Sum-coloring a tournament 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 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?


*

*for all $v\in V$ we have $c(v) = \sum_{w\in \text{In}(v)} c(w)$, and

*$c$ is not the constant $0$-map.

 A: 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 $0$.
