Let w = e^{2*pi*i/17}. Define S = w^1 + w^2 + w^4 + w^8 + w^9 + w^13…

Question

Let w = e^{2*pi*i/17}. Define S = w^1 + w^2 + w^4 + w^8 + w^9 + w^13 + w^15 + w^16, where the exponents are exactly the quadratic residues modulo 17. Compute the exact value of (2S + 1)^2. Report the integer.

Accepted Answer

S is the quadratic Gauss period for the prime 17: the sum of w^a over the eight quadratic residues a modulo 17. The non-residues give the complementary sum S'. Since sum_{a=1}^{16} w^a = -1 (sum of all non-trivial 17th roots), we have S + S' = -1. The classical Gauss-period identity gives S*S' = (1 - p)/4 = (1 - 17)/4 = -4 for p = 17 (since p = 1 mod 4). Then S and S' are roots of t^2 + t - 4 = 0, so S = (-1 + sqrt(17))/2 (the residue sum being the larger real root). Therefore 2S + 1 = sqrt(17) and (2S + 1)^2 = 17.
Final answer: 17

Final answer: 17

Let w = e^{2pii/17}. Define S = w^1 + w^2 + w^4 + w^8 + w^9 + w^13…

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