plugging $x_{i,j}=1$ for every $x_{i,j}$ is $(-1)^{n-1} (n+1) \ne 0$. This completes the proof of (i).

\(ii) The case $n=2$, i.e., the identity (\[Heron-M\]), is easy to check. So we assume that $$n\ge 3.$$

Let $V_0$, $V_1$, and $V$ be the sets of indeterminates defined by $$\begin{aligned}
V_1 = \{ x_{i,j} : 1 \le i < j \le n\},~V_0 = \{ x_{n+1,j} : 1 \le j \le n\},~
V = V_0 \cup V_1, \label{VV0V1}\end{aligned}$$ and let $R_0$ and $R$ be the polynomial rings $$\begin{aligned}
R_0 = k[V_0],~R = k[V]. \label{R0R}\end{aligned}$$ For any non-zero $\phi \in R$, let the total degree of $\phi$ be denoted by $\deg (\phi)$, and let the degree of $\phi$ in the indeterminates $V_0$ be denoted by $\deg_0 (\phi)$. Since the indeterminates $V_0$ appear only in the last row and last column of $M$, it follows that $$\begin{aligned}
\deg_0 (M) &\le& 4. \label{delta}\end{aligned}$$

Now suppose that $M$ 