jmc

Question

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 thatr_1+r_2=r_1^2+r_2^2=r_1^3+r_2^3==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 of1r_1^2008+1r_2^2008.Help Michael by computing this maximum.

Accepted Answer

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 = s^2 - s2. We also know that s = s^3 - 3ps, so substituting for p results in align* s &= s^3 - 3s s^2 - s2 s &= s^3 - 32 s^3 + 32 s^2 0 &= -12 s^3 + 32 s^2 - s 0 &= s^3 - 3s^2 + 2s &= s(s-2)(s-1) 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 1r_1^2008+1r_2^2008 is undefined. If s = 2, then p = 1, making both roots equal to 1. Since 1^n = 1 for 1 n 2007, this result satisfies all conditions. Thus, 1r_1^2008+1r_2^2008 = 1+1 = 2.