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From Kirchhoff matrix tree theorem we know that the number of spanning trees in a graph is equal to the product of the combinatorial Laplacian eigenvalues (removing eigenvalue 0) divided by the number of vertices. Taking a square grid of size $n_1\times n_2$, it means $$\frac{1}{n_1n_2}\prod_{\substack{0\leqslant k_1\leqslant n_1,\ 0\leqslant k_2\leqslant n_2\\(k_1,k_2)\neq0}}\left(4-2\cos\left(\frac{\pi k_1}{n_1}\right)-2\cos\left(\frac{\pi k_2}{n_2}\right)\right)\in\mathbb{N}.$$ I have always been fascinated that those products are integers, and I am wondering whether there is an algebraic (or geometric) reason for this fact?

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    $\begingroup$ The product is the free term of an integer matrix characteristic polynomial. (Arguably that's an algebraic reason, unless you meant something else?) $\endgroup$ Commented May 29 at 15:45
  • $\begingroup$ @MikhailTikhomirov Yes indeed, thanks for your reply. The motivation of my question is that I want to show that some kinds of similar products involving cosines are also integers. So in case anyone has another point of view, I would be happy to read. $\endgroup$ Commented May 29 at 21:04
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    $\begingroup$ Another argument that goes most of the way is that $\cos(\pi k / n)$ are roots of $P_{2n}(x) = 1$ (where $P_n$ is the Chebyshev polynomial), and from arithmetic properties of algebraic integers (so that if collections $\{\lambda_i\}$ and $\{\mu_j\}$ are roots of integer polynomials, then so is $\{\lambda_i + \mu_j\}$). $\endgroup$ Commented May 30 at 6:56
  • $\begingroup$ @MikhailTikhomirov This could help me, but i don't see why is it true? $\endgroup$ Commented Jun 16 at 9:22
  • $\begingroup$ I'm not sure which part is unclear? Standard references (or wiki pages) about Chebyshev polynomials and algebraic numbers should be helpful. $\endgroup$ Commented Jun 20 at 13:13

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