2008 iTest Problems/Problem 76

Problem

During the car ride home, Michael looks back at his recent math exams. A problem on Michael's calculus mid-term gets him starting thinking about a particular quadratic, $x^2-sx+p,$ with roots $r_1$ and $r_2$ . He notices that $r_1+r_2=r_1^2+r_2^2=r_1^3+r_2^3=\cdots=r_1^{2007}+r_2^{2007}.$ He wonders how often this is the case, and begins exploring other quantities associated with the roots of such a quadratic. He sets out to compute the greatest possible value of $\dfrac1{r_1^{2008}}+\dfrac1{r_2^{2008}}.$ Help Michael by computing this maximum.

Solution

By Vieta's Formulas, $r_1 + r_2 = s$ . That means $r_1^2 + r_2^2 = s^2 - 2p = s$ and $r_1^3 + r_1^3 = (r_1 + r_2)^3 - 3r_1^2r_2 - 3r_1r_2^2 = s^3 - 3ps$ .

Note that $s = s^2 - 2p$ , so $p = \frac{s^2 - s}{2}$ . We also know that $s = s^3 - 3ps$ , so substituting for $p$ results in

$\begin{align*} s &= s^3 - 3s \cdot \frac{s^2 - s}{2} \\ s &= s^3 - \tfrac32 s^3 + \tfrac32 s^2 \\ 0 &= -\tfrac12 s^3 + \tfrac32 s^2 - s \\ 0 &= s^3 - 3s^2 + 2s \\ &= s(s-2)(s-1) \end{align*}$

Thus, $s = 0,1,2$ . If $s = 1$ or $s = 0$ , then $p = 0$ . However, both cases result in one root being zero, so $\dfrac1{r_1^{2008}}+\dfrac1{r_2^{2008}}$ is undefined. If $s = 2$ , then $p = 1$ , making both roots equal to $1$ . Since $1^n = 1$ for $1 \le n \le 2007$ , this result satisfies all conditions. Thus, $\dfrac1{r_1^{2008}}+\dfrac1{r_2^{2008}} = 1+1 = \boxed{2}$ .

2008 iTest (Problems)
Preceded by: Problem 75	Followed by: Problem 77
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2008 iTest Problems/Problem 76

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