Can we fill the table such that the number in each field is equal to the sum of the numbers in all the fields with adjacent edges?

Question

Can you fill $n ^ 2$ real numbers in the table (grid) of size $n \times n$, which are not all zero, and the number in each field is equal to the sum of the numbers in all the fields with adjacent edges?

What conditions does $n$ have to meet, so that there is such a scheme? If there is a corresponding scheme, how many free variables will be among these $n^2$ numbers?

For example

$$ \begin{array}{|c|c|c|c|} \hline 2 & 1 & 1 & 2 \\ \hline 1 & -2 & -2 & 1 \\ \hline 1 & -2 & -2 & 1 \\ \hline 2 & 1 & 1 & 2 \\ \hline \end{array}$$ or $$\begin{array}{|c|c|c|c|} \hline 0 & 1 & 1 & 0 \\ \hline -1 & 0 & 0 & -1 \\ \hline -1 & 0 & 0 & -1 \\ \hline 0 & 1 & 1 & 0 \\ \hline \end{array}$$

Answer 1

Summary: I have some partial general results, consisting of a lower bound on the dimension of the solution space depending on the value of $n$ modulo $30$. I conjecture that lower bound actually holds with equality.

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Determining the complete set of solutions to this problem is equivalent to finding the null space of a certain matrix. Let $M_n$ be the matrix with dimensions $n^2\times n^2$, where $$ M_n(i,j)= \begin{cases} 1 & \text{$i$ is adjacent to $j$} \\ -1 & i=j \\ 0 & \text{otherwise} \end{cases}\hspace{2cm} i,j\in \{1,\dots,n^2\} $$ You need to choose some mapping from $\{1,\dots,n^2\}$ to the grid $\{1,\dots,n\}\times \{1,\dots,n\}$ for the condition "$i$ is adjacent to $j$" to make sense.

Any vector in $\ker M_n$ represents a labeling of the $n\times n$ grid where every entry equals the sum of its neighbors. It is simple enough to use Mathematica to compute this null space for $n\le 30$. This table summarizes the dimensions of the results. You can see that more often than not, there is no nontrivial solution.

Null space has dimension	when $n=$
$0$	$1, 2, 3, 6, 7, 8, 10, 12, 13, 15, 16, 18, 20, 21, 22, 25, 26, 27, 28, 30$
$2$	$4, 5, 9, 11, 14, 17, 19, 23, 24$
$4$	$29$

Naturally, one would hope for some sort of pattern that allows you to determine the dimension of the null space for any $n\ge 1$. From this limited data, I see no pattern appearing.

Edit: With a bit more thought, we can glean a lot from this data. One of the basis vectors for $n=4$ below has a nice circular pattern, which we see repeated again in $n=9$. In fact, it is clear that this pattern can be extended to all $n$ of the form $5k+4$. Similarly, the pattern in $n=5$ reappears in the basis for $n=11$, and this extends to any $n$ of the form $6h+5$. We have shown that

As long as $n\equiv 4\pmod 5$ or $n\equiv 5\pmod 6$, the solution space has at least two degrees of freedon.

If $n\equiv 29\pmod{30}$, so both these conditions occur, then the solution space is at least $4$ dimensional.

The question is, are these two conditions also necessary? From the data, there is nothing to suggest otherwise, so I conjecture the following:

Conjecture: $$\dim(\ker M_n)\stackrel{?}= \begin{cases} 2 & \text{ if }n\equiv 4\pmod 5,n\not\equiv 5\pmod 6\\ 2 & \text{ if }n\not\equiv 4\pmod 5,n\equiv 5\pmod 6\\ 4 & \text{ if }n\equiv 29\pmod{30} \\ 0 & \text{otherwise} \end{cases}$$

Basis for $n=4$: $$\begin{bmatrix} 1& 0& 0& 1\\ 1& -1& -1& 1\\ 1& -1& -1& 1\\ 1& 0& 0& 1\\ \end{bmatrix} \begin{bmatrix} 0& 1& 1& 0\\ -1& 0& 0& -1\\ -1& 0& 0& -1\\ 0& 1& 1& 0\\ \end{bmatrix}$$
Basis for $n=5$: $$\begin{bmatrix} -1& 0& 1& 0& -1\\ -1& 0& 1& 0& -1\\ 0& 0& 0& 0& 0\\ 1& 0& -1& 0& 1\\ 1& 0& -1& 0& 1\\ \end{bmatrix} \begin{bmatrix} 0& -1& -1& 1& 2\\ 1& 0& -1& 0& 1\\ 1& 1& 0& -1& -1\\ -1& 0& 1& 0& -1\\ -2& -1& 1& 1& 0\\ \end{bmatrix}$$
Basis for $n=9$: $$\begin{bmatrix} 1& 0& 0& 1& 0& -1& 0& 0& -1\\ 1& -1& -1& 1& 0& -1& 1& 1& -1\\ 1& -1& -1& 1& 0& -1& 1& 1& -1\\ 1& 0& 0& 1& 0& -1& 0& 0& -1\\ 0& 0& 0& 0& 0& 0& 0& 0& 0\\ -1& 0& 0& -1& 0& 1& 0& 0& 1\\ -1& 1& 1& -1& 0& 1& -1& -1& 1\\ -1& 1& 1& -1& 0& 1& -1& -1& 1\\ -1& 0& 0& -1& 0& 1& 0& 0& 1\\ \end{bmatrix} \begin{bmatrix} 0& 1& 1& 0& 0& 0& -1& -1& 0\\ -1& 0& 0& -1& 0& 1& 0& 0& 1\\ -1& 0& 0& -1& 0& 1& 0& 0& 1\\ 0& 1& 1& 0& 0& 0& -1& -1& 0\\ 0& 0& 0& 0& 0& 0& 0& 0& 0\\ 0& -1& -1& 0& 0& 0& 1& 1& 0\\ 1& 0& 0& 1& 0& -1& 0& 0& -1\\ 1& 0& 0& 1& 0& -1& 0& 0& -1\\ 0& -1& -1& 0& 0& 0& 1& 1& 0\\ \end{bmatrix}$$
Basis for $n=11$: $$\begin{bmatrix} 1& 0& -1& 0& 1& 0& -1& 0& 1& 0& -1\\ 1& 0& -1& 0& 1& 0& -1& 0& 1& 0& -1\\ 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0\\ -1& 0& 1& 0& -1& 0& 1& 0& -1& 0& 1\\ -1& 0& 1& 0& -1& 0& 1& 0& -1& 0& 1\\ 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0\\ 1& 0& -1& 0& 1& 0& -1& 0& 1& 0& -1\\ 1& 0& -1& 0& 1& 0& -1& 0& 1& 0& -1\\ 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0\\ -1& 0& 1& 0& -1& 0& 1& 0& -1& 0& 1\\ -1& 0& 1& 0& -1& 0& 1& 0& -1& 0& 1\\ \end{bmatrix} \begin{bmatrix} 0& 1& 1& -1& -2& 0& 2& 1& -1& -1& 0\\ -1& 0& 1& 0& -1& 0& 1& 0& -1& 0& 1\\ -1& -1& 0& 1& 1& 0& -1& -1& 0& 1& 1\\ 1& 0& -1& 0& 1& 0& -1& 0& 1& 0& -1\\ 2& 1& -1& -1& 0& 0& 0& 1& 1& -1& -2\\ 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0\\ -2& -1& 1& 1& 0& 0& 0& -1& -1& 1& 2\\ -1& 0& 1& 0& -1& 0& 1& 0& -1& 0& 1\\ 1& 1& 0& -1& -1& 0& 1& 1& 0& -1& -1\\ 1& 0& -1& 0& 1& 0& -1& 0& 1& 0& -1\\ 0& -1& -1& 1& 2& 0& -2& -1& 1& 1& 0\\ \end{bmatrix}$$

Finally, here is the Mathematica code I used to find this data.

Adj[ x1_, y1_, x2_, y2_] := Boole[Abs[x1 - x2] + Abs[y1 - y2] == 1];
M[n_] := Table[Adj[Mod[k, n], Floor[k/n], Mod[h, n], Floor[h/n]] - Boole[k == h],
               {k, 0, n^2 - 1}, {h, 0, n^2 - 1}]
For[n = 1, n < 20, n++,
  V = NullSpace[M[n]];
  Print["There are ", ToString[Length[V]], " patterns when n = ", n];
  For [i = 1, i <= Length[V], i++,
    pattern = ArrayReshape[V[[i]], {n, n}];
    For[r = 1, r <= n, r++, 
      Print[pattern[[r]]]
      ]
    ]
  ]