# Proving n∑l=1n−1∑k=1tanlkπ2n+1tanl(k+1)π2n+1=0\sum\limits_{l=1}^n \sum\limits _{k=1}^{n-1}\tan \frac {lk\pi }{2n+1}\tan \frac {l(k+1)\pi }{2n+1}=0

Prove that

It is very easy to prove this identity for each fixed $n$ . For example let $n = 6$; writing out all terms in a $5 \times 6$ matrix, we obtain:

$\begin{matrix} \tan \dfrac {\pi } {13}\tan \dfrac {2\pi } {13} & \tan \dfrac {2\pi } {13}\tan \dfrac {3\pi } {13} & \tan \dfrac {3\pi } {13}\tan \dfrac {4\pi } {13} & \tan \dfrac {4\pi } {13}\tan \dfrac {5\pi } {13} & \tan \dfrac {5\pi } {13}\tan \dfrac {6\pi } {13} \\ \tan \dfrac {2\pi } {13}\tan \dfrac {4\pi } {13} & \tan \dfrac {4\pi } {13}\tan \dfrac {6\pi } {13} & \tan \dfrac {6\pi } {13}\tan \dfrac {8\pi } {13} & \tan \dfrac {8\pi } {13}\tan \dfrac {10\pi } {13} & \tan \dfrac {10\pi } {13}\tan \dfrac {12\pi } {13} \\ \tan \dfrac {3\pi } {13}\tan \dfrac {6\pi } {13} & \tan \dfrac {6\pi } {13}\tan \dfrac {9\pi } {13} & \tan \dfrac {9\pi } {13}\tan \dfrac {12\pi } {13} & \tan \dfrac {12\pi } {13}\tan \dfrac {15\pi } {13} & \tan \dfrac {15\pi } {13}\tan \dfrac {18\pi } {13} \\ \tan \dfrac {4\pi } {13}\tan \dfrac {8\pi } {13} & \tan \dfrac {8\pi } {13}\tan \dfrac {12\pi } {13} & \tan \dfrac {12\pi } {13}\tan \dfrac {16\pi } {13} & \tan \dfrac {16\pi } {13}\tan \dfrac {20\pi } {13} & \tan \dfrac {20\pi } {13}\tan \dfrac {24\pi } {13} \\ \tan \dfrac {5\pi } {13}\tan \dfrac {10\pi } {13} & \tan \dfrac {10\pi } {13}\tan \dfrac {15\pi } {13} & \tan \dfrac {15\pi } {13}\tan \dfrac {20\pi } {13} & \tan \dfrac {20\pi } {13}\tan \dfrac {25\pi } {13} & \tan \dfrac {25\pi } {13}\tan \dfrac {30\pi } {13} \\ \tan \dfrac {6\pi } {13}\tan \dfrac {12\pi } {13} & \tan \dfrac {12\pi } {13}\tan \dfrac {18\pi } {13} & \tan \dfrac {18\pi } {13}\tan \dfrac {24\pi } {13} & \tan \dfrac {24\pi } {13}\tan \dfrac {30\pi } {13} & \tan \dfrac {30\pi } {13}\tan \dfrac {36\pi } {13} \end{matrix}$

one can notice then, that the first column vanish the fourth one :

$\tan \dfrac {\pi } {13}\tan \dfrac {2\pi } {13}=-\tan \dfrac {12\pi } {13}\tan \dfrac {15\pi } {13}$

$\tan \dfrac {2\pi } {13}\tan \dfrac {4\pi } {13}=-\tan \dfrac {24\pi } {13}\tan \dfrac {30\pi } {13}$

$\tan \dfrac {3\pi } {13}\tan \dfrac {6\pi } {13}=-\tan \dfrac {16\pi } {13}\tan \dfrac {20\pi } {13}$

$\tan \dfrac {4\pi } {13}\tan \dfrac {8\pi } {13}=-\tan \dfrac {4\pi } {13}\tan \dfrac {5\pi } {13}$

$\tan \dfrac {5\pi } {13}\tan \dfrac {10\pi } {13}=-\tan \dfrac {8\pi } {13}\tan \dfrac {10\pi } {13}$

$\tan \dfrac {6\pi } {13}\tan \dfrac {12\pi } {13}=-\tan \dfrac {20\pi } {13}\tan \dfrac {25\pi } {13}$

and the third column vanish the fifth one :

$\tan \dfrac {3\pi } {13}\tan \dfrac {4\pi } {13}=-\tan \dfrac {30\pi } {13}\tan \dfrac {36\pi } {13}$

$\tan \dfrac {6\pi } {13}\tan \dfrac {8\pi } {13}=-\tan \dfrac {5\pi } {13}\tan \dfrac {6\pi } {13}$

$\tan \dfrac {9\pi } {13}\tan \dfrac {12\pi } {13}=-\tan \dfrac {25\pi } {13}\tan \dfrac {30\pi } {13}$

$\tan \dfrac {12\pi } {13}\tan \dfrac {16\pi } {13}=-\tan \dfrac {10\pi } {13}\tan \dfrac {12\pi } {13}$

$\tan \dfrac {15\pi } {13}\tan \dfrac {20\pi } {13}=-\tan \dfrac {20\pi } {13}\tan \dfrac {24\pi } {13}$

$\tan \dfrac {18\pi } {13}\tan \dfrac {24\pi } {13}=-\tan \dfrac {15\pi } {13}\tan \dfrac {18\pi } {13}$

while the second column is self-vanishing:

$\tan \dfrac {2\pi } {13}\tan \dfrac {3\pi } {13}=-\tan \dfrac {10\pi } {13}\tan \dfrac {15\pi } {13}$

$\tan \dfrac {4\pi } {13}\tan \dfrac {6\pi } {13}=-\tan \dfrac {6\pi } {13}\tan \dfrac {9\pi } {13}$

$\tan \dfrac {8\pi } {13}\tan \dfrac {12\pi } {13}=-\tan \dfrac {12\pi } {13}\tan \dfrac {18\pi } {13}$ .

So the equality occurs.
But how to generalize the proof?

Micah already pointed the way in a comment: The identity

causes the inner sum to telescope. To make full use of this, let’s extend the inner sum to $k=2n$:

This sum contains each of the terms we want to sum twice, with mirror symmetry, and it contains one additional term in the middle for $k=n$. Thus, for our sum to vanish, we need

The arguments of the two factors add to $l\pi$, so they’re negatives of each other, so we’re looking for

We can again extend the sum to $2n$ to double it, since the arguments form pairs that add up to $n\pi$; then, since $n$ and $2n+1$ are coprime, we can replace $ln$ by $l$ while traversing the same arguments; and then we can set the upper limit back to $n$, since the arguments still add up to $\pi$ in pairs. Thus, what we need is

How to find this sum is shown at Prove that $\sum\limits_{k=1}^{n-1}\tan^{2}\frac{k \pi}{2n} = \frac{(n-1)(2n-1)}{3}$. Adapting the argument in the accepted answer there for our odd denominator, we obtain

taking the imaginary part,

dividing by $\left(\cos\frac{k\pi}{2n+1}\right)^{2n+1}$,

and dividing by $\tan\frac{k\pi}{2n+1}$ and letting $x=\tan^2\frac{k\pi}{2n+1}$,

Then Vieta’s formula shows that the sum of the roots of this equation is

as required.

[Update:]

This answer suggests an alternative, more elementary way to calculate the sum of the squares of the tangents: On the grid $\frac{l\pi}{2n+1}$, the tangent decomposes into $n$ mutually orthogonal sines, each of whose dot product with itself is $2n+1$, so the dot product of the tangent with itself is $n(2n+1)$.