# 2015 AMC 10A Problems/Problem 23

## Problem

The zeroes of the function $f(x)=x^2-ax+2a$ are integers. What is the sum of the possible values of $a?$ $\textbf{(A)}\ 7\qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ 16\qquad\textbf{(D)}\ 17\qquad\textbf{(E)}\ 18$

## Solution 1

By Vieta's Formula, $a$ is the sum of the integral zeros of the function, and so $a$ is integral.

Because the zeros are integral, the discriminant of the function, $a^2 - 8a$, is a perfect square, say $k^2$. Then adding 16 to both sides and completing the square yields $$(a - 4)^2 = k^2 + 16.$$ Therefore $(a-4)^2 - k^2 = 16$ and $$((a-4) - k)((a-4) + k) = 16.$$ Let $(a-4) - k = u$ and $(a-4) + k = v$; then, $a-4 = \dfrac{u+v}{2}$ and so $a = \dfrac{u+v}{2} + 4$. Listing all possible $(u, v)$ pairs (not counting transpositions because this does not affect ( $u + v$), $(2, 8), (4, 4), (-2, -8), (-4, -4)$, yields $a = 9, 8, -1, 0$. These $a$ sum to $16$, so our answer is $\boxed{\textbf{(C) }16}$.

## Solution 2

Let $r_1$ and $r_2$ be the integer zeroes of the quadratic. Since the coefficient of the $x^2$ term is $1$, the quadratic can be written as $$(x - r_1)(x - r_2)=x^2 - (r_1 + r_2)x + r_1r_2$$

By comparing this with $x^2 - ax + 2a$, $$r_1 + r_2 = a\text{ and }r_1r_2 = 2a.$$

Plugging the first equation in the second, $$r_1r_2 = 2 (r_1 + r_2).$$ Rearranging gives $$r_1r_2 - 2r_1 - 2r_2 = 0\implies (r_1 - 2)(r_2 - 2) = 4.$$ These factors can be $(1, 4), (-1, -4), (4, 1), (-4, -1), (2, 2),$ or $(-2, -2).$

We want the number of distinct $a = r_1 + r_2$, and these factors gives $a = -1, 0, 8, 9$. So the answer is $-1 + 0 + 8 + 9 = \boxed{\textbf{(C) }16}$.

## Video Solution

~savannahsolver

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