Problem 1675

To prove the given statement, we need to show that it is possible to write a real number in each case of the cylindrical grid such that each number is equal to the sum of the numbers of the neighboring cases and not all the numbers are zero if and only if there exist integers $k$ and $l$ such that $n+1$ does not divide $k$ and
$$\cos \frac{2l\pi}{m} + \cos \frac{k\pi}{n+1} = \frac{1}{2}.$$

1. Define the grid and the problem:
Denote the entries of the cylindrical grid by $\{x_{i,j}\}_{1 \le i \le n, 1 \le j \le m}$. Assume that the second indices are always taken $\pmod m$. For all $(i,j) \in [n] \times [m]$, we have:
$$x_{i,j} = x_{i-1,j} + x_{i,j-1} + x_{i,j+1} + x_{i+1,j},$$
where $x_{0,j} = x_{n+1,j} = 0$ for all $j \in [m]$.

2. Matrix representation:
Given a pair of adjacent columns $\vec{x}_i = (x_{1,i}, x_{2,i}, \ldots, x_{n,i})^T$ and $\vec{x}_{i+1} = (x_{1,i+1}, x_{2,i+1}, \ldots, x_{n,i+1})^T$, 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:
$$A_n \vec{y}_i = \vec{y}_{i+1}.$$

3. Eigenvalues and eigenvectors:
Since $A_n^m \vec{y}_i = \vec{y}_i$, it follows that $(A_n^m - I)\vec{y}_i = \vec{0}$. Thus, $1$ is an eigenvalue of $A_n^m$. Given a polynomial $p(x)$ and a matrix $A$ with eigenvalues $E(A)$, we have $E(p(A)) \subseteq p(E(A))$. Therefore, $1 \in \{\lambda^m \mid \lambda \in E(A_n)\}$, implying that there exists $\lambda \in E(A_n)$ such that $\lambda^m = 1$. This eigenvalue $\lambda$ is of the form $e^{2\pi i l / m}$ for some $l \in \{0, 1, \ldots, m-1\}$.

4. Recurrence relation:
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 $\lambda = e^{2\pi i l / m}$ for $A_n$. Then:
$$v_i^{(1)} = \lambda v_i^{(2)} \quad \text{and} \quad -v_{i-1}^{(1)} + v_i^{(1)} - v_{i+1}^{(1)} - v_i^{(2)} = \lambda v_i^{(1)}.$$
This leads to the recurrence relation:
$$v_{i+1}^{(1)} = (1 - \lambda - \frac{1}{\lambda}) v_i^{(1)} - v_{i-1}^{(1)}.$$
Let $b = 1 - \lambda - \frac{1}{\lambda} = 1 - 2\cos(2\pi l / m)$. Then:
$$v_{i+1}^{(1)} = b v_i^{(1)} - v_{i-1}^{(1)}.$$

5. Non-zero eigenvector:
If $v_1^{(1)} = 0$, then $v_i^{(1)} = v_i^{(2)} = 0$ for all $i$, contradicting the fact that $\vec{v}$ is an eigenvector. Thus, $v_1^{(1)} \neq 0$.

6. **Bound on $b$:**
Assume for contradiction that $b \ge 2$. By induction, $v_{i+1}^{(1)} \ge v_i^{(1)} \ge 0$, leading to $v_1^{(1)} = v_2^{(1)} = \cdots = v_n^{(1)} = 0$, which is a contradiction. Therefore, $b < 2$.

7. Conclusion:
Since $b = 1 - 2\cos(2\pi l / m)$ and $b < 2$, we can express $b$ as $2\cos \theta$ for some $\theta \in \left(0, \frac{2\pi}{3}\right]$. Then:
$$v_t^{(1)} = 2 \sin(t\theta).$$
Since $0 = v_{n+1}^{(1)} = 2 \sin((n+1)\theta)$, we have $(n+1)\theta = \pi k$ for some integer $k$. Thus, $n+1 \nmid k$ and:
$$\cos \left(\frac{2l\pi}{m}\right) + \cos \left(\frac{k\pi}{n+1}\right) = \frac{1}{2}.$$