How to obtain the complete factorization of $P_n(X)=\frac{(X+i)^{2n+1} - (X-i)^{2n+1}}{2i} = 0$ I need to give the complete factorization of $P_n(X)=\frac{(X+i)^{2n+1} - (X-i)^{2n+1}}{2i} = 0$ and show that $P_n(X)=(2n+1)\prod_{k=1}^{n}(X^2-\frac{1}{t_{n,k}^{2}})$ with $t_{n,k}^{2}=\tan(\frac{k\pi}{2n+1})$.
Here is what I have done so far:
$$P_n(X)=\sum_{k=0}^{n}(-1)^{n-k}\binom{2n+1}{2k}X^{2k}=(2n+1)\left(X^{2n}+(-1)\frac{(1-2n)}{3}X^{2n-2}+...+(-1)^{n-1}nX^2+\frac{(-1)^n}{2n+1}\right)$$
But now I don't know how to progress further to get the asked result. I know that i can factorize a polynomial using the dominant coefficient $(2n+1)$ followed by the products of $(X-t_1)(X-t_2)...$ where $t_i$ are the roots, but how can i get the factorization with $X^2$ ?
 A: In this answer to your question we showed that
$x= 1/\tan \theta_k$ where $\theta_k=\frac{k\pi}{2n+1}$ are the solutions to $P_n(x)=0.$  There we showed $k\ne0$ and suggested $k\in\{1,2,\cdots,2n\}$.  Another set of $k$'s that work are $\{ \mp1,\mp2,\cdots,\mp n\}.$  We also showed that $P_n(x) = (2n+1) (x^{2n} + \text{lower order terms}).$
Thus
$$\begin{aligned} 
P_n(x) &= (2n+1) \prod_{k=1}^n \left(x- \frac{1}{\tan \theta_k}\right)\left(x+ \frac{1}{\tan \theta_k}\right)\\
&=(2n+1) \prod_{k=1}^n \left( x^2-\frac{1}{\tan^2 \theta_k}\right)\\
&=(2n+1) \prod_{k=1}^n \left( x^2-\frac{1}{\tan^2 \frac{k\pi}{2n+1}}\right)\\
&=(2n+1) \prod_{k=1}^n \left( x^2-{\cot^2 \frac{k\pi}{2n+1}} \right).
\end{aligned}$$
A: You already found out that $P_n$ has degree $2n$ and got the dominant coefficient. It remains to find the roots.
First, notice that $i$ is not a root of $P_n$. If $z$ is a root of $P_n$, then we necessarily have that $\left(\frac{z+i}{z-i}\right)^{2n+1}=1$ hence $\frac{z+i}{z-i}$ is a $(2n+1)$-th root of unity which is not one. For each of such roots of unity $\omega_{n,k}$, $1\leqslant k\leqslant 2n$, we can solve $\frac{z+i}{z-i}=\omega_{n,k}$.
