2015 AMC 12A Problems/Problem 18

The problem asks us to find the sum of every integer value of $a$ such that the roots of $x^2 - ax + 2a = 0$ are both integers.

The quadratic formula gives the roots of the quadratic equation: $x = \frac{a \± \sqrt{a^2 - 8a}}{2}$ (Error compiling LaTeX. Unknown error_msg)

As long as the numerator is an even integer, the roots are both integers. But first of all, the radical term in the numerator needs to be an integer; that is, the discriminant $a^2 - 8a$ equals $k^2$ , for some nonnegative integer $k$ .

$a^2 - 8a = k^2$

$a(a - 8) = k^2$

$((a - 4) + 4)((a - 4) - 4) = k^2$

$(a - 4)^2 - 4^2 = k^2$

$(a - 4)^2 = k^2 + 4^2$

From this last equation, we are given a hint of the Pythagorean theorem. Thus, $(k, 4, |a - 4|)$ must be a Pythagorean triple unless $k = 0$ .

In the case $k = 0$ , the equation simplifies to $|a - 4| = 4$ . From this equation, we have $a = 0, 8$ . For both $a = 0$ and $a = 8$ , $\frac{a \± \sqrt{a^2 - 8a}}{2}$ (Error compiling LaTeX. Unknown error_msg) yields two integers, so these values satisfy the constraints from the original problem statement. (Note: the two zero roots count as "two integers.")

If $k$ is a positive integer, then only one Pythagorean triple could match the triple $(k, 4, |a - 4|)$ because the only Pythagorean triple with a $4$ as one of the values is the classic $(3, 4, 5)$ triple. Here, $k = 3$ and $|a - 4| = 5$ . Hence, $a = -1, 9$ . Again, $\frac{a \± \sqrt{a^2 - 8a}}{2}$ (Error compiling LaTeX. Unknown error_msg) yields two integers for both $a = -1$ and $a = 9$ , so these two values also satisfy the original constraints.

There are a total of four possible values for $a$ : $-1, 0, 8,$ and $9$ . Hence, the sum of all of the possible values of $a$ is 16 (C).

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2015 AMC 12A Problems/Problem 18