TITLE: How to evaluate $\sum_{n=1}^{38}\sin\left(\frac{n^8\pi}{38}\right)$
QUESTION [12 upvotes]: Evaluate $$\sum_{n=1}^{38}\sin\left(\frac{n^8\pi}{38}\right)$$
I have found the problem on this page.
I have no idea how to do it. Thank you very much.

REPLY [2 votes]: Again not a solution, but I noted that
$$
\left( \sum_{n=1}^{14} n^8 \mod (2 \times 14) \right) \mod (2 \times 14) = 7,\\
\sum_{n=1}^{14} \sin \left( \frac{\pi n^8}{14} \right) = \sqrt{7}.
$$
$$
\left( \sum_{n=1}^{22} n^8 \mod (2 \times 22) \right) \mod (2 \times 22) = 11,\\
\sum_{n=1}^{22} \sin \left( \frac{\pi n^8}{22} \right) = \sqrt{11}.
$$
$$
\left( \sum_{n=1}^{38} n^8 \mod (2 \times 38 ) \right) \mod (2 \times 38) = 19,\\
\sum_{n=1}^{38} \sin \left( \frac{\pi n^8}{38} \right) = \sqrt{19}.
$$

And other strange things:
$$
\left( \sum_{n=1}^{10} n^2 \mod (2 \times 10) \right) \mod (2 \times 10) = 5,\\
\sum_{n=1}^{10} \sin \left( \frac{\pi n^2}{10} \right) = \sqrt{5}.
$$
$$
\left( \sum_{n=1}^{14} n^2 \mod (2 \times 14) \right) \mod (2 \times 14) = 5,\\
\sum_{n=1}^{14} \sin \left( \frac{\pi n^2}{14} \right) = \sqrt{7}.
$$
$$
\left( \sum_{n=1}^{22} n^2 \mod (2 \times 22) \right) \mod (2 \times 22) = 11,\\
\sum_{n=1}^{22} \sin \left( \frac{\pi n^2}{22} \right) = \sqrt{11}.
$$
$$
\left( \sum_{n=1}^{26} n^2 \mod (2 \times 26) \right) \mod (2 \times 26) = 13,\\
\sum_{n=1}^{26} \sin \left( \frac{\pi n^2}{26} \right) = \sqrt{13}.
$$

Let us write
$$
Q(p,q) = \left( \sum_{n=1}^{2p} n^q \mod (4p) \right) \mod (4p),\\
S(p,q) = \sum_{n=1}^{2p} \sin \left( \frac{\pi n^q}{2p} \right).
$$
Then we find
$$
\begin{array}{cccc}
p & q & Q(p,q) & S(p,q)\\
\hline
5 & 2 & 5 & \sqrt{5}\\
7 & 2 & 7 & \sqrt{7}\\
11 & 2 & 11 & \sqrt{11}\\
13 & 2 & 13 & \sqrt{13}\\
17 & 2 & 17 & \sqrt{17}\\
19 & 2 & 19 & \sqrt{19}\\
23 & 2 & \color{red}{21} & \sqrt{23}\\
\hline
7 & 4 & 7 & \sqrt{7}\\
11 & 4 & 11 & \sqrt{11}\\
38 & 4 & 19 & \sqrt{19}\\
\hline
7 & 8 & 7 & \sqrt{7}\\
11 & 8 & 11 & \sqrt{11}\\
38 & 8 & 19 & \sqrt{19}\\
\end{array}
$$
Hope this gives some insight for others...