= Solution
Number the vertices consecutively along the underlying chain. Orientation does not affect the quadratic <Tits form of a quiver>:
$$
q_Q(\mathbf n)=\sum_{i=1}^m n_i^2-\sum_{i=1}^{m-1}n_in_{i+1}=\frac12\sum_{i=0}^m(n_{i+1}-n_i)^2,\qquad n_0=n_{m+1}=0.
$$
If $\mathbf n$ has nonnegative integer coordinates and $q_Q(\mathbf n)=1$, the sum of integer squares on the right is $2$. There are therefore exactly two nonzero consecutive differences, each of absolute value one. Their sum is zero, so one is $+1$ and the other $-1$. Nonnegativity forces the $+1$ to occur first. Thus the <positive roots of type A> are exactly the vectors with \b[a single nonempty interval of ones and zeros elsewhere].
There is one such vector for every pair of endpoints $1\leq a\leq b\leq m$, giving
$$
\boxed{\#\{\text{positive roots of type }A_m\}=\frac{m(m+1)}2.}
$$
For $A_3$, the full list is
$$
\boxed{(1,0,0),\ (0,1,0),\ (0,0,1),\ (1,1,0),\ (0,1,1),\ (1,1,1).}
$$
The difference-of-coordinates proof includes $m=1$ and is independent of the chosen arrow orientation.
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