Problem: z 1 , z 2 , … , z 673 z 1 , z 2 , … , z 6 7 3
( x − z 1 ) 3 ( x − z 2 ) 3 ⋯ ( x − z 673 ) 3 ( x − z 1 ) 3 ( x − z 2 ) 3 ⋯ ( x − z 6 7 3 ) 3
can be expressed as x 2019 + 20 x 2018 + 19 x 2017 + g ( x ) x 2 0 1 9 + 2 0 x 2 0 1 8 + 1 9 x 2 0 1 7 + g ( x ) g ( x ) g ( x ) 2016 2 0 1 6
∣ ∑ 1 ≤ j < k ≤ 673 z j z k ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ 1 ≤ j < k ≤ 6 7 3 ∑ z j z k ∣ ∣ ∣ ∣ ∣ ∣ ∣
can be expressed in the form m n n m m m n n m + n m + n
Solution:
Because each root of the polynomial appears with multiplicity 3, Viète's Formulas show that
z 1 + z 2 + ⋯ + z 673 = − 20 3 z 1 + z 2 + ⋯ + z 6 7 3 = − 3 2 0
and
z 1 2 + z 2 2 + ⋯ + z 673 2 = 1 3 ( ( − 20 ) 2 − 2 ⋅ 19 ) = 362 3 z 1 2 + z 2 2 + ⋯ + z 6 7 3 2 = 3 1 ( ( − 2 0 ) 2 − 2 ⋅ 1 9 ) = 3 3 6 2
Then the identity
( ∑ i = 1 673 z i ) 2 = ∑ i = 1 673 z i 2 + 2 ( ∑ 1 ≤ j < k ≤ 673 z j z k ) ( i = 1 ∑ 6 7 3 z i ) 2 = i = 1 ∑ 6 7 3 z i 2 + 2 ⎝ ⎛ 1 ≤ j < k ≤ 6 7 3 ∑ z j z k ⎠ ⎞
shows that
∑ 1 ≤ j < k ≤ 673 z j z k = ( − 20 3 ) 2 − 362 3 2 = − 343 9 1 ≤ j < k ≤ 6 7 3 ∑ z j z k = 2 ( − 3 2 0 ) 2 − 3 3 6 2 = − 9 3 4 3
The requested sum is 343 + 9 = 352 3 4 3 + 9 = 3 5 2 z 673 = − 20 3 z 6 7 3 = − 3 2 0 i = 1 , 2 , 3 , … , 336 i = 1 , 2 , 3 , … , 3 3 6
z i = 343 i 9 ∑ j = 1 336 j and z i + 336 = − z i z i = 9 ∑ j = 1 3 3 6 j 3 4 3 i and z i + 3 3 6 = − z i
Then
∑ i = 1 673 z i = − 20 3 and ∑ i = 1 673 z i 2 = 2 ∑ i = 1 336 343 i 9 ∑ j = 1 336 j + ( 20 3 ) 2 = 362 3 i = 1 ∑ 6 7 3 z i = − 3 2 0 and i = 1 ∑ 6 7 3 z i 2 = 2 i = 1 ∑ 3 3 6 9 ∑ j = 1 3 3 6 j 3 4 3 i + ( 3 2 0 ) 2 = 3 3 6 2
as required.
OR OR
There are constants a a b b
( x − z 1 ) ( x − z 2 ) ( x − z 3 ) ⋯ ( x − z 673 ) = x 673 + a x 672 + b x 671 + ⋯ . ( x − z 1 ) ( x − z 2 ) ( x − z 3 ) ⋯ ( x − z 6 7 3 ) = x 6 7 3 + a x 6 7 2 + b x 6 7 1 + ⋯ .
Then
( x 673 + a x 672 + b x 671 + ⋯ ) 3 = x 2019 + 20 x 2018 + 19 x 2017 + ⋯ ( x 6 7 3 + a x 6 7 2 + b x 6 7 1 + ⋯ ) 3 = x 2 0 1 9 + 2 0 x 2 0 1 8 + 1 9 x 2 0 1 7 + ⋯
Comparing the x 2018 x 2 0 1 8 x 2017 x 2 0 1 7 3 a = 20 3 a = 2 0 3 a 2 + 3 a 2 + 3 b = 19 3 b = 1 9 a = 20 3 a = 3 2 0 b = − 343 9 b = − 9 3 4 3 ∣ ∑ 1 ≤ j < k ≤ 673 z j z k ∣ = ∣ b ∣ = 343 9 ∣ ∣ ∣ ∣ ∑ 1 ≤ j < k ≤ 6 7 3 z j z k ∣ ∣ ∣ ∣ = ∣ b ∣ = 9 3 4 3
The problems on this page are the property of the MAA's American Mathematics Competitions