1. Find all such positive integers $a$ such that the equation $a x^{2}+2(2 a-1) x+4(a-3)=0$ has at least one real root. (3rd Zu Chongzhi Cup Junior High School Competition)
Analysis 1 Since the equation has integer solutions, then its discriminant $\Delta=4(2 a-1)^{2}-16 a(a-3)=4(8 a+1)$ is a perfect square, thus $8 a+1$ is a perfect square.

Since $8 a+1$ is odd. Therefore, we can set $8 a+1$ $=(2 m+1)^{2}\left(m\right.$ is a positive integer), then $a=\frac{1}{2} m(m+$ 1).
Hence $x_{1,2}=\frac{1-2 a \pm \sqrt{8 a}+1}{a}$
$$
=-2+\frac{4}{m} \text { or }-2-\frac{4}{m+1} \text {. }
$$

Since $x_{1}, x_{2}$ have at least one integer, then $\frac{4}{m}$ or $\frac{4}{m+1}$ has at least one integer. Therefore, $m=1$, 2,4 or $m=1,3$.
Thus we get $a=1,3,6,10$.
Proof 2 In the given problem, the parameter $a$ is linear. We express $a$ in terms of $x$, i.e., $a=\frac{2(x+6)}{(x+2)^{2}}$.

Since $a$ is a positive integer, then $\frac{2(x+6)}{(x+2)^{2}} \geq 1$. Solving this, we get -4 $\leqslant x \leqslant 2$, and $x \neq \cdots 2$, i.e., $x=-4,-3, \cdots 1$, 0 , 1.2. Substituting these into (1), we get $a=1,3,6.10$.