Problem:
The polynomial
P(x)=(1+x+x2+β¦+x17)2βx17
has 34 complex zeros of the form zkβ=rkβ[cos(2ΟΞ±kβ)+isin(2ΟΞ±kβ)], k=1,2,3,β¦,34, with 0<Ξ±1ββ€Ξ±2ββ€Ξ±3ββ€β¦β€Ξ±34β<1 and rkβ>0. Given that Ξ±1β+Ξ±2β+Ξ±3β+Ξ±4β+Ξ±5β=m/n, where m and n are relatively prime positive integers, find m+n.
Solution:
Note that for xξ =1,
P(x)(xβ1)2P(x)P(x)β=(xβ1x18β1β)2βx17 so =(x18β1)2βx17(xβ1)2=x36β2x18+1βx19+2x18βx17=x36βx19βx17+1=x19(x17β1)β(x17β1)=(x19β1)(x17β1), and so =(xβ1)2(x19β1)(x17β1)ββ
Thus the zeros of P(x) are the 34 complex numbers other than 1 which satisfy x17=1 or x19=1. It follows that Ξ±1β=1/19,Ξ±2β=1/17,Ξ±3β=2/19,Ξ±4β=2/17, and Ξ±5β=3/19, so Ξ±1β+Ξ±2β+Ξ±3β+Ξ±4β+Ξ±5β=159/323, and m+n=482β.
The problems on this page are the property of the MAA's American Mathematics Competitions