Recently, I formulated the following conjecture which seems novel.


**Conjecture**. For any positive odd integer $n$, we have the identity
$$\sum\_{j,k=0}^{n-1}\frac1{\cos 2\pi j/n+\cos 2\pi k/n}=\frac{n^2}2.\tag{1}$$


Using Galois theory, I see that the sum is a rational number. The identity $(1)$ has some equivalent versions, for example,
$$\sum\_{0\le j<k\le n-1}\frac1{\cos 2\pi j/n+\cos 2\pi k/n}=\frac n4\left(n-(-1)^{(n-1)/2}\right)\tag{2}$$
and
$$\sum\_{1\le j<k\le n-1}\frac1{\cos 2\pi j/n+\cos 2\pi k/n}=-\frac{n-(-1)^{(n-1)/2}}4\left(n+(-1)^{(n-1)/2}2\right).\tag{3}$$
It is easy to check $(1)$-$(3)$ numerically.


**Question.** How to prove the conjecture?


Your comments are welcome!



A conjectural trigonometric identity
The following argument is very short, but bit tricky, so I remain it along with the previous answer.


Actually this sum quickly factorizes: using $\cos (x+y)+\cos (x-y)=2\cos x\cos y$ and denoting $(j+k)\cdot \frac{n+1}2=a,(j-k)\cdot \frac{n+1}2=b$ we get $$\cos 2\pi j/n+\cos 2\pi k/n=2\cos 2\pi a/n \cdot \cos 2\pi b/n.$$
Here $j, k, a, b$ may be considered as residues modulo $n$ ($\cos 2\pi a/n$ depends only on $a$ modulo $n$), and the map $(j,k)\mapsto (a,b)$ is bijective on $(\mathbb{Z}/n\mathbb{Z})^2$, thus
$$
2\sum\_{j,k=0}^{n-1}\frac1{\cos 2\pi j/n+\cos 2\pi k/n}=\sum\_{a,b=0}^{n-1}\frac1{\cos 2\pi a/n\cdot \cos 2\pi b/n}=\left(\sum\_{a=0}^{n-1}\frac1{\cos 2\pi a/n}\right)^2.
$$
It remains to calculate the sum $\sum\_{a=0}^{n-1}\frac1{\cos 2\pi a/n}$. Denoting $\omega=e^{2\pi i/n}$ we get 
$$
\sum\_{a=0}^{n-1}\frac1{\cos 2\pi a/n}=\sum\_{a=0}^{n-1}\frac{2\omega^a}{1+\omega^{2a}}=
\sum\_{a=0}^{n-1}\frac{\omega^a+\omega^{(2n+1)a}}{1+\omega^{2a}}=\\
\sum\_{a=0}^{n-1} (\omega^a-\omega^{3a}+\dots+(-1)^{(n-1)/2}\omega^{na}+\dots+\omega^{(2n-1)a})=(-1)^{(n-1)/2}n,
$$
since the geometric progressions $\sum\_{a=0}^{n-1} \omega^{ma}$ sums up to $$\sum\_{a=0}^{n-1} \omega^{ma}=\begin{cases}0& \text{for}\, m\, \text{not divisible by}\, n\\
n& \text{for}\, m\, \text{divisible by}\, n.
\end{cases}$$


