Prove that $\displaystyle \sum_{1\leq k<j\leq n} \tan^2\left(\frac{k\pi}{2n+1}\right)\tan^2\left(\frac{j\pi}{2n+1}\right)=\binom{2n+1}{4} $

Question

Prove that $$\sum_{1\leq k < j\leq n}\tan^2\left(\frac{k\pi}{2n+1}\right)\tan^2\left(\frac{j\pi}{2n+1}\right)=\binom{2n+1}{4}$$

Answer 1

It is enough to recall Cauchy's proof of the Basel problem, relying on the identity:

$$ \binom{2n+1}{1}t^n-\binom{2n+1}{3}t^{n-1}+\ldots+(-1)^n\binom{2n+1}{2n+1}=\prod_{k=1}^{n}\left(t-\cot^2\frac{k\pi}{2n+1}\right)\tag{1} $$ If we consider the "reciprocal polynomial" $$ \binom{2n+1}{2n+1}t^n-\binom{2n+1}{2n-1}t^{n-1}+\ldots+(-1)^n\binom{2n+1}{1}=\prod_{k=1}^{n}\left(t-\tan^2\frac{k\pi}{2n+1}\right)\tag{2} $$ the original sum is just the second elementary symmetric polynomial of the roots, $e_2$, that by Vieta's theorem is given by the coefficient of $t^{n-2}$ in the LHS of $(2)$, so: $$ \sum_{1\leq j<k\leq n}\tan^2\left(\frac{j\pi}{2n+1}\right)\tan^2\left(\frac{k\pi}{2n+1}\right)=\binom{2n+1}{2n-3}=\color{red}{\binom{2n+1}{4}}\tag{3}$$ as wanted.