= Solution
If $Q_0$ is empty, there are no nonzero indecomposables and the conclusion is immediate. Otherwise positive definiteness of the <Tits form of a quiver> gives a constant $c>0$ with $q_Q(v)\geq c\|v\|^2$ for every real vector $v$: take the minimum of $q_Q$ on the compact unit sphere. Every <positive root of a quiver> therefore satisfies $\|\mathbf n\|^2\leq1/c$. Only finitely many nonnegative integer vectors lie in this bounded set. Parts (ii) and (iii) give exactly one indecomposable isomorphism class for each such root, and no others. Consequently \b[$Q$ has finite representation type].
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