Prove that $$\sum_{1\leq k < j\leq n}\tan^2\left(\frac{k\pi}{2n+1}\right)\tan^2\left(\frac{j\pi}{2n+1}\right)=\binom{2n+1}{4}$$
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1$\begingroup$ Where does this problem come from ? What have you attempted ? $\endgroup$– Jean MarieCommented Aug 2, 2016 at 6:32
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$\begingroup$ It's nice identity, I have just detected it 😉 $\endgroup$– hxthanhCommented Aug 2, 2016 at 6:51
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$\begingroup$ Related math.stackexchange.com/questions/217240/… . This identity can be proved by combining the sums for squares and fourth powers of tangents from this link functions.wolfram.com/ElementaryFunctions/Tan/23/01 $\endgroup$– NemoCommented Aug 2, 2016 at 7:10
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$\begingroup$ See math.stackexchange.com/questions/1562037/… $\endgroup$– lab bhattacharjeeCommented Aug 6, 2016 at 5:21
1 Answer
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.
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1$\begingroup$ Clair et net ! (the way I would have answered it). +1 $\endgroup$ Commented Aug 2, 2016 at 18:57