Waiting for 2016! - 4

Let $w=e^{2\pi i/n}$ be a n-th primitive root of unity and $P(x)=x^n-1$ . Then we know that $\cot\left(\dfrac{\pi k}{n}\right)=i\dfrac{w^k+1}{w^k-1}$ . Let $y=\dfrac{x+1}{x-1}$ for $x \neq 1$ , so $x=\dfrac{y+1}{y-1}$ .

We want to find $\displaystyle \mathcal{S}=\sum_{k=1}^{(n-1)/2} \cot^4 \left(\dfrac{\pi k}{n}\right)=\dfrac{1}{2}\sum_{k=1}^{n-1} \cot^4 \left(\dfrac{\pi k}{n}\right)=\dfrac{1}{2}\sum_{k=1}^{n-1} y_k^4$ for odd $n$ and for every $y$ .

So, $P(x)=\left(\dfrac{y+1}{y-1}\right)^n-1 \implies (y-1)^nP(x)=(y+1)^n-(y-1)^n$

Let $(y-1)^nP(x)=Q(y)$ . This new polynomial is of degree $n-1$ as desired (because we discard $x=1$ ):

$\displaystyle Q(y)=2ny^{n-1}+2\binom{n}{3}y^{n-3}+2\binom{n}{5}y^{n-5}+\cdots$

Finally we'll use Netwon's Sums to obtain $\mathcal{S}$ . Let $P_m$ be the m-powers sum of the roots of $Q(y)$ and $S_1,S_2,\cdots,S_{n-1}$ be the elementary symmetric sums of the same roots. Then $\mathcal{S}=\dfrac{1}{2}P_4$ and by Vieta's formulas:

$S_1=0,S_2=\dfrac{\binom{n}{3}}{n},S_3=0,S_4=\dfrac{\binom{n}{5}}{n}$

$P_1=S_1=0$

$P_2=S_1P_1-2S_2=-\dfrac{2\binom{n}{3}}{n}$

$P_3=S_1P_2-S_2P_1+3S_3=0$

$P_4=S_1P_3-S_2P_2+S_3P_1-4S_4=2\left(\dfrac{\binom{n}{3}}{n}\right)^2-4\left(\dfrac{\binom{n}{5}}{n}\right)$

Expanding we get:

$P_4=\dfrac{2(n-1)(n-2)(n^2+3n-13)}{90}$

Then $\mathcal{S}=\dfrac{(n-1)(n-2)(n^2+3n-13)}{90}$ .

Let $n=4033$ , then $\lfloor 10^{-7} \mathcal{S} \rfloor=\boxed{293947}$ .