The number of ways to tile an $m \times n$ rectangle with dominos is $$S_{m,n} = \prod_{j=1}^{\lceil \frac{m}{2} \rceil} \prod_{k=1}^{\lceil \frac{n}{2} \rceil}\left( 4 \cos^2 \frac{\pi j}{m+1} + 4 \cos^2 \frac{\pi k}{n+1} \right) $$
How can we prove that this expression is an integer without appeal to its combinatorial interpretation? It's been suggested to me that since the expression above is an algebraic integer, it suffices to show that it is rational using Galois theory, but I am not familiar enough with Galois theory to produce such an argument.
Edit: Following the hint below, using $\cos(\theta) = \frac{e^{i\theta} + e^{-i\theta}}{2}$, we have that $\cos^2(\theta) = \frac{e^{2i\theta} + e^{-2i\theta} + 2}{4}$ with which we can rewrite $$S_{m,n} = \prod_{j=1}^{\lceil \frac{m}{2} \rceil}\prod_{k=1}^{\lceil \frac{n}{2} \rceil} \left(e^{\frac{2\pi i j}{m+1}} + e^{-\frac{2\pi i j}{m+1}} + e^{\frac{2 \pi i k}{n+1}} + e^{-\frac{2 \pi i k}{n+1}} +4 \right) =\prod_{j=1}^{\lceil \frac{m}{2} \rceil}\prod_{k=1}^{\lceil \frac{n}{2} \rceil} \left(\zeta_{m+1}^j + \zeta_{m+1}^{-j} + \zeta_{n+1}^k + \zeta_{n+1}^{-k} +4 \right) $$
Clearly, $S_{m,n} \in \mathbb{Q}[\zeta_{m+1},\zeta_{n+1}]$ where $\zeta_k = e^\frac{2\pi i }{k}$.
Then $\Gamma(\mathbb{Q}[\zeta_{m+1},\zeta_{n+1}]/\mathbb{Q}) \subseteq (\mathbb{Z}/m \mathbb{Z})^\times \times (\mathbb{Z}/n \mathbb{Z})^\times$ generated by the automorphisms $\zeta_{m+1} \mapsto \zeta_{m+1}^a$ for $1 \leq a < m+1$ relatively prime to $m+1$ and $\zeta_{n+1} \mapsto \zeta_{n+1}^b$ for $1 \leq b < n+1$ relatively prime to $n+1$. From here, I don't know how to show that $S_{m,n}$ is invariant under this action, as (for example) $\zeta_{m+1}^{\lceil \frac{m}{2} \rceil -1} \mapsto \left(\zeta_{m+1}^{\lceil \frac{m}{2} \rceil -1}\right)^2$ gives rise to a term outside the index of the product.
We can rewrite $$S_{m,n} = 2\prod_{j=1}^m \prod_{k=1}^n \left(\cos \frac{\pi j}{m+1} + i \cos \frac{\pi k}{n+1} \right) $$ using the properties of the cosine. We further expand this as above: $$S_{m,n} = \prod_{j=1}^m \prod_{k=1}^n \left(e^{i\frac{\pi j}{m+1}} + e^{-i\frac{\pi j}{m+1}} + e^{i\frac{\pi k}{n+1}} + e^{-i\frac{\pi k}{n+1}} \right) = \prod_{j=1}^m \prod_{k=1}^n (\zeta_{2m+2}^j + \zeta_{2m+2}^{-j} + \zeta_{2n+2}^k + \zeta_{2n+2}^{-k})$$ but we have the same problem here as above, there are too many automorphisms.