5. Find all triples of natural numbers for which the following condition is satisfied: the product of any two of them, increased by one, is divisible by the remaining number.
Solution. Only the triplet $(1 ; 1 ; 1)$ works. First, note that among these numbers there cannot be any even numbers: if $a$ is even, then $bc+1$ is odd. Now let's prove that these numbers must be coprime. If this is not the case, assume $\text{GCD}(a, b)=m>1$, then $ac+1$ gives a remainder of 1 when divided by $m$ and, consequently, cannot be divisible by $b$. Consider the number $ab + ac + bc + 1$. It is divisible by the coprime numbers $a, b$, and $c$ (since by the condition, the product of any two numbers, increased by one, is divisible by the third number), hence it is divisible by $abc$. Then the expression $\frac{1}{c} + \frac{1}{b} + \frac{1}{a} + \frac{1}{abc}$ must be an integer. But this is impossible, since the sum of the fractional parts in this sum is strictly less than 1: if any of the numbers $a, b, c$ are greater than one, then they are at least three. Answer: $(1 ; 1 ; 1)$.