2015 AMC 10A Problem 23

Let the zeros be integers $r$ and $s$ (a repeated zero is allowed). Vieta's formulas give
$$r + s = a, \qquad rs = 2a.$$
Eliminate $a$: $rs = 2(r + s)$, so $rs - 2r - 2s = 0$. Add $4$ to both sides to factor:
$$(r - 2)(s - 2) = 4.$$

Now list the unordered ways to write $4$ as a product of two integers, negatives included:

Each value really works: $x^2 - 9x + 18 = (x-3)(x-6)$, $x^2 - 8x + 16 = (x-4)^2$, $x^2 + x - 2 = (x+2)(x-1)$, and $x^2$.

The sum of the possible values is $9 + 8 + (-1) + 0 = 16$.

The answer is $\boxed{\textbf{(C)}\ 16}$.

"Integer roots" plus Vieta turns a quadratic into a Diophantine equation linking the sum and product of the roots. Whenever that equation has the shape $rs = m(r + s)$, adding $m^2$ factors it as $(r - m)(s - m) = m^2$, and the finite list of divisors of $m^2$ gives every solution. Always include the negative divisor pairs and the perfect-square (repeated root) pair.

Work with the discriminant: $a^2 - 8a$ must be a perfect square $k^2$. Completing the square, $(a - 4)^2 - k^2 = 16$, so $(a - 4 - k)(a - 4 + k) = 16$ with both factors of the same parity: $(2, 8)$, $(4, 4)$, $(-8, -2)$, $(-4, -4)$ give $a - 4 = 5, 4, -5, -4$, i.e. $a = 9, 8, -1, 0$, sum $16$.

• Using only the positive factor pairs of $4$, which yields $a = 9$ and $8$ and the sum $17$ (choice D).
• Discarding $a = 0$ or $a = 8$ on the grounds that the zeros are repeated; a double zero is still an integer zero.
• Counting $a = 9$ twice from the ordered pairs $(3, 6)$ and $(6, 3)$.