Proving $\sum\left[\sin\frac{2\pi(i-j)}{n}+\sin\frac{2\pi(j-k)}{n}+\sin\frac{2\pi(k-i)}{n}\right]^2=\frac{n^3}{4}$

Question

How to prove this interesting identity?

$$\sum_{1\leq i<j<k\leq n}\left[ \sin\left(\frac{2\pi(i-j)}{n}\right)+\sin\left(\frac{2\pi(j-k)}{n}\right)+\sin\left(\frac{2\pi(k-i)}{n}\right) \right]^2= \frac{n^3}{4}$$

Answer 1

Clearly we must have $n>2$ for this to hold. Let's assume that.

We have $\sum\limits_{1\leqslant i,j,k\leqslant n}=6\sum\limits_{1\leqslant i<j<k\leqslant n}$ since the terms on the LHS with pairwise distinct $i,j,k$ can be grouped into $6$ equal sums, and the rest — with at least two of $i,j,k$ equal — are all zeros. Writing $$\sin\frac{2\pi m}{n}=\frac{\zeta^m-\zeta^{-m}}{2\mathrm{i}},\qquad\zeta=\exp\frac{2\pi\mathrm{i}}{n},$$ and simplifying the expression under the square, we see that the given sum is equal to $$S(n)=-\frac{1}{24}\sum_{i,j,k=1}^{n}\big[(1-\zeta^{i-j})(1-\zeta^{j-k})(1-\zeta^{k-i})\big]^2.$$ The $[\ldots]^2$ is the sum of terms of the form $c_{\alpha,\beta,\gamma}\zeta^{\alpha i+\beta j+\gamma k}$ where $|\alpha|,|\beta|,|\gamma|\leqslant 2$.

As we assume $n>2$, $$\sum_{i,j,k=1}^{n}\zeta^{\alpha i+\beta j+\gamma k}=\begin{cases}n^3,&\alpha=\beta=\gamma=0\\0,&\text{otherwise}\end{cases}$$ (indeed, if, say, $\alpha\neq 0$, then the summation over $i$ gives $0$), so we have $$S(n)=-\frac{c_{0,0,0}}{24}n^3.$$ And immediate evaluation shows that $c_{0,0,0}=-6$.