Can four positive integers be found such that the greatest common divisor of any two of them is greater than 1, and the greatest common divisor of any three of them is 1?
Solution. The following procedure can provide a quadruple of positive integers that meet the conditions. Let the four numbers be $a, b, c$, and $d$. The prime factorization of these numbers is shown in the following table:

| $a$ | $b$ | $c$ | $d$ |
| :---: | :---: | :---: | :---: |
| 2 | 2 |  |  |
|  | 3 | 3 |  |
|  |  | 5 | 5 |
| 7 |  | 7 |  |
|  | 11 |  | 11 |
| 13 |  |  | 13 |

$$
a=2 \cdot 7 \cdot 13=182, \quad b=2 \cdot 3 \cdot 11=66, \quad c=3 \cdot 5 \cdot 7=105, \quad d=5 \cdot 11 \cdot 13=715
$$

The table was constructed such that for any two columns chosen, there is exactly one prime number that appears only in these two columns, ensuring that no prime number appears in at least three columns. Thus, since the prime factorization is unique, the quadruple $\{66 ; 105 ; 182 ; 715\}$ meets the conditions.

Remarks. 1. Since the number of primes is infinite, there are infinitely many such quadruples.

2. The "tabular method" described above can also be used to solve the problem in general: if $n ; k \in \mathbb{Z}^{+}, n \geq k+1$, then it is possible to provide $n$ positive integers such that the greatest common divisor of any $k$ of them is greater than 1, and the greatest common divisor of any $(k+1)$ of them is 1. The appropriate table will have $\binom{n}{k}$ rows, and we will use as many primes: for every $k$-tuple of columns, we open a new row, choose a new prime number, and write it in the appropriate $k$ positions in the row.