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 $x^{}_{}$ ?

Solution

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

$x = \frac{-a \pm \sqrt{a^2 - 24a}}{2}$

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

$(a - 12)^2 = b^2 + 144$

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

$c^2 - b^2 = 144 \Longrightarrow (c+b)(c-b) = 144$

The pairs of factors of $144$ are $(\pm1,\pm144),( \pm 2, \pm 72),( \pm 3, \pm 48),( \pm 4, \pm 36),( \pm 6, \pm 24),( \pm 8, \pm 18),( \pm 9, \pm 16),( \pm 12, \pm 12)$ ; 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.

1991 AIME (Problems • Answer Key • Resources)
Preceded by Problem 7	Followed by Problem 9
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
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1991 AIME Problems/Problem 8

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