Denote the entries of the cylindrical grid by $\{x_{i,j}\}_{1 \le i \le n, 1 \le j \le m}$. Throughout we will assume that the second indices are always taken $\pmod m$. Then $\forall (i,j) \in [n] \times [m]$, we have that $x_{i,j} = x_{i-1,j} + x_{i,j-1} + x_{i,j+1} + x_{i+1,j}$ where $\forall j \in [m]$, $x_{0,j} = x_{n+1,j} = 0$, which can be naturally assumed as this makes the math remain valid. We then formally define these as the values of $x_{0,j}$ and $x_{n+1,j}$. We also have that $\exists (i,j) \in [n] \times [m]$ such that $x_{i,j} \neq 0$. Observe that given the complete list of entries of any two adjacent columns of the cylindrical grid, all entries of the grid are determined, as knowing $x_{i-1,j}, x_{i,j-1}, x_{i,j}, x_{i,j+1}$ implies the value of $x_{i+1,j}$ through $x_{i+1,j} = x_{i,j} - x_{i-1,j} - x_{i,j-1} - x_{i,j+1}$. Thus, the entries of any column of the cylindrical grid can be represented as a linear combination of the entries of the previous two columns (WLOG going counterclockwise when viewed from above). Formally, given a pair of adjacent columns $\vec{x}_i = (x_{1,i}, x_{2,i}, \ldots, x_{n,i})^T$ and $\vec{x}_{i+1} = (x_{1,i+1}, x_{2,i+1},\ldots, x_{n,i+1})^T$, we define $\vec{y}_i = (x_{1,i+1}, x_{2,i+1}, \ldots, x_{n,i+1}, x_{1,i}, x_{2,i}, \ldots, x_{n,i})^T$ and the $2n \times 2n$ matrix $A_n$ such that all main diagonal entries in the top left quadrant are $1$ and all entries $1$ away from the main diagonal in the top left quadrant are $-1$. In the top right quadrant, all main diagonal entries are $-1$. In the bottom left quadrant, all main diagonal entries are $1$. All other entries are $0$. As an example, this is $A_5$: $\begin{bmatrix} 1 & -1 & 0 & 0 & 0 & -1 & 0 & 0 & 0 & 0\\ -1 & 1 & -1 & 0 & 0 & 0 & -1 & 0 & 0 & 0\\ 0 & -1 & 1 & -1 & 0 & 0 & 0 & -1 & 0 & 0\\ 0 & 0 & -1 & 1 & -1 & 0 & 0 & 0 & -1 & 0\\ 0 & 0 & 0 & -1 & 1 & 0 & 0 & 0 & 0 & -1\\ 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0\\ \end{bmatrix}$

Proof: The relevance of $A_n$ is that $\forall i$, $A_n \vec{y}_i = \vec{y}_{i+1}$ by the definition of the addition property of the cylindrical grid. Then $A_n^m \vec{y}_i = \vec{y}_i$, i.e. $(A_n^m - I)\vec{y}_i = \vec{0}$, so by definition, $1$ is an eigenvalue of $A_n^m$. It is well known that given a polynomial $p(x) = a_s x^s + \cdots + a_1 x + a_0$ and a matrix $A$ with its multiset of eigenvalues being $E(A)$, then $E(p(A)) \subseteq p(E(A)) := \{p(\lambda) \mid \lambda \in E(A)\}$ where $p(A) := a_s A^s + \cdots + a_1 A + a_0I$. In fact, $E(p(A)) = p(E(A))$ as multisets but we only need the left to right inclusion just mentioned, which can be proven in this wayConsider $\mu \in E(p(A))$. By definition then, there exists $\vec{w} \neq \vec{0}$ such that $p(A)\vec{w} = \mu \vec{w}$. The polynomial $p(x) - \mu$ can be factored over $\mathbb{C}$ as: \begin{align*} p(x) - \mu = a(x-s_1)(x-s_2) \cdots (x-s_r) \end{align*} Then: \begin{align*} p(A) - \mu I = a(A - s_1I)(A - s_2I) \cdots (A - s_rI) \end{align*} As $\mu$ is an eigenvalue of $p(A)$, $p(A) - \mu I$ is not invertible so there exists some $i$ so that $A - s_i I$ is non-invertible, i.e. $s_i$ is an eigenvalue of $A$. Since $p(s_i) - \mu = 0$, $\mu \in p(E(A))$, so $E(p(A)) \subseteq p(E(A))$, as witnessed by the eigenvalue $s_i$ of $A$. $\blacksquare$. Thus $1 \in \{\lambda^m \mid \lambda \in E(A_n)\} \Rightarrow \exists \lambda \in E(A_n)$ such that $\lambda^m = 1$. This eigenvalue $\lambda$ of $A_n$ is then of the form $cis(2\pi l / m)$ for some $l \in \{0, 1, \ldots, m-1\}$. $\blacksquare$

Let $\vec{v} = (v_1^{(1)}, v_2^{(1)}, \ldots, v_n^{(1)}, v_1^{(2)}, v_2^{(2)}, \ldots, v_n^{(2)})^T$ be an eigenvector corresponding to $\lambda = cis(2\pi l / m)$ for $A_n$. Then $A_n \vec{v} = \lambda \vec{v}$. We then have $\forall i \in [n]$: \begin{align*} v_i^{(1)} &= \lambda v_i^{(2)} \text{ and}\\ -v_{i-1}^{(1)} + v_i^{(1)} - v_{i+1}^{(1)} - v_i^{(2)} &= \lambda v_i^{(1)}\\ \Longrightarrow v_{i+1}^{(1)} &= (1-\lambda)v_{i}^{(1)} - v_{i-1}^{(1)} - v_i^{(2)}\\ \Longrightarrow v_{i+1}^{(1)} &= \left(1-\lambda-\frac{1}{\lambda}\right)v_{i}^{(1)} - v_{i-1}^{(1)} \end{align*} Let $b = 1 - \lambda - \frac{1}{\lambda} = 1 - 2\cos(2\pi l / m)$. Then we have that for $i \in [n]$, $v_{i+1}^{(1)} = b v_i^{(1)} - v_{i-1}^{(1)}$, where, like above, we may take $v_0^{(1)} = v_{n+1}^{(1)} = 0$, as the calculations remain correct when calculating the value of $v_n^{(1)}$ from only the previous entry. $\blacksquare$

From claim 2, $v_{i+1}^{(1)} = b v_i^{(1)} - v_{i-1}^{(1)}$ with $v_{0}^{(1)} = 0$ and $v_i^{(2)} = \frac{1}{\lambda} v_i^{(1)}$. If $v_1^{(1)} = 0$, then it is straightforward to see that $\forall i$, $v_i^{(1)} = v_i^{(2)} = 0$, so $\vec{v} = \vec{0}$, contradicting the fact that $\vec{v}$ is an eigenvector of $A_n$. $\blacksquare$

From claim 3, $v_1^{(1)} \neq 0$ and we can scale $\vec{v}$ so that $v_1^{(1)} > 0$. Assume for contradiction that $b \ge 2$. We can then prove by induction that $v_{i+1}^{(1)} \ge v_i^{(1)} \ge 0$. The base follows from the initial conditions of $v_{1}^{(0)}$ and $v_{1}^{(1)}$ and the choice WLOG of sign of $v_1^{(1)}$. Now assume $v_{i}^{(1)} \ge v_{i-1}^{(1)} \ge 0$. Then: \begin{align*} v_{i+1}^{(1)} &= b v_i^{(1)} - v_{i-1}^{(1)} \\ &\ge 2 v_i^{(1)} - v_{i-1}^{(1)}\\ &= v_i^{(1)} + (v_i^{(1)} - v_{i-1}^{(1)})\\ &\ge v_i^{(1)} \ge 0 \end{align*} This implies that $0 \le v_1^{(1)} \le v_2^{(1)} \le \cdots \le v_n^{(1)} \le 0$, so $v_1^{(1)} = v_2^{(1)} = \cdots = v_n^{(1)} = 0$ and by the condition $v_i^{(1)} = \lambda v_i^{(2)}$, $\forall i, v_i^{(2)} = 0$ as well, so $\vec{v} = \vec{0}$, which again cannot follow since $\vec{v}$ is an eigenvector of $A_n$. $\blacksquare$

Recall that $b = 1 - 2\cos(2\pi l/m)$ so $b \ge -1$. From claim 4, $b < 2$ as well, so we can express $b$ as $2\cos \theta$ for some $\theta \in \left(0, \frac{2\pi}{3}\right]$. We may then scale $v_1^{(1)} > 0$ to be $2 \sin \theta > 0$. It then follows that $\forall t$, $v_t^{(1)} = 2 \sin(t\theta)$. Indeed, proceeding by induction, if $v_{t-1}^{(1)} = 2 \sin((t-1)\theta)$ and $v_t^{(1)} = 2 \sin(t \theta)$, we have: \begin{align*} v_{t+1}^{(1)} &= b v_t^{(1)} - v_{t-1}^{(1)}\\ &= 2 \cos (\theta) v_t^{(1)} - v_{t-1}^{(1)}\\ &= 2 \cos (\theta) \cdot 2 \sin (t \theta) - 2 \sin ((t-1) \theta) \\ &= 2\sin((t+1)\theta) \end{align*}where the last equality is an identity which can be derived as done hereWe know: \begin{align*} \sin(A+B) &= \sin A \cos B + \cos A \sin B \text{ and}\\ \sin(A-B) &= \sin A \cos B - \cos A \sin B\\ \end{align*} Summing these and re-arranging: \begin{align*} \sin(A+B) &= 2 \sin A \cos B - \sin(A-B) \end{align*} Taking $A = t\theta$ and $B = \theta$, this becomes: \begin{align*} \sin((t+1)\theta)& = 2 \sin(t \theta) \cos \theta - \sin ((t-1) \theta) \\ \Longrightarrow 2\sin((t+1)\theta) &= 2 \sin(t \theta) \cdot 2 \cos \theta - 2\sin ((t-1) \theta) \end{align*}. Then $0 = v_{n+1}^{(1)} = 2 \sin((n+1)\theta)$ so $(n+1)\theta = \pi k$ for some integer $k$. Since $\sin \theta \neq 0$, we also have that $n+1 \nmid k$. Thus, there are integers $k, l$ with $n+1 \nmid k$ such that: \begin{align*} b = 1 - 2\cos\left(\frac{2l\pi}{m}\right) &= 2 \cos\left(\frac{k\pi}{n+1}\right) \\ \Longrightarrow \cos\left(\frac{2l\pi}{m}\right) + \cos\left(\frac{k\pi}{n+1}\right) &= \frac{1}{2} \end{align*}