SOLUTION. Let $a, b, c, d, e, f, g, h, i$ be the numbers inscribed in the left upper $3 \times 3$ square of the table (Fig. 1). By comparing the products for pentads of the form $\mathbb{\text { a }}$ and placed in this part of the table, we must have $a b c d e=b d e f g$, or $a c=f g$. Similarly, for pentads $\mathrm{a}$ and $\mathrm{a}$, we get $a h f d i=c i g d h$, or $a f=c g$. Since all these are positive numbers, it follows from both equalities that $f=c$ and $g=a$. At the same time, we realize that this property (i.e., the equality of numbers in opposite corners of the $3 \times 3$ square) must hold for each of the four such squares that exist in the table. We will use this property for further completion of the given table.

![](https://cdn.mathpix.com/cropped/2024_04_17_fca565adf729549b0cd4g-01.jpg?height=260&width=263&top_left_y=973&top_left_x=367)

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

Fig. 1
![](https://cdn.mathpix.com/cropped/2024_04_17_fca565adf729549b0cd4g-01.jpg?height=260&width=534&top_left_y=972&top_left_x=1164)

Fig. 2

Consider again the placement $\mathbb{\mathrm { v }} \mathrm{v}$ in the upper left corner of the given table $\mathrm{s}$ with the inscribed numbers $a, b, c, d, e$, and complete the other numbers according to the property just proven, and let $x$ be the missing number in the pentad 团 (Fig. 2). By comparing both identical products, we get $a b c d e=a b d e x$, or $x=c$. If we perform the same reasoning for the pentads of fields $\mathbb{Q}$ and $\boxminus$, which we get by flipping the considered pentads along the vertical axis of the table, we get an analogous equality for two more pairs of fields in the table (Fig. 3).

![](https://cdn.mathpix.com/cropped/2024_04_17_fca565adf729549b0cd4g-01.jpg?height=261&width=266&top_left_y=1691&top_left_x=512)

Fig. 3

| $a$ | $b$ | $c$ | $e$ |
| :--- | :--- | :--- | :--- |
| $b$ | $d$ | $y$ | $c$ |
| $c$ | $e$ | $a$ | $b$ |
| $y$ | $c$ | $b$ | $d$ |

Fig. 4

Now we have the table almost completely filled, except for two cells, into which we will write the number $y$ (Fig. 4). By comparing the products in both marked pentads, we get $a b c d e=a b c d y$, or $y=e$. The same equality must, however, also hold for the other two central fields of the table lying on the other diagonal, i.e., $d=a$. It is enough to repeat the entire reasoning for the pentads of fields that arise from the considered pentads by flipping them along the vertical axis of the table.

Now notice in the filled table the pentads of fields marked in Fig. 5. Clearly, we must have $a^{2} b c e=a b c e^{2}$, or $a=e$. We see that the table contains at most three different numbers $a, b, c$ (Fig. 6), with $a^{3} b c=1$. Now it remains to verify that the same product $a^{3} b c$ has every pentad of fields of the form $\$$, which can be placed in the table. Since the filled table is symmetric with respect to both diagonals, and thus also centrally symmetric, it is enough to verify this for only four possible positions of identically oriented pentads (e.g., 母 in the usual position of the letter $\mathrm{T}$ ).

| $a$ | $b$ | $c$ | $e$ |
| :--- | :--- | :--- | :--- |
| $b$ | $a$ | $e$ | $c$ |
| $c$ | $e$ | $a$ | $b$ |
| $e$ | $c$ | $b$ | $a$ |

![](https://cdn.mathpix.com/cropped/2024_04_17_fca565adf729549b0cd4g-02.jpg?height=252&width=263&top_left_y=427&top_left_x=714)

Fig. 5

| $a$ | $b$ | $c$ | $a$ |
| :--- | :--- | :--- | :--- |
| $b$ | $a$ | $a$ | $c$ |
| $c$ | $a$ | $a$ | $b$ |
| $a$ | $c$ | $b$ | $a$ |

Fig. 6

Answer. The table contains at most three different positive numbers $a, b, c$, with $a^{3} b c=1$.
1. In a $4 \times 4$ table, positive real numbers are written such that the product in each pentad of the shape 母 is equal to 1. Determine the maximum number of different numbers written in the table.

Note: The shape 母 refers to a specific arrangement of five cells in the table, resembling the Chinese character 母. For clarity, it typically involves a central cell surrounded by four adjacent cells.