Question

There are nonzero integers $a,$ $b,$ $r,$ and $s$ such that the complex number $r + si$ is a zero of the polynomial $P(x) = x^3 - ax^2 + bx - 65.$ 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 $b.$

答案

Solution:

Since $P$ has real coefficients, $r - si$ is also a zero, and the third zero $q$ is real. The product of the zeros is $q(r^2 + s^2) = 65,$ so $q$ is a nonzero integer and $r^2 + s^2$ is a factor of $65.$ With $r, s$ nonzero, the possibilities are $r^2 + s^2 = 5 = 1^2 + 2^2$ (with $q = 13$ ), $13 = 2^2 + 3^2$ (with $q = 5$ ), and $65 = 1^2 + 8^2 = 4^2 + 7^2$ (with $q = 1$ ).

For each representation $\{u, v\},$ the zero $r + si$ can have $r = \pm u$ or $\pm v,$ giving $4$ distinct polynomials (the sign of $s$ changes nothing). The sum of the zeros is $p_{a,b} = q + 2r,$ and over the four choices the $2r$ terms cancel, leaving $4q$ from each representation.

The total is $4 \cdot 13 + 4 \cdot 5 + 4 \cdot 1 + 4 \cdot 1 = 80.$

2013 AIME I Problem 10

Solution:

Problem 10 in Other Years