Funny Polynomial!

Algebra Level 4

$P(x) = 24x^{24} +\displaystyle \sum_{j = 1}^{23}(24 - j)(x^{24 - j} + x^{24 + j}).$ Let $z_{1},z_{2},\ldots,z_{r}$ be the distinct zeros of $P(x)$ and let $z_{k}^{2} = a_{k} + b_{k}i$ for real $a_k ,b_k$ , $k = 1,2,\ldots,r,$ Let

$\displaystyle\sum_{k = 1}^{r}|b_{k}| = m + n\sqrt {p},$ where $m,n,$ and $p$ are integers and $p$ is not divisible by the square of any prime. Find $m + n + p.$

The answer is 15.

1 solution

Kushal Dey
Dec 20, 2020

Note that (1+x+x²+...x^n)²=(n+1)x^n+sum((n+1-j)(x^(n+1-j)+x^(n+1+j)) where I varies from 1 to n. Thus p(x)=x(1+x+x²+...x^23)². Thus distinct roots of p(x) are 0 and 24th roots of 1. Thus r varies from 1 to 25, let z(r)=e^(2i r pi/24) for r varying from 1 to 24 and z(25)=0. Hence it's easy to see our summation is |sin(r*pi/6)| r varying from 1 to 24 which is 4(2+root(3)). Thus m=8,n=4,p=3.

Funny Polynomial!

The answer is 15.

1 solution

0 pending reports