Four-by-four square

Question

Consider the sixteen cells in a $4\times4$ square. Two (distinct) cells are neighbors, if they share a horizontal or vertical edge; note that every cell has two or three or four neighbors. Now each of the sixteen cells contains an integer, so that for every cell the integers in the neighboring cells add up to the sum $120$.

(a) What are the possible values for the sum of the sixteen integers?

(b) Does there exist such a $4\times4$ square in which the sixteen integers are pairwise distinct?

Answer 1

Coloring the 16 cells like a checkerboard, it follows that the cells of one color can be represented as:

With $a+b=120$. This guarantees that the neighborhood of each grey cell sums to 120.

As multiple cells share the same value, it is not possible to have all cell values pairwise distinct.

The sum total for the 8 cells of given color equals $3a+3b=360$, so that all cells together add up to 720.

Answer 2

The solution to part (a):

The integers must sum to 720

Reason:

Let us number the cells like in a matrix. Consider the neighbours of the cells $(0, 0), (0, 1), (1, 3), (2, 3), (3, 1)$ and $(3, 0)$. The sets of neighbours for each of these cells are pairwise disjoint, and their union covers all the cells. The sum of the neighbouring cells for a particular cell is $120$, and there are 6 cells above, therefore the sum is $6 \times 120 = 720$.

The solution to part (b):

No.

Reason:

Consider cells $(0, 1)$ and $(1, 0)$. They have common neighbours $(0, 0)$ and $(1, 1)$. Cell $(0, 1)$ has a third neighbour $(0, 2)$, and cell $(1, 0)$ has a third neighbour $(2, 0)$. For the sums of the neighbours of each of the two cells to equal $120$ (and therefore equal each other), cells $(0, 2)$ and $(2, 0)$ must contain the same integer (as the other neighbours are common), so therefore there does not exist a $4 \times 4$ square such that the sixteen integers are pairwise distinct.