= Solution
If $X$ is indecomposable, part (i) makes it a <brick module>. For its nonzero <dimension vector of a quiver representation> $\mathbf n$,
$$
0<q_Q(\mathbf n)=1-\dim\operatorname{Ext}^1_Q(X,X)\leq1.
$$
The <Tits form of a quiver> takes integer values on integer vectors, so \b[$q_Q(\mathbf n)=1$ and $X$ has no self-extensions].
Conversely, suppose $\mathbf n$ has nonnegative integer entries and $q_Q(\mathbf n)=1$. Choose a representation $M$ of that dimension vector whose orbit has maximal dimension. Such an orbit exists because dimensions are integers bounded by $\dim\operatorname{Rep}_Q(\mathbf n)$. If $M=U\oplus V$ with nonzero $U,V$, a nonzero extension in either direction would, by part 5(c), produce a middle representation of the same dimension vector with larger orbit. Hence $\operatorname{Ext}^1_Q(U,V)=\operatorname{Ext}^1_Q(V,U)=0$.
Writing $\mathbf u=\dim U$ and $\mathbf v=\dim V$, the <Ringel form> identity then gives
$$
\begin{aligned}
1=q_Q(\mathbf u+\mathbf v)&=q_Q(\mathbf u)+q_Q(\mathbf v)+\langle\mathbf u,\mathbf v\rangle_Q+\langle\mathbf v,\mathbf u\rangle_Q\\
&=q_Q(\mathbf u)+q_Q(\mathbf v)+\dim\operatorname{Hom}_Q(U,V)+\dim\operatorname{Hom}_Q(V,U)\geq2.
\end{aligned}
$$
The last inequality uses positive integral values of $q_Q$ at both nonzero vectors. This contradiction makes $M$ indecomposable. Thus \b[the indecomposable dimension vectors are exactly the positive roots $q_Q(\mathbf n)=1$].
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