= Solution
Use the allowed <Ringel lemma on bricks> in the following precise form: a finite-dimensional indecomposable <quiver representation> that is not a <brick module> contains a nonzero brick $B$ with $\operatorname{Ext}^1_Q(B,B)\ne0$.
Suppose such a nonbrick indecomposable existed, and let $\mathbf b=\dim B$. The <Tits form of a quiver> is $q_Q(\mathbf b)=\langle\mathbf b,\mathbf b\rangle_Q$, so the <Ringel form> identity would give
$$
q_Q(\mathbf b)=\dim\operatorname{End}_Q(B)-\dim\operatorname{Ext}^1_Q(B,B)=1-\dim\operatorname{Ext}^1_Q(B,B)\leq0.
$$
Positive definiteness gives $q_Q(\mathbf b)>0$ for nonzero $\mathbf b$, a contradiction. Therefore \b[every indecomposable is a brick].
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