Annoying Reciprocals
If is a complex number satisfying , then for all positive integers :
where and are positive integers. Find .
The answer is 6.
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1 solution
The complex roots of x 2 + x + 1 = 0 are the third root of unity ω . ⟹ ω 2 + ω + 1 = 0 and ω 3 = 1 . Therefore,
S = k = 1 ∑ n ( x k + x k 1 ) 2 = k = 1 ∑ n ( ω k + ω k 1 ) 2 = k = 1 ∑ n ( ω k + ω k ω 3 k ) 2 = k = 1 ∑ n ( ω k + ω 2 k ) 2 = k = 1 ∑ n ( ω 2 k + 2 ω 3 k + ω 4 k ) = k = 1 ∑ n ( ω 2 k + ω k + 2 )
= = = = ω 2 + ω + 2 ω 2 + ω + 2 − 1 + 2 1 + ω 4 + ω 2 + 2 + ω + ω 2 + 2 − 1 + 2 + 1 + ω 6 + ω 3 + 2 + 1 + 1 + 2 + 2 + 2 + 4 + ω 8 + ω 4 + 2 + ω 2 + ω + 2 − 1 + 2 + 1 + ω 1 0 + ω 5 + 2 + ω + ω 2 + 2 − 1 + 2 + 1 + ω 1 2 + ω 6 + 2 + 1 + 1 + 2 + 2 + 2 + 4 + ⋯ + ⋯ + ⋯ + ⋯ + ⋯ + ⋯ + ⋯ + ⋯ + ⋯ + ω 2 n + ω n + 2 + ω 2 n + ω n + 2 + ω 2 n + ω n + 2 + ω 2 n + ω n + 2 Note that ω + ω 2 = − 1
\(\begin{array} {} & = n + 3 \left \lfloor \dfrac n3 \right \rfloor \end{array}\)
⟹ a + b = 3 + 3 = 6