Product of real numbers is an integer Let $n \in \mathbb{N}$ and $c_{i,j}(n) = 1+2\cos(\frac{i \pi}{n+1}) + 2\cos(\frac{j \pi}{n+1})$. How can one show that the product $$\prod_{{1 \leq i,j \leq n} \atop {i < j}} c_{i,j}$$ is always an integer?
 A: Below is not a complete answer yet; but I have found the idea to compute it and will finish everything once I got enough time.
Let us find the following product first :$$P =\prod_{1\leq k,j\leq n}(1+2\cos(\dfrac{ k\pi}{n+1})+2\cos\big(\dfrac{j\pi}{n+1}\big)).$$
Using the identity $\cos a+\cos b = 2\cos(\frac{a+b}{2})\cos(\frac{a-b}{2})$, we can write $c_{k,j} = 1+4\cos(\frac{(k+j)\pi}{2n+2})\cos(\frac{(k-j)\pi}{2n+2}).$ Let $k+j = s, k-j = p$, so that we get new $c_{s,p} = 1+4\cos(\frac{s\pi}{2n+2})\cos(\frac{p\pi}{2n+2}).$
Now notice that for a fixed index $p$, we can calculate the product $A_p$ as follows: 
$$A_p = \Big(4\cos(\frac{p\pi}{2n+2}\Big)^nf_p\Big(\dfrac{1}{4\cos(\frac{p\pi}{2n+2})}\Big).$$
Where $$f_p(t) = \prod_{l=1}^n(t+\cos(\dfrac{(p+2l)\pi}{2n+2}))$$
For example, when $p=0$, $$f_0(t) = \prod_{l=1}^n(t+\cos(\dfrac{l\pi}{n+1})) =(-1)^nU_n(-t)$$, where $U_n$ is the second type Chebyshev polynomial given by the formula: $$U_n(x) = \dfrac{(x+\sqrt{x^2-1})^{n+1} - (x-\sqrt{x^2-1})^{n+1}}{2\sqrt{x^2-1}}$$. 
Furthermore, one can observe that this works for all even $p$, and for odd $p$ we will get the first type Chevyshev polynomials. Therefore, we can compute this product for each fixed $p$ and obtain $P$. 
Once we know $P$, it is easy to find the original product.
EDIT:
I am providing a different solution than what I had in mind when I wrote the above piece. 
Let $a_i = 2\cos\dfrac{\pi i}{n+1}$ so $c_{ij} = 1+a_i+a_j.$ If we let $$P(x_1,x_2,...x_n) =\prod_{1\leq i<j\leq n}(1+x_i+x_j)\in\mathbb{Z}[x_1, x_2,...x_n]$$, then by The fundamental theorem of symmetric polynomials, we can write $$P(x_1, x_2,...x_n) = Q(e_1, e_2, ...e_n)\in\mathbb{Z}[x_1, x_2,...x_n]$$
where $e_k$ is the k-th elementary symmetric polynomial $$e_k(x_1,x_2,...x_n) = \sum_{1\leq j_1<j_2<...j_k\leq n}x_{j_1}x_{j_2}...x_{j_k}$$. 
Now we claim that each of $e_k(a_1, a_2,...a_n)$ is integer. But this follows from the fact that $\frac{a_k}{2}$ are the roots of the Chebyshev polynomial $U_n$ of second type. It is well-known that $U_n$ have integer coefficients with its leading coefficient is exactly $2^n$. To be precise, the following expression of $U_n$ is used here: $$U_n(x) = \sum_{k=0}^{\lceil{\frac{n}{2}}\rceil}(-1)^k\binom{n-k}{k}(2x)^{n-2k}$$ 
