Matrix with each cell equal to sum of its adjacent cells Consider a $m \times n$ matrix $M$ such that each cell in $M$ is equal to the sum of its adjacent cells (sharing either an edge or a corner with this cell). What are the values in this matrix. 
I am trying to find a non-zero matrix satisfying this condition. Does there exist such a matrix, or if not, is there a proof for it?
I know that when the matrix values are equal to average of surrounding cells, then all values in the matrix are equal. Is there a similar conclusion we can reach to in this case?
 A: For sure such matrix could be found for $m \in 2\mathbb{N}+1, n \in 3\mathbb{N}+2$:
$$\left[\begin{matrix}
a&a&0&-a&-a&0&a&a&0&\cdots & 0 & a &a &  0 &-a &-a\\
0&0&0&0&0&0&0&0&0&\cdots&0&0&0&0&0&0\\
-a&-a&0&a&a&0&-a&-a&0&\cdots & 0 & -a &-a &  0 &a &a\\
0&0&0&0&0&0&0&0&0&\cdots&0&0&0&0&0&0\\
\cdots&&&&&&&&&\cdots\\
0&0&0&0&0&0&0&0&0&\cdots&0&0&0&0&0&0\\
a&a&0&-a&-a&0&a&a&0&\cdots & 0 & a &a &  0 &-a &-a\\
\end{matrix}\right]$$
Of course depending on the value of $n$ and $m$ last row and two last columns might be different (last row might start with $-a, -a$ and last columns might start with $a,a$), but general idea is the same.
A: The point of this answer is to show that the only solutions are Jaroslaw Matlak's solution (when $m \equiv 1 \bmod 2$ and $n \equiv 2 \bmod 3$), the transpose of his solution, and linear combinations of the above (when $m \equiv n \equiv 5 \bmod 6$). 
Let's first avoid boundary effects by studying the problem on an $M \times N$ torus, so indices wrap around. Let $\phi$ be the linear map from $M \times N$ matrices which replaces each entry of the matrix with the sum of its $8$ neighbors. We want to understand the $1$-eigenspace of $\phi$.
In fact, it is easy to diagonalize $\phi$ completely. For any $0 \leq a < M$ and $0 \leq b < N$, the matrix 
$$A_{jk} = \exp\left( (2 \pi i) \left( \tfrac{aj}{M} + \tfrac{bk}{N} \right) \right)$$
is an eigenvector of $\phi$ with eigenvalue $(2 \cos \tfrac{a \pi}{M} + 1)(2 \cos \tfrac{b \pi}{N} + 1)-1$. That's $MN$ eigenvectors in an $MN$ dimensional vector space, and it is easy to show they are linearly independent. So we have found all the eigenvectors.
We want 
$$(2 \cos \tfrac{a \pi}{M} + 1)(2 \cos \tfrac{b \pi}{N} + 1)=2 \quad (\ast).$$ I will now show that the only solutions to $(\ast)$ in integers are $(\cos \tfrac{a \pi}{M}, \cos \tfrac{b \pi}{N}) = (1/2,0)$ and $=(0, 1/2)$.
Lemma Let $M > 6$. Then there is some $a$ with $M/4 < a < 3M/4$ and $\mathrm{GCD}(a,M)=1$.
Proof: We break into cases according to $M \bmod 4$. If $M$ is odd, $\tfrac{M-1}{2}$ does the trick. If $M \equiv 2 \bmod 4$, then $\tfrac{M-4}{2}$ works; if $M \equiv 0 \bmod 4$, then $\tfrac{M-2}{2}$ does. $\square$.
Now, suppose we have a solution to $(\ast)$ in lowest terms with $M>6$. The Galois group of $\mathbb{Q}(\cos \pi/M, \cos \pi/N)$ acts transitively on the values $\cos \tfrac{a \pi}{M}$ with $\mathrm{GCD}(a,M)=1$, so we can apply a Galois automorphism to make $a$ be as in the lemma, with $M/4 < a < 3M/4$. 
But then $-1 < 2 \cos \tfrac{\pi a}{M} +1 < 0$, which means that $2 \cos \tfrac{\pi b}{N} +1 < -2$, an impossibility. We have a contradiction and deduce that $M \leq 6$; then we just check cases.
Finally, we need go from the torus case back to the rectangle. Given an $m \times n$ solution to the rectangular problem, we can build a $(2m+2) \times (2n+2)$ solution to the torus problem as shown in the example below for $(m,n) = (2,3)$.
$$\begin{bmatrix}  u&v&w\\x&y&z \\ \end{bmatrix} \leadsto
\begin{bmatrix}
0&0&0&0&0&0&0&0 \\
0&u&v&w&0&-w&-v&-u \\
0&x&y&z&0&-z&-y&-x \\
0&0&0&0&0&0&0&0 \\
0&-x&-y&-z&0&z&y&x \\
0&-u&-v&-w&0&w&v&u \\
\end{bmatrix}.$$
So our previous solutions with $(\tfrac{a}{M}, \tfrac{b}{N}) = (\tfrac{1}{4}, \tfrac{1}{6})$ turn into $(\tfrac{a}{2m+2}, \tfrac{b}{2n+2}) = (\tfrac{1}{4}, \tfrac{1}{6})$ etcetera and, after a bit of cleaning up, we see that Jaroslaw Matlak has found them all.
