An algebra problem by arko roychoudhury

Let the zeroes be $m,n$ . Then we require that $m + n = a$ and $mn = 2a$ . Now if either of $m,n$ equals $0$ then since $mn = 2a$ we would have $a = 0$ , which in turn, since $m + n = a$ , would imply that both of $m,n$ equal $0$ . So ruling out this possibility, we have that

$\dfrac{mn}{m + n} = \dfrac{2a}{a} = 2 \Longrightarrow mn = 2(m + n) \Longrightarrow mn - 2m - 2n = 0 \Longrightarrow (m - 2)(n - 2) = 4$ .

Now $4$ can be factored in four (unordered) ways, namely $-1 \times -4, -2 \times -2, 2 \times 2$ and $1 \times 4$ . Assigning these pairs to $m - 2$ and $n - 2$ , and solving for $(m,n)$ , we can have

$(m,n) = (1, -2), (0,0), (4,4)$ or $(3,6)$ , giving us values for $a = m + n$ of $-1, 0, 8$ or $9$ .

Thus the sum of the possible values of $a$ is $-1 + 0 + 8 + 9 = \boxed{16}$ .