975. Natural numbers from 1 to 9 are placed in the cells of a third-order magic square. What number is in the center of the square? Where can 1 be placed? And 8? Restore at least one such magic square.
$\triangle$ First, let's find the sum of the magic square. It is equal to the sum of all natural numbers from 1 to 9, divided by 3.

$$
S=(1+2+3+\ldots+9): 3=45: 3=15
$$

Now let's find all the ways in which the number 15 can be represented as the sum of three different natural numbers from 1 to 9.

$$
\begin{gathered}
15=1+5+9=1+6+8=2+4+9=2+5+8=2+6+7= \\
=3+4+8=3+5+7=4+5+6
\end{gathered}
$$

There turned out to be 8 such ways. But the total number of rows, columns, and diagonals of the square is also 8. Therefore, each of the obtained representations of the number 15 must enter the desired square, and only once.

Which number is in the center of the square? This number is included in four sums: in a row, in a column, and along two diagonals. Let's look at all the representations of the number 15 as a sum: which number is included in four such sums? Only the number 5. It will be in the center.

Let's deal with the number 1: in how many sums does it appear? It turns out, in two. But which numbers of the square are included in two such sums? Only those that are in the middle of the extreme row or extreme column. Let's write 1 in one of such cells.

| 8 | 1 | 6 |
| :--- | :--- | :--- |
| 3 | 5 | 7 |
| 4 | 9 | 2 |

The number 8 is included in three sums. Therefore, 8 stands in one of the corners of the square, but on our diagram, not in the lower, but in the upper corner (think about why).

The remaining cells of the square are filled in unambiguously. The magic square is restored.