Problem:
P ( x ) = 24 x 24 + ∑ j = 1 23 ( 24 − j ) ( x 24 − j + x 24 + j ) P ( x ) = 2 4 x 2 4 + j = 1 ∑ 2 3 ( 2 4 − j ) ( x 2 4 − j + x 2 4 + j )
Let z 1 , z 2 , … , z r z 1 , z 2 , … , z r P ( x ) P ( x ) z k 2 = a k + b k i z k 2 = a k + b k i k = 1 , 2 , … , r k = 1 , 2 , … , r i = − 1 i = − 1 a k a k b k b k
∑ k = 1 r ∣ b k ∣ = m + n p k = 1 ∑ r ∣ b k ∣ = m + n p
where m , n m , n p p p p m + n + p m + n + p
Solution:
Note that
P ( x ) = x + 2 x 2 + 3 x 3 + ⋯ + 24 x 24 + 23 x 25 + 22 x 26 + ⋯ + 2 x 46 + x 47 P ( x ) = x + 2 x 2 + 3 x 3 + ⋯ + 2 4 x 2 4 + 2 3 x 2 5 + 2 2 x 2 6 + ⋯ + 2 x 4 6 + x 4 7
and
x P ( x ) = x 2 + 2 x 3 + 3 x 4 + ⋯ + 24 x 25 + 23 x 26 + ⋯ + 2 x 47 + x 48 x P ( x ) = x 2 + 2 x 3 + 3 x 4 + ⋯ + 2 4 x 2 5 + 2 3 x 2 6 + ⋯ + 2 x 4 7 + x 4 8
So
( 1 − x ) P ( x ) = x + x 2 + ⋯ + x 24 − ( x 25 + x 26 + ⋯ + x 47 + x 48 ) = ( 1 − x 24 ) ( x + x 2 + ⋯ + x 24 ) ( 1 − x ) P ( x ) = x + x 2 + ⋯ + x 2 4 − ( x 2 5 + x 2 6 + ⋯ + x 4 7 + x 4 8 ) = ( 1 − x 2 4 ) ( x + x 2 + ⋯ + x 2 4 )
Then, for x ≠ 1 x = 1
P ( x ) = x 24 − 1 x − 1 x ( 1 + x + ⋯ + x 23 ) = x ( x 24 − 1 x − 1 ) 2 (*) P ( x ) = x − 1 x 2 4 − 1 x ( 1 + x + ⋯ + x 2 3 ) = x ( x − 1 x 2 4 − 1 ) 2 ( * )
One zero of P ( x ) P ( x ) 0 0 P ( x ) P ( x ) ( x 24 − 1 ) 2 ( x 2 4 − 1 ) 2 1 1 ( x 24 − 1 ) 2 ( x 2 4 − 1 ) 2 x 24 − 1 x 2 4 − 1 P ( x ) P ( x ) z k = z k = 15 k ∘ 1 5 k ∘ k = 1 , 2 , 3 , ⋯ , 23 k = 1 , 2 , 3 , ⋯ , 2 3 30 k ∘ 3 0 k ∘
∑ k = 1 23 ∣ sin 30 k ∘ ∣ = 4 ∑ k = 1 5 ∣ sin 30 k ∘ ∣ = 4 ( 2 ⋅ ( 1 / 2 ) + 2 ⋅ ( 3 / 2 ) + 1 ) = 8 + 4 3 k = 1 ∑ 2 3 ∣ sin 3 0 k ∘ ∣ = 4 k = 1 ∑ 5 ∣ sin 3 0 k ∘ ∣ = 4 ( 2 ⋅ ( 1 / 2 ) + 2 ⋅ ( 3 / 2 ) + 1 ) = 8 + 4 3
Thus m + n + p = 15 m + n + p = 1 5
Note: the expression ( ∗ ) ( ∗ )
( 1 + x + x 2 + ⋯ + x n ) 2 = 1 + 2 x + 3 x 2 + ⋯ + ( n + 1 ) x n + ⋯ + 3 x 2 n − 2 + 2 x 2 n − 1 + x 2 n ( 1 + x + x 2 + ⋯ + x n ) 2 = 1 + 2 x + 3 x 2 + ⋯ + ( n + 1 ) x n + ⋯ + 3 x 2 n − 2 + 2 x 2 n − 1 + x 2 n
The problems on this page are the property of the MAA's American Mathematics Competitions