In evaluating the sum:
$$\tag{1}\label{e1}\sum\limits_{j = 1 < h}^N {\left( {{e^{\frac{{i\pi }}{N}\left( {{s_1}j + {s_2}h} \right)}} - {e^{\frac{{i\pi }}{N}\left( {{s_1}h + {s_2}j} \right)}}} \right)} $$
where both ${s_1}$ and ${s_2}$ are fixed odd numbers not equal to each other, I found that the sum in (\ref{e1}) is always zero unless ${s_1} + {s_2} = 2mN$. I can prove this result by hard calculation described below, but I wonder if there is a more elegant/insightful way of proving it, which may involve some algebraic properties of these roots of unity.
Hard calculation by converting it into a geometric series:
$$\tag{2}\label{e2}\displaylines{
\sum\limits_{j = 1 < h}^N {{e^{i\frac{\pi }{N}\left[ {{s_1}j + {s_2}h} \right]}}} = \sum\limits_{k = 1}^{N - 1} {\sum\limits_{n = 1}^{N - k} {{e^{i\frac{\pi }{N}\left[ {\left( {{s_1} + {s_2}} \right)k + {s_2}n} \right]}}} } \cr
= 2{e^{i\frac{{\left( {{s_1} + {s_2}} \right)\pi }}{N}}} \cdot {\left( {1 - {e^{i\frac{{{s_1}\pi }}{N}}}} \right)^{ - 1}} \cdot {\left( {1 - {e^{i\frac{{{s_2}\pi }}{N}}}} \right)^{ - 1}} \cr} $$
when ${s_1}$ and ${s_2}$ are fixed odd numbers with ${s_1} + {s_2} \ne 2mN$. I notice the result in (\ref{e2}) is invariant with ${s_1}$ and ${s_2}$ exchanged and hence the original sum in (\ref{e1}) equals to zero. If indeed ${s_1} + {s_2} = 2mN$ then
$$\tag{3}\label{e3}\displaylines{
\sum\limits_{j = 1 < h}^N {{e^{i\frac{\pi }{N}\left[ {{s_1}j + {s_2}h} \right]}}} = \sum\limits_{k = 1}^{N - 1} {\sum\limits_{n = 1}^{N - k} {{e^{i\frac{\pi }{N}\left[ {\left( {{s_1} + {s_2}} \right)k + {s_2}n} \right]}}} } \cr
= \sum\limits_{k = 1}^{N - 1} {\sum\limits_{n = 1}^{N - k} {{e^{i\frac{{{s_2}\pi }}{N}n}}} } \cr
= \left( {N{e^{i\frac{{{s_2}\pi }}{N}}} + 2 - N} \right){\left( {2 - 2\cos \frac{{{s_2}\pi }}{N}} \right)^{ - 1}} \cr} $$
and therefore the original sum in (\ref{e1}) is not zero.
