Modulli Roots

If the integer roots are $u,v$ , then $a(u+v) = 4$ and $auv = 9$ . Thus $a = \tfrac{9}{uv}$ is a rational which (in lowest terms) has numerator dividing $9$ . Similarly $a = \tfrac{4}{u+v}$ has (in lowest terms) numerator dividing $4$ . Thus the numerator of $a$ must be equal to $1$ , and hence $a = c^{-1}$ for some integer $c$ . This means that $u+v = 4c$ and $uv = 9c$ . Since $u^2 - 4uc + 9c = v^2 - 4vc + 9c = 0$ , we see that $c$ divides both $u^2$ and $v^2$ . Thus $u^2 = c\alpha$ and $v^2 = c\beta$ , where $81c^2 = u^2v^2 = c^2\alpha\beta$ , so that $\alpha,\beta$ are integers such that $\alpha\beta = 81$ . Note also that $\alpha - 4u + 9 = 0$ , so that $u = \tfrac14(\alpha + 9)$ , which means that $\alpha \equiv 3 \pmod{4}$ . Similarly, $\beta \equiv 3 \pmod{4}$ . Thus we have the possible solutions $\begin{array}{|c|c|c|c|c|} \hline \alpha & \beta & u & v & c \\ \hline 27 & 3 & 9 & 3 & 3 \\ 3 & 27 & 3 & 9 & 3 \\ -1 & -81 & 2 & -18 & -4 \\ -81 & -1 & -18 & 2 & -4 \\ \hline \end{array}$ The case $\alpha = \beta =-9$ leads $u=v=0$ , which is not a solution. Thus the possible values of $a$ are $c^{-1} = \tfrac13,-\tfrac14$ , making the answer $3+4 = \boxed{7}$ .

The roots add up to $4/a$ and multiply to $9/a,$ so these are both integers. The second integer is $9/4$ times the first, so the first one must be divisible by $4.$ Since $4/a$ is divisible by $4,$ this implies $a$ is the reciprocal of a (nonzero) integer. Say that integer is $n.$

The equation becomes $x^2-4nx+9n=0,$ with roots $2n \pm \sqrt{4n^2-9n}$ by the quadratic formula. So we get $4n^2-9n=b^2$ for some integer $b.$ Multiply by $16$ and complete the square to get $\begin{aligned} (8n-9)^2-81 &= 16b^2 \\ (8n-9-4b)(8n-9+4b) &= 81. \end{aligned}$ Assuming without loss of generality that $b$ is nonnegative, there are six possible factorizations of $81$ where the first term is not larger than the second: $-81 \cdot -1, -27 \cdot -3, -9 \cdot -9, 1 \cdot 81, 3 \cdot 27, 9 \cdot 9.$ These lead to the solutions $n=-4,-3/4,0,25/4,3,9/4,$ of which only two are nonzero integers: $-4$ and $3.$ The sum of their absolute values is $\fbox{7}.$

@Mark Hennings , can you please post the solution