# 2013 AIME I Problems/Problem 10

## Problem

There are nonzero integers $a$, $b$, $r$, and $s$ such that the complex number $r+si$ is a zero of the polynomial $P(x)={x}^{3}-a{x}^{2}+bx-65$. For each possible combination of $a$ and $b$, let ${p}_{a,b}$ be the sum of the zeros of $P(x)$. Find the sum of the ${p}_{a,b}$'s for all possible combinations of $a$ and $b$.

## Solution

Since $r+si$ is a root, by the Complex Conjugate Root Theorem, $r-si$ must be the other imaginary root. Using $q$ to represent the real root, we have $(x-q)(x-r-si)(x-r+si) = x^3 -ax^2 + bx -65$

Applying difference of squares, and regrouping, we have $(x-q)(x^2 - 2rx + (r^2 + s^2)) = x^3 -ax^2 + bx -65$

So matching coefficients, we obtain $q(r^2 + s^2) = 65$ $b = r^2 + s^2 + 2rq$ $a = q + 2r$

By Vieta's each ${p}_{a,b} = a$ so we just need to find the values of $a$ in each pair. We proceed by determining possible values for $q$, $r$, and $s$ and using these to determine $a$ and $b$.

If $q = 1$, $r^2 + s^2 = 65$ so (r, s) = $(\pm1, \pm 8), (\pm8, \pm 1), (\pm4, \pm 7), (\pm7, \pm 4)$

Similarly, for $q = 5$, $r^2 + s^2 = 13$ so the pairs $(r,s)$ are $(\pm2, \pm 3), (\pm3, \pm 2)$

For $q = 13$, $r^2 + s^2 = 5$ so the pairs $(r,s)$ are $(\pm2, \pm 1), (\pm1, \pm 2)$

Now we can disregard the plus minus signs for s because those cases are included as complex conjugates of the counted cases. The positive and negative values of r will cancel, so the sum of the ${p}_{a,b} = a$ for $q = 1$ is $q$ times the number of distinct $r$ values (as each value of $r$ generates a pair $(a,b)$). Our answer is then $(1)(8) + (5)(4) + (13)(4) = \boxed{080}$.

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 