SOLUTIONS

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26. [1990, 140; 1991, 152] Proposed by Stanley Rabinowitz, Westford, Massachusetts.

Prove that

∑_{k=1}^{38} sin \frac{k^{8}π}{38} = \sqrt{19}.

Solution by Les Reid, Southwest Missouri State University, Springfield, Missouri. We will show more generally that if p is a prime number, p ≡ 3 mod 4, and m ≥ 1, then

∑_{k=1}^{2p} sin \frac{k^{2^{m}}π}{2p} = ∑_{k=1}^{2p} cos \frac{k^{2^{m}}π}{2p} = \sqrt{p}.

Let ω = cos π/(2p) + i sin π/(2p) and consider ∑_{k=1}^{4p} ω^{k^{2^{m}}}. Since ω is a primitive 4p^{th} root of unity, it suffices to compute the exponents modulo 4p.

We claim that ∑_{k=1}^{4p} ω^{k^{2^{m}}} is independent of m for m ≥ 1. To show this it suffices to show that {k^{2}|k ∈ ℤ_{4p}} = {k^{4}|k ∈ ℤ_{4p}} and the claim will follow by induction. Since {k^{4}|k ∈ ℤ_{4p}} ⊆ {k^{2}|k ∈ ℤ_{4p}}, we will be done if we can construct a bijection from the superset to the subset. We claim that the map f(x) = x^{2} does the trick. Now ℤ_{4p} ≅ ℤ_{4} × ℤ_{p}, by the Chinese Remainder Theorem. There are four types of squares in ℤ_{4} × ℤ_{p}: (0, 0), (1, 0), (0, u^{2}), and (1, u^{2}) where u ≠ 0 ∈ ℤ_{p}. Squaring clearly leaves the first two types invariant. For the second two types, the first coordinate is unchanged by squaring. The second coordinate is from the multiplicative abelian group (ℤ_{p}^{×})^{2} which has order (p − 1)/2 (since p is an (odd) prime). This is odd since p ≡ 3 mod 4. Therefore squaring yields a group isomorphism (on the second coordinate), and hence f is a bijection.

A classic result of Gauss [for example, see Lang’s Algebraic Number Theory, pp.85-87] states that if α = cos 2π/b + i sin 2π/b for b > 0, then

∑_{k=1}^{b} α^{k^{2}} = \{ \begin{matrix}(1 + i)\sqrt{b} & if b ≡ 0 mod 4 \\ \sqrt{b} & if b ≡ 1 mod 4 \\ 0 & if b ≡ 2 mod 4 \\ i\sqrt{b} & if b ≡ 3 mod 4 \end{matrix}

In our case, b = 4p, so ∑_{k=1}^{4p} ω^{k^{2}} = (1+i)\sqrt{4p}. Finally ∑_{k=1}^{4p} ω^{k^{2^{m}}} = 2 ∑_{k=1}^{2p} ω^{k^{2^{m}}}, so

\begin{matrix}∑_{k=1}^{2p}(cos \frac{k^{2^{m}}π}{2p} + i sin \frac{k^{2^{m}}π}{2p}) = ∑_{k=1}^{2p} ω^{k^{2^{m}}} = \frac{1}{2} ∑_{k=1}^{4p} ω^{k^{2^{m}}} \\ = \frac{1}{2} ∑_{k=1}^{4p} ω^{k^{2}} = \frac{1}{2}(1 + i)\sqrt{4p} = (1 + i)\sqrt{p}. \end{matrix}

Comparing real and imaginary parts, the result follows.