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 1

Let $x^2 + ax + 6a = (x - s)(x - r)$ . Vieta's yields $s + r = - a, sr = 6a$ . $\begin{eqnarray*}sr + 6s + 6r &=& 0\\ sr + 6s + 6r + 36 &=& 36\\ (s + 6)(r + 6) &=& 36 \end{eqnarray*}$

Without loss of generality let $r \le s$ .

The possible values of $(r + 6,s + 6)$ are: $( - 36, - 1),( - 18, - 2),( - 12, - 3),( - 9, - 4),( - 6, - 6),(1,36),(2,18),(3,12),(4,9),(6,6)$ $\Rightarrow \boxed{10}\ \text{values of } a$ .

Solution 2

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,

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

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
All AIME Problems and Solutions

Page

Toolbox

Search

1991 AIME Problems/Problem 8

Problem

Contents

Solution 1

Solution 2

See also