I found a strange identity in a system of polynomial equations I am studying, and I do not know how to verify it. The system has $n+3\binom{n}{2}$ unknowns, represented by the variables $(\{x_j\}_{j=1}^n, \{p_{ij}\}_{i<j}, \{\theta_{ij}\}_{i<j}, \{s_{ij}\}_{i<j})$. The $x$ variables represent "positions", the $p$ variables represent "directions", the $\theta$ variables are "strengths", and the $s$ variables are "lengths".
There are $n+\binom{n}{2}$ equations. Each vertex $i$ has an equation representing a "force balance condition" $$ \sum_{j<i} \theta_{ji}p_{ji} = \sum_{j>i} \theta_{ij} p_{ij}, $$ and each edge $(i,j)$ with $i<j$ comes with a "compatibility condition" $$ x_j = x_i + s_{ij} p_{ij}. $$
I have verified for $n\leq 7$ (using sage) that when $(x_j, p_{ij}, \theta_{ij}, s_{ij})$ satisfy the above equations, the identity \begin{equation} \tag{*} \sum_{i<j} s_{ij}\theta_{ij} p_{ij}^2 = 0 \end{equation} holds. This seems like a simple enough identity that I can hope for a proof, but I have not managed to find the right manipulations.
As alluded to earlier, these equations are a rough model for a system of "springs" that are attached to the vertices of the complete graph, with variable stiffnesses. The conditions enforce that the forces on the vertices are balanced and that the vertices of the graph can be placed in a consistent way. It seems that the identity (*) looks like it says that the "potential energy" of the system is zero, so it seems like it might have a relatively simple proof. That said, I don't know why I should expect the potential energy to be zero.
The case $n=3$
For concreteness I just wanted to show how to work out this identity in th case $n=2$. In this case the equations are \begin{align*} x_1&=x_0+s_{01}p_{01}\\ x_2&=x_1+s_{12}p_{12}\\ x_2&=x_0+s_{02}p_{02} \end{align*} on the edges and \begin{align*} 0 &= \theta_{01}p_{01} + \theta_{02}p_{02} \\ \theta_{01}p_{01} &= \theta_{12}p_{12} \\ \theta_{02}p_{02} + \theta_{12}p_{12} &= 0 \end{align*} on the vertices.
We can first eliminate the $x$ variables by summing the edge equations to obtain $$ s_{01}p_{01} + s_{12}p_{12} = s_{02}p_{02}. $$ Then, from the vertex equations it seems reasonable to define a new variable $Q=\theta_{01}p_{01}$, and then we have $$ Q = \theta_{01}p_{01} = \theta_{12}p_{12} = -\theta_{02}p_{02}. $$ Now we can write $$ \sum_{i<j}s_{ij}\theta_{ij}p_{ij}^2 = \theta_{01}s_{01}p_{01}^2 + \theta_{02}s_{02}p_{02}^2 + \theta_{12}s_{12}p_{12}^2 = Q (s_{01}p_{01} + s_{12}p_{12}-s_{02}p_{02}) = 0. $$ Unfortunately from here I don't see how to generalize the argument. In this case we were lucky that it was so easy to eliminate the $x$ variables. In general I don't think it is so simple.