Manuel Kauers Hagenberg, December 17, 2002 RISC-Linz

Algorithmic Algebraic Geometry

Solution 1

a) Let ω be a primitive n-th root of unity, if n is odd and a 2n-th root of unity, if n is even. Then, I claim that

(x_{1}, x_{2}, ..., x_{n}) := \{ \begin{matrix}(ω^{1}, ω^{2}, ..., ω^{n}) & n odd \\ (ω^{1}, ω^{3}, ..., ω^{n−1}) & n even \end{matrix}

solves the system. The first and the last equation can be verified immediately using well-known facts about roots of unity:

• n odd:

∑_{i=1}^{n} ω^{i} = \frac{ω^{n+1} − ω}{ω − 1} = \frac{ω − ω}{ω − 1} = 0.

∏_{i=1}^{n} ω^{i} = ω^{∑_{i=1}^{n} i} = ω^{\frac{n(n−1)}{2}} = ω^{0} = 1.

• n even:

∑_{i=1}^{n} ω^{2i−1} = ω^{−1} ∑_{i=1}^{n} ω^{2i} = ω^{−1} \frac{ω^{2(n+1)} − ω^{2}}{ω^{2} − 1} = ω \frac{\overbrace{ω^{2n}}^{=1} −1}{ω^{2} − 1} = 0.

∏_{i=1}^{n} ω^{2i−1} = ω^{−n} ∏_{i=1}^{n} ω^{2i} = − ∏_{i=1}^{n}(ω^{2})^{i} = −ω^{n(n−1)} = −1.

In a similar (but lengthier) fashion, I was able to prove also the second equation,

∑_{i=1}^{n−1} ∑_{j=i+1}^{n} ω^{i+j} = 0,

but unfortunately, I did not succeed to prove the validity of the other equations in general. :-(

However, if we accept my claim then every permutation of the ω^{i} is another solution (since ω^{i} ≠ ω^{j} for i ≠ j) so that we have n! solution altogether, as desired.