If $G=(V,E)$ is a loopless finite directed graph and $v\in V$, we set $\text{In}(v) = \{(w,v): w\in V \land (w,v) \in E\}$. 


Let $T=(V,E)$ be a [tournament](https://en.wikipedia.org/wiki/Tournament_(graph_theory)) such that for every $v\in V$ the set $\text{In}(v)$ contains at least $2$ elements. Is there a map $c: V\to \mathbb{Z}$ with the following properties?


1. for all $v\in V$ we have $c(v) = \sum\_{w\in \text{In}(v)} c(w)$, and
2. $c$ is not the constant $0$-map.



Sum-coloring a tournament
No, there isn't. For a counterexample, let $T$ be the tournament with vertex
set $V=\left\{ 1,2,3,4,5\right\} $ and arc set
\begin{align\*}
E & =\left\{ \left( 1,2\right) ,\left( 1,3\right) ,\left( 2,3\right)
,\left( 2,4\right) ,\left( 3,4\right) ,\left( 3,5\right) ,\left(
4,5\right) ,\left( 4,1\right) ,\left( 5,1\right) ,\left( 5,2\right)
\right\} \\
& =\left\{ \left( i,i+1\right) \ \mid\ 1\leq i\leq5\right\} \cup\left\{
\left( i,i+2\right) \ \mid\ 1\leq i\leq5\right\}
\end{align\*}
(where indices are modulo $5$). Each veretx of $T$ has indegree $2$ and
outdegree $2$. Your map $c$ must thus satisfy the following system of linear
equations:
\begin{equation}
\left\{
\begin{array}
[c]{c}
c\left( 1\right) =c\left( 4\right) +c\left( 5\right) ;\\
c\left( 2\right) =c\left( 5\right) +c\left( 1\right) ;\\
c\left( 3\right) =c\left( 1\right) +c\left( 2\right) ;\\
c\left( 4\right) =c\left( 2\right) +c\left( 3\right) ;\\
c\left( 5\right) =c\left( 3\right) +c\left( 4\right)
\end{array}
\right. .
\end{equation}
But the only solution to this system is when all $c\left( v\right) $ are $0$.


