Inspired by Otto Bretscher
If ω is a primitive 1729th root of unity. Then find the value of
k = 1 ∑ 1 7 2 8 1 + ω k + ω 2 k + ω 3 k 1 .
The answer is 432.
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2 solutions
Moderator note:
Oh wow, that's an interesting conversion of the expression.
To what extent is it relevant that 1 7 2 8 = 4 × 4 3 2 ?
interesting easy solution. mine was to break down the down into ( w k + 1 ) ( w k + i ) ( w k − i ) , make fractions inform of ( x − w k ) 1 and using p ( x ) p ′ ( x ) for p ( x ) = x − 1 x 1 7 2 9 − 1
how did w^k change to w^(1+3*1729) in second step
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I need a j so that 1 + j ∗ 1 7 2 9 is divisible by 4, to make the next two steps work.
Oh wow, that's an interesting conversion of the expression.
To what extent is it relevant that 1 7 2 8 = 4 × 4 3 2 ?
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In the second step I'm using the fact that 1 7 2 9 ≡ 1 ( m o d 4 ) , so, yes, it is relevant that 1 7 2 8 is divisible by 4.
just for the sake of answer you can use omega=1 and find the solution
since 1729 is congruent to 1 modulo 4, the sum is equal to (1729-1)/4 which is equal to 432
All we need are finite geometric series: k = 1 ∑ 1 7 2 8 w 4 k − 1 w k − 1 = k = 1 ∑ 1 7 2 8 w 4 k − 1 w ( 1 + 3 ∗ 1 7 2 9 ) k − 1 = k = 1 ∑ 1 7 2 8 w 4 k − 1 w 5 1 8 8 k − 1 = k = 1 ∑ 1 7 2 8 j = 0 ∑ 1 2 9 6 w 4 k j = 1 7 2 8 + j = 1 ∑ 1 2 9 6 k = 1 ∑ 1 7 2 8 ( w 4 j ) k = 1 7 2 8 + j = 1 ∑ 1 2 9 6 ( − 1 ) = 1 7 2 8 − 1 2 9 6 = 4 3 2