1991 AIME Problems/Problem 8

Problem

For how many real numbers $a^{}_{}$ does the quadratic equation $x^2 + ax^{}_{} + 6a=0$ have only integer roots for $x^{}_{}$ ?

By Vieta's formulas, $x_1 + x_2 = -a$ where $x_1, x_2$ are the roots of the quadratic, and since $x_1,x_2$ are integers, $a$ must be an integer. Applying the quadratic formula,

Since $-a$ is an integer, we need $\sqrt{a^2-24a}$ to be an integer (let this be $b$ ): $b^2 = a^2 - 24a$ . Completing the square, we get

Which implies that $b^2 + 144$ is a perfect square also (let this be $c^2$ ). Then

The pairs of factors of $144$ are $(\pm1,\pm144),( \pm 2, \pm 72),( \pm 3, \pm 48),( \pm 4, \pm 36),( \pm 6, \pm 24),( \pm 8, \pm 18),( \pm 9, \pm 16),( \pm 12,...$ ; since $c$ is the average of each respective pair and is also an integer, the pairs that work must have the same parity. Thus we get $\boxed{10}$ pairs (counting positive and negative) of factors that work, and substituting them backwards show that they all work.