Show that $\prod\limits_{k=1}^n{\left(1+\tan^4{\left(\frac{k\pi}{2n+1}\right)}\right)}$ is the sum of two perfect squares

Question

Let $n$ be a positive integer. Show that :

$$\prod_{k=1}^n{\left(1+\tan^4{\left(\frac{k\pi}{2n+1}\right)}\right)}$$

is the sum of two perfect squares.

Answer 1

Let $ n $ be a positive integer.

Consider the Following polynomial : $ P_{n}=\left(X-\mathrm{i}\right)^{2n+1}+\left(X+\mathrm{i}\right)^{2n+1} $, whose zeros can be given by : \begin{aligned} P_{n}\left(z\right)=0\iff \left(z-\mathrm{i}\right)^{2n+1}=\left(-z-\mathrm{i}\right)^{2n+1}&\iff \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \tiny \ \normalsize z-\mathrm{i}=\mathrm{e}^{\mathrm{i}\frac{2k\pi}{2n+1}}\left(-z-\mathrm{i}\right),\ \ \ \ 0\leq k\leq 2n \\ &\iff z\left(1+\mathrm{e}^{\mathrm{i}\frac{2k\pi}{2n+1}}\right)=\mathrm{i}\left(1-\mathrm{e}^{\mathrm{i}\frac{2k\pi}{2n+1}}\right), \ \ \ \ \ 0\leq k\leq 2n\\ &\iff \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \tiny \ \normalsize z=\tan{\left(\frac{k\pi}{2n+1}\right)}, \ \ \ 0\leq k\leq 2n\end{aligned}

Since $ P_{n} $ can be developed as following :

$ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ P_{n}=\sum\limits_{k=0}^{2n+1}{\binom{2n+1}{k}z^{2n+1-k}\left(-\mathrm{i}\right)^{k}}+\sum\limits_{k=0}^{2n+1}{\binom{2n+1}{k}z^{2n+1-k}\,\mathrm{i}^{k}}=2\sum\limits_{k=0}^{n}{\left(-1\right)^{k}\binom{2n+1}{2k}z^{2n+1-2k}} $

It can be factored as $ P_{n}=2\prod\limits_{k=1}^{2n}{\left(X-\tan{\left(\frac{k\pi}{2n+1}\right)}\right)} $ and thus, $$ \frac{\left(X-\mathrm{i}\right)^{2n+1}+\left(X+\mathrm{i}\right)^{2n+1}}{2X}=\prod\limits_{k=1}^{2n}{\left(X-\tan{\left(\frac{k\pi}{2n+1}\right)}\right)} $$ But, \begin{aligned} \prod\limits_{k=1}^{2n}{\left(X-\tan{\left(\frac{k\pi}{2n+1}\right)}\right)}&=\prod\limits_{k=1}^{n}{\left(X-\tan{\left(\frac{k\pi}{2n+1}\right)}\right)}\prod\limits_{k=n+1}^{2n}{\left(X-\tan{\left(\frac{k\pi}{2n+1}\right)}\right)}\\ &=\prod\limits_{k=1}^{n}{\left(X-\tan{\left(\frac{k\pi}{2n+1}\right)}\right)}\prod\limits_{k=1}^{n}{\left(X-\tan{\left(\frac{\left(2n+1-k\right)\pi}{2n+1}\right)}\right)}\\ &=\prod\limits_{k=1}^{n}{\left(X^{2}-\tan^{2}{\left(\frac{k\pi}{2n+1}\right)}\right)} \end{aligned}

Meaning, $$ \fbox{$\begin{array}{rcl}\displaystyle\frac{\left(X-\mathrm{i}\right)^{2n+1}+\left(X+\mathrm{i}\right)^{2n+1}}{2X}=\prod\limits_{k=1}^{n}{\left(X^{2}-\tan^{2}{\left(\frac{k\pi}{2n+1}\right)}\right)}\end{array}$} $$

Setting $ X\leftarrow\mathrm{e}^{\mathrm{i}\frac{\pi}{4}} $, observing that $ \mathrm{e}^{\mathrm{i}\frac{\pi}{4}}-\mathrm{i}=\sqrt{2-\sqrt{2}}\,\mathrm{e}^{-\mathrm{i}\frac{\pi}{8}} $, and that $ \mathrm{e}^{\mathrm{i}\frac{\pi}{4}}+\mathrm{i}=\sqrt{2+\sqrt{2}}\,\mathrm{e}^{\mathrm{i}\frac{3\pi}{8}} $, we get : $$ \displaystyle\frac{\sqrt{2-\sqrt{2}}\left(2-\sqrt{2}\right)^{n}\mathrm{e}^{-\mathrm{i}\frac{2n+3}{8}\pi}+\sqrt{2+\sqrt{2}}\left(2+\sqrt{2}\right)^{n}\mathrm{e}^{\mathrm{i}\frac{6n+1}{8}\pi}}{2}=\prod\limits_{k=1}^{n}{\left(\mathrm{i}-\tan^{2}{\left(\frac{k\pi}{2n+1}\right)}\right)} $$

Setting $ X\leftarrow\mathrm{e}^{-\mathrm{i}\frac{\pi}{4}} $, observing that $ \mathrm{e}^{-\mathrm{i}\frac{\pi}{4}}-\mathrm{i}=\sqrt{2+\sqrt{2}}\,\mathrm{e}^{-\mathrm{i}\frac{3\pi}{8}} $, and that $ \mathrm{e}^{-\mathrm{i}\frac{\pi}{4}}+\mathrm{i}=\sqrt{2-\sqrt{2}}\,\mathrm{e}^{\mathrm{i}\frac{\pi}{8}} $, we get : $$ \displaystyle\frac{\sqrt{2+\sqrt{2}}\left(2+\sqrt{2}\right)^{n}\mathrm{e}^{-\mathrm{i}\frac{6n+1}{8}\pi}+\sqrt{2-\sqrt{2}}\left(2-\sqrt{2}\right)^{n}\mathrm{e}^{\mathrm{i}\frac{2n+3}{8}\pi}}{2}=\prod\limits_{k=1}^{n}{\left(-\mathrm{i}-\tan^{2}{\left(\frac{k\pi}{2n+1}\right)}\right)} $$

Multiplying everything together gives us : $$ \frac{\left(2-\sqrt{2}\right)^{2n+1}+\left(2+\sqrt{2}\right)^{2n+1}}{4}=\prod\limits_{k=1}^{n}{\left(1+\tan^{4}{\left(\frac{k\pi}{2n+1}\right)}\right)} $$

I think I would let you continue.