Problem:  
Let
P ( x ) = 24 x 24 + β j = 1 23 ( 24 β j ) ( x 24 β j + x 24 + j )  P(x)=24 x^{24}+\sum_{j=1}^{23}(24-j)\left(x^{24-j}+x^{24+j}\right)
P ( x ) = 2 4 x 2 4 + j = 1 β 2 3 β ( 2 4 β j ) ( x 2 4 β j + x 2 4 + j ) 
Let z 1 , z 2 , β¦ , z r  z_{1}, z_{2}, \ldots, z_{r}z 1 β , z 2 β , β¦ , z r β   be the distinct zeros of P ( x )  P(x)P ( x )  , and let z k 2 = a k + b k i  z_{k}^{2}=a_{k}+b_{k} iz k 2 β = a k β + b k β i   for k = 1 , 2 , β¦ , r  k=1,2, \ldots, rk = 1 , 2 , β¦ , r  , where i = β 1  i=\sqrt{-1}i = β 1 β  , and a k  a_{k}a k β   and b k  b_{k}b k β   are real numbers. Let
β k = 1 r β£ b k β£ = m + n p  \sum_{k=1}^{r}\left|b_{k}\right|=m+n \sqrt{p}
k = 1 β r β β£ b k β β£ = m + n p β 
where m , n  m, nm , n  , and p  pp   are integers and p  pp   is not divisible by the square of any prime. Find m + n + p  m+n+pm + n + p  .
Solution: 
Note that
P ( x ) = x + 2 x 2 + 3 x 3 + β― + 24 x 24 + 23 x 25 + 22 x 26 + β― + 2 x 46 + x 47  P(x)=x+2 x^{2}+3 x^{3}+\cdots+24 x^{24}+23 x^{25}+22 x^{26}+\cdots+2 x^{46}+x^{47}
P ( x ) = x + 2 x 2 + 3 x 3 + β― + 2 4 x 2 4 + 2 3 x 2 5 + 2 2 x 2 6 + β― + 2 x 4 6 + x 4 7 
and
x P ( x ) = x 2 + 2 x 3 + 3 x 4 + β― + 24 x 25 + 23 x 26 + β― + 2 x 47 + x 48  x P(x)=\quad x^{2}+2 x^{3}+3 x^{4}+\cdots \quad+24 x^{25}+23 x^{26}+\cdots+\quad 2x^{47}+x^{48}
x P ( x ) = x 2 + 2 x 3 + 3 x 4 + β― + 2 4 x 2 5 + 2 3 x 2 6 + β― + 2 x 4 7 + x 4 8 
So
( 1 β x ) P ( x ) = x + x 2 + β― + x 24 β ( x 25 + x 26 + β― + x 47 + x 48 ) = ( 1 β x 24 ) ( x + x 2 + β― + x 24 )  \begin{aligned}
(1-x) P(x)=x+x^{2}+\cdots+x^{24}-\left(x^{25}+x^{26}+\cdots+x^{47}+x^{48}\right) \\ \quad=\left(1-x^{24}\right)\left(x+x^{2}+\cdots+x^{24}\right)
\end{aligned}
( 1 β x ) P ( x ) = x + x 2 + β― + x 2 4 β ( x 2 5 + x 2 6 + β― + x 4 7 + x 4 8 ) = ( 1 β x 2 4 ) ( x + x 2 + β― + x 2 4 ) β 
Then, for x β  1  x \neq 1x ξ  = 1  ,
P ( x ) = x 24 β 1 x β 1 x ( 1 + x + β― + x 23 ) = x ( x 24 β 1 x β 1 ) 2 (*)  \begin{aligned}P(x) =\frac{x^{24}-1}{x-1} x\left(1+x+\cdots+x^{23}\right) \\=x\left(\frac{x^{24}-1}{x-1}\right)^{2} \tag{*}\end{aligned}
P ( x ) = x β 1 x 2 4 β 1 β x ( 1 + x + β― + x 2 3 ) = x ( x β 1 x 2 4 β 1 β ) 2 β ( * ) 
One zero of P ( x )  P(x)P ( x )   is 0  00   , which does not contribute to the requested sum. The remaining zeros of P ( x )  P(x)P ( x )   are the same as those of ( x 24 β 1 ) 2  \left(x^{24}-1\right)^{2}( x 2 4 β 1 ) 2  , excluding 1  11  . Because ( x 24 β 1 ) 2  \left(x^{24}-1\right)^{2}( x 2 4 β 1 ) 2   and x 24 β 1  x^{24}-1x 2 4 β 1   have the same distinct zeros, the remaining zeros of P ( x )  P(x)P ( x )   can be expressed as z k =  z_{k}=z k β =   cis 15 k β  15 k^{\circ}1 5 k β   for k = 1 , 2 , 3 , β― β , 23  k=1,2,3, \cdots, 23k = 1 , 2 , 3 , β― , 2 3  . The squares of the zeros are therefore of the form cis 30 k β  30 k^{\circ}3 0 k β  , and the requested sum is
β k = 1 23 β£ sin β‘ 30 k β β£ = 4 β k = 1 5 β£ sin β‘ 30 k β β£ = 4 ( 2 β
 ( 1 / 2 ) + 2 β
 ( 3 / 2 ) + 1 ) = 8 + 4 3  \sum_{k=1}^{23}\left|\sin 30 k^{\circ}\right|=4 \sum_{k=1}^{5}\left|\sin 30 k^{\circ}\right|=4(2 \cdot(1 / 2)+2 \cdot(\sqrt{3} / 2)+1)=8+4 \sqrt{3}
k = 1 β 2 3 β β£ sin 3 0 k β β£ = 4 k = 1 β 5 β β£ sin 3 0 k β β£ = 4 ( 2 β
 ( 1 / 2 ) + 2 β
 ( 3 β / 2 ) + 1 ) = 8 + 4 3 β 
Thus m + n + p = 15  m+n+p=\boxed{15}m + n + p = 1 5 β  .
Note: the expression ( β )  (*)( β )   can also be obtained using the identity
( 1 + x + x 2 + β― + x n ) 2 = 1 + 2 x + 3 x 2 + β― + ( n + 1 ) x n + β― + 3 x 2 n β 2 + 2 x 2 n β 1 + x 2 n  \left(1+x+x^{2}+\cdots+x^{n}\right)^{2}=1+2 x+3 x^{2}+\cdots+(n+1) x^{n}+\cdots+3 x^{2 n-2}+2 x^{2 n-1}+x^{2 n}
( 1 + x + x 2 + β― + x n ) 2 = 1 + 2 x + 3 x 2 + β― + ( n + 1 ) x n + β― + 3 x 2 n β 2 + 2 x 2 n β 1 + x 2 n 
 
The problems on this page are the property of the MAA's American Mathematics Competitions