Problem:
The polynomial
P(x)=(1+x+x2+…+x17)2−x17
has 34 complex zeros of the form zk=rk[cos(2παk)+isin(2παk)], k=1,2,3,…,34, with 0<α1≤α2≤α3≤…≤α34<1 and rk>0. Given that α1+α2+α3+α4+α5=m/n, where m and n are relatively prime positive integers, find m+n.
Solution:
Note that for x=1,
P(x)(x−1)2P(x)P(x)=(x−1x18−1)2−x17 so =(x18−1)2−x17(x−1)2=x36−2x18+1−x19+2x18−x17=x36−x19−x17+1=x19(x17−1)−(x17−1)=(x19−1)(x17−1), and so =(x−1)2(x19−1)(x17−1)
Thus the zeros of P(x) are the 34 complex numbers other than 1 which satisfy x17=1 or x19=1. It follows that α1=1/19,α2=1/17,α3=2/19,α4=2/17, and α5=3/19, so α1+α2+α3+α4+α5=159/323, and m+n=482.
The problems on this page are the property of the MAA's American Mathematics Competitions