Problem 1629

1. Define the magic square properties:
A 3x3 magic square with magic number $m$ is a matrix where the sum of the numbers in each row, each column, and each diagonal is equal to $m$.

2. Sum of all entries:
Let the entries of the 3x3 magic square be $a_{ij}$ where $i, j \in \{1, 2, 3\}$. The sum of all entries in the magic square is:
$$\sum_{i=1}^3 \sum_{j=1}^3 a_{ij}$$
Since each row sums to $m$ and there are 3 rows, the total sum is:
$$3m$$

3. Sum of the four corner cells, edge cells, and middle cell:
Let the sum of the four corner cells be $c$, the sum of the four edge cells be $e$, and the middle cell be $g$. Then:
$$c + e + g = 3m$$

4. Sum of the diagonals:
Each diagonal sums to $m$. The two diagonals include the middle cell twice, so:
$$c + 2g = 2m$$

5. Sum of the middle row and middle column:
The middle row and middle column each include the middle cell once, so:
$$e + 2g = 2m$$

6. Combine the equations:
Adding the equations for the diagonals and the middle row/column, and then subtracting the total sum equation:
$$(c + 2g) + (e + 2g) - (c + e + g) = 2m + 2m - 3m$$
Simplifying, we get:
$$3g = m$$
Therefore, $m = 3g$.

7. Divisibility by 3:
Since $m = 3g$, $m$ is divisible by 3.

8. Inequality for each entry:
We need to show that each entry is at most $2g - 1$. Since $m = 3g$, we have $g = \frac{m}{3}$.

- For the corner cells: Consider a diagonal. The sum of the diagonal is $m$, and it includes two corner cells and the middle cell $g$. Therefore, each corner cell is at most $m - g = 2g$. Since the entries are positive integers, the maximum value for each corner cell is $2g - 1$.

- For the edge cells: Consider the middle row or column. The sum of the middle row or column is $m$, and it includes two edge cells and the middle cell $g$. Therefore, each edge cell is at most $m - g = 2g$. Since the entries are positive integers, the maximum value for each edge cell is $2g - 1$.

Thus, each entry in the magic square is at most $2g - 1$.

$\blacksquare$