Given a graph $G = (V, E)$, we can calculate its chromatic polynomial $P(G, k)$, and it has $n$ (complex) roots, also known as chromatic roots. It is a well-known fact that the sum of chromatic roots of the graph $G$ is equal to the number of edges $e(G)$.
I am working on a problem and it somehow is related to the following pattern, which I also experiment with SageMath:
Want to prove: Given a bipartite graph $G$ of minimum degree $2$ and $G$ is not a forest, with its chromatic polynomial $P(G, k)$ and chromatic roots $r_{1}, r_{2}, \ldots, r_{n}$, then we have
$$\sum_{i = 1}^{n} r_{i}^{2} = e(G)$$
However, I could not seem to prove it. I tried some Cauchy-Schwarz inequality calculations. I also couldn't find relevant literatures containing this result, or relevant research on (complex) chromatic roots of bipartite graphs.
Thank you!
The property seems to be true for any bipartite graph unless I missed something in my proof.
Let $P = X^n+c_1X_{n-1}+c_2X_{n-2}+...$ be the chromatic polynomial of $G$.
Properties of roots allow to write: $$c_1 = -\sum r_i$$ $$c_2 = \sum_{i\neq j} r_ir_j$$
So $\sum r_i^2 = c_1^2-2c_2$
We already know by deletion contraction that $c_1 = -e$ Lets show by induction using the deletion contraction that $c_2=\frac{e^2-e}{2}$.
This is true for a graph with no edges, since $c_2=0$.
Suppose this is true for bipartite graphs with $e$ edges. Let $G$ be a bipartite graph with $e+1$ edges.
Let $uv$ be an edge. We know that $G/uv$ (the graph obtained by contracting $uv$) has $e$ edges because the contraction deletes exactly one edge (the contracted one). A contraction can delete more edges by replacing the multi-edges formed by the contraction with a single edge. However, in this case, it does not lose any other edge, otherwise, that would mean that $G$ had a triangle, which can't be possible.
Using the contraction-deletion formula, $c_2(G) = c_2(G-uv) - c_1(G/uv) = \frac{e^2-e}{2} + e = \frac{(e+1)^2-(e+1)}{2}$, which prove the property is hereditary and finish the induction.
It follows that $\sum r_i^2 = c_1^2-2c_2 = e^2 - 2\frac{e^2-e}{2} = e$
Edit: As suggested by @MishaLavrov, the result holds more generally for triangle-free graphs. For a general graph, we have $c_2 = \frac{e^2-e}{2}-t$ where $t$ is the number of triangles because we lose one edge for each triangle in the induction. So $\sum r_i^2 = e+2t$