Quadratic Diophantine Equation

If the integer solutions are $m$ and $n$ , then $m+n = \tfrac{4}{a}$ and $mn = \tfrac{9}{a}$ , so that $\begin{array}{rcl} 4mn & = & 9(m+n) \\ 16mn - 36(m+n) & = & 0 \\ (4m-9)(4n-9) & = & 81 \end{array}$ and hence $4m-9$ must be one of $\pm1,\pm3,\pm9,\pm27,\pm81$ , with $4n-9$ chosen to match. The only options for the ordered triple $(4m-9,4n-9)$ for which both $m$ and $n$ are integers are $(3,27)$ , $(27,3)$ , $(-1,-81)$ , $(-81,-1)$ and $(-9,-9)$ . The first two yield $\{m,n\} = \{3,9\}$ , and $a=\tfrac13$ . The third and fourth yield $\{m,n\} = \{2,-18\}$ , and $a=-\tfrac14$ . The fifth option yields $m=n=0$ which is impossible.

The only possible values of $a$ are $\tfrac13$ and $-\tfrac14$ , and the sum of these numbers is $\tfrac{1}{12}$ .