2008 AIME I Problems/Problem 13

Problem

Let

Suppose that

There is a point $\left(\frac {a}{c},\frac {b}{c}\right)$ for which $p\left(\frac {a}{c},\frac {b}{c}\right) = 0$ for all such polynomials, where $a$ , $b$ , and $c$ are positive integers, $a$ and $c$ are relatively prime, and $c > 1$ . Find $a + b + c$ .

$\begin{align*}p(0,0) &= a_0 = 0\\p(1,0) &= a_0 + a_1 + a_3 + a_6 = a_1 + a_3 + a_6 = 0\\p(-1,0) &= -a_1 + a_3 - a...$

Adding the above two equations gives $a_3 = 0$ , and so we can deduce that $a_6 = -a_1$ .

Similarly, plugging in $(0,1)$ and $(0,-1)$ gives $a_5 = 0$ and $a_9 = -a_2$ . Now,

$\begin{align*}p(1,1) &= a_0 + a_1 + a_2 + a_3 + a_4 + a_5 + a_6 + a_7 + a_8 + a_9\\&= 0 + a_1 + a_2 + 0 + a_4 + 0 - a...$

Therefore $a_8 = 0$ and $a_7 = -a_4$ . Finally,

$p(2,2) = 0 + 2a_1 + 2a_2 + 0 + 4a_4 + 0 - 8a_1 - 8a_4 +0 - 8a_2 = -6 a_1 - 6 a_2 - 4 a_4 = 0$

In order for the above to be zero, we must have

$x(1-x)(1+x) = y(1-y)(1+y)$

and

$x(1-x)(1+x) = 1.5 xy (1-x).$

Canceling terms on the second equation gives us $1+x = 1.5 y \Longrightarrow x = 1.5 y - 1$ . Plugging that into the first equation and solving yields $x = 5/19, y = 16/19$ , and $5+16+19 = \boxed{040}$ .