1. A magic square is a square table of natural numbers in which the sum of all numbers in each row, each column, and both diagonals is the same. Find all $3 \times 3$ magic squares for which the product of the four numbers in the corner cells is equal to 3465.
Let's denote the natural numbers in the magic square by the letters $a, b, c, d, e$, $f, g, h, i$ as shown in Fig. 1.1, and let $S$ be the sum of the three numbers in each row, column, and diagonal. We will show that $S=3e$: If we add the numbers in the first and third rows and subtract the numbers in the middle column, we get the equation

$$
S+S-S=a+c+g+i-e \text { . }
$$

| $a$ | $b$ | $c$ |
| :--- | :--- | :--- |
| $d$ | $e$ | $f$ |
| $g$ | $h$ | $i$ |

Fig. 1.1

From this, considering the equations $a+i=c+g=S-e$, we get

$$
S=(S-e)+(S-e)-e, \quad \text { or } \quad S=3 e \text { . }
$$

As a consequence, we have the equations

$$
a+i=c+g=2 e \text { . }
$$

We need to find four natural numbers $a, i, c, g$ whose product is equal to 3465, and such that $a+i=c+g$. To find the finite set of solutions to the equation $a \cdot i \cdot c \cdot g = 3465$, we can first list all possible factorizations of the number $3465 = 3^2 \cdot 5 \cdot 7 \cdot 11$ into the product of two factors $M$ and $N$ (which should correspond to the products $a \cdot i$ and $c \cdot g$):

$$
\begin{aligned}
3465 & =1 \cdot 3465=3 \cdot 1155=5 \cdot 693=7 \cdot 495=9 \cdot 385= \\
& =11 \cdot 315=15 \cdot 231=21 \cdot 165=33 \cdot 105=35 \cdot 99= \\
& =45 \cdot 77=55 \cdot 63 .
\end{aligned}
$$

Now, for each pair $M, N$, we can easily find the factorizations $M=a \cdot i$ and $N=c \cdot g$ with the property $a+i=c+g$ (for the first eight pairs, such factorizations clearly do not exist). The only two valid factorizations are

$$
3465=(5 \cdot 11) \cdot(7 \cdot 9)=(3 \cdot 15) \cdot(7 \cdot 11) .
$$

In the first case, $2e=16$, so $e=8$; in the second case, $2e=18$, so $e=9$. We can easily calculate the other numbers in the magic square (Fig. 1.2).

| 5 | 12 | 7 |
| :---: | :---: | :---: |
| 10 | 8 | 6 |
| 9 | 4 | 11 |


| 3 | 17 | 7 |
| :---: | :---: | :---: |
| 13 | 9 | 5 |
| 11 | 1 | 15 |

Fig. 1.2

Since the four corner numbers can be placed in the table in eight ways, each table in Fig. 1.2 represents eight tables that can be derived from it by "flipping" along the axes of symmetry of the square. No other solutions to the problem exist.

SUPPLEMENTARY AND COMPLEMENTARY PROBLEM: