3A. Find all positive numbers $a$ for which both roots of the equation $a^{2} x^{2}+a x+1-7 a^{2}=0$ are integers.
Solution. The equation will have real solutions if $D \geq 0$, i.e.

$$
0 \leq a^{2}-4 a^{2}\left(1-7 a^{2}\right)=a^{2}\left(28 a^{2}-3\right)
$$

from which we get $a^{2} \geq \frac{3}{28}$, i.e. $\frac{1}{a} \leq \sqrt{\frac{28}{3}}$.

From Vieta's formulas, we have $x_{1}+x_{2}=-\frac{1}{a}$. Since $x_{1}$ and $x_{2}$ are integers, it follows that $-\frac{1}{a}$ is an integer, and from $a>0$ we conclude that $\frac{1}{a}$ is a natural number. The only natural numbers for which $\frac{1}{a} \leq \sqrt{\frac{28}{3}}$ are the numbers 1, 2, and 3. Therefore, $a \in\left\{1, \frac{1}{2}, \frac{1}{3}\right\}$. For these values of $a$, the corresponding values of the discriminants are $D_{1}=25, D_{\frac{1}{2}}=1, D_{\frac{1}{3}}=\frac{1}{81}$, and the corresponding solutions of the equations are: -3 and 2; -3 and 1; -2 and -1.