Is there an identity for $\sum_{k=0}^{n-1}\tan^4\left({k\pi\over n}\right)$? Is there a simple relation for $\sum_{k=0}^{n-1}\tan^4\left({k\pi\over n}\right)$
like there is for $\sum_{k=0}^{n-1}\tan^2\left({k\pi\over n}\right)$?
Looking at Jolley, Summation of Series, formula 445:
$\sum_{k=0}^{n-1}\tan^2\left(\theta+{k\pi\over n}\right)=n^2\cot^2\left({n\pi\over2}+n\theta\right)+n(n-1)$
or more simply:
$\sum_{k=0}^{n-1}\tan^2\left({k\pi\over n}\right)=n(n-1)$
Does $\sum_{k=0}^{n-1}\tan^4\left({k\pi\over n}\right)$ possibly equals some integer values?
The numerical table seems to suggest so even for higher even powers of $\tan$:
$\left(
\begin{array}{ccccc}
n & \sum_{k=0}^{n-1}\tan^2\left({k\pi\over n}\right)&\sum_{k=0}^{n-1}\tan^4\left({k\pi\over n}\right)&\sum_{k=0}^{n-1}\tan^6\left({k\pi\over n}\right)&\sum_{k=0}^{n-1}\tan^8\left({k\pi\over n}\right)\\
 3. & 6. & 18. & 54. & 162. \\
 5. & 20. & 180. & 1700. & 16100. \\
 7. & 42. & 742. & 14154. & 271558. \\
 9. & 72. & 2088. & 66600. & 2.14049\times 10^6 \\
 11. & 110. & 4730. & 226622. & 1.09528\times 10^7 \\
 13. & 156. & 9308. & 624780. & 4.2335\times 10^7 \\
 15. & 210. & 16590. & 1.48533\times 10^6 & 1.34314\times 10^8 \\
\end{array}
\right)$
I had tried to use residue methods and I particularly like falager's method in this answer using Vieta's formula but kept getting stuck.
If one could make progress with the quartic, one could possibly find for higher even powers of tan too.
 A: A possible approach is to use (here $n$ is still odd!) $$\prod_{k=1}^{n-1}\left(1+x^2\tan^2\frac{k\pi}{n}\right)=\left(\frac{(1+x)^n+(1-x)^n}{2}\right)^2$$ obtained by factoring the RHS over $\mathbb{C}$, or as $P_n(1+x,x-1)/P_n(1,-1)$ where $$P_n(a,b)=\prod_{k=0}^{n-1}\left(a^2+b^2-2ab\cos\frac{2k\pi}{n}\right)=(a^n-b^n)^2$$ comes from factoring $z^n-1$ over $\mathbb{C}$. Denoting $S_{2m}:=\sum_{k=1}^{n-1}\tan^{2m}(k\pi/n)$, we get
\begin{align}
\sum_{m=1}^\infty\frac{(-1)^m}{m}S_{2m}x^{2m}&=-\sum_{k=1}^{n-1}\log\left(1+x^2\tan^2\frac{k\pi}{n}\right)
\\&=-2\log\frac{(1+x)^n+(1-x)^n}{2}
\\&=-2\log\sum_{k=0}^{\lfloor n/2\rfloor}\binom{n}{2k}x^{2k}
\\&=2\sum_{m=1}^\infty\frac{(-1)^m}{m}\left(\sum_{k=1}^{\lfloor n/2\rfloor}\binom{n}{2k}x^{2k}\right)^m.
\end{align}
Comparing the individual coefficients, we get
\begin{align}
S_2&=2\binom{n}{2}&&=n^2-n,\\
S_4&=2\binom{n}{2}^2-4\binom{n}{4}&&=\color{blue}{\frac{n^4-4n^2+3n}{3}},\\
S_6&=2\binom{n}{2}^3-6\binom{n}{2}\binom{n}{4}+6\binom{n}{6}&&=\frac{2n^6-10n^4+23n^2-15n}{15},
\end{align}
and so on; say, $S_8=(17n^8-112n^6+308n^4-528n^2+315n)/315$.
