2. Determine all triples $(p, q, r)$ of prime numbers for which

$$
(p+1)(q+2)(r+3)=4 p q r
$$

(Jaromír Šimša)
Solution. We will show that the given equation is satisfied by exactly three triples of prime numbers $(p, q, r)$, namely $(2,3,5)$, $(5,3,3)$, and $(7,5,2)$.

First, we rewrite the given equation in the form

$$
\left(1+\frac{1}{p}\right)\left(1+\frac{2}{q}\right)\left(1+\frac{3}{r}\right)=4
$$

Since $3^{3}<4 \cdot 2^{3}$, at least one of the three factors on the left side of the modified equation must be greater than $\frac{3}{2}$. For prime numbers $p, q, r$, this necessarily means $p<2$ or $q<4$ or $r<6$. Given that there is no prime number less than 2, we need to investigate the following five possibilities: $q \in\{2,3\}$ and $r \in\{2,3,5\}$. We will now analyze each of these cases individually, substituting the assumed value of $q$ or $r$ into the given equation and solving for the remaining two unknowns (in the domain of prime numbers).

$\triangleright$ For $q=2$, we get $(p+1)(r+3)=2 p r$, from which it follows that $r=3+6 /(p-1)$, which is an integer only for prime numbers $p \in\{2,3,7\}$. However, these correspond to $r \in\{9,6,4\}$, which are not prime numbers.

$\triangleright$ For $q=3$, we get $5(p+1)(r+3)=12 p r$, from which it follows that $p=5$ or $r=5$. For $p=5$, we get the solution $(5,3,3)$, and for $r=5$, the solution $(2,3,5)$.

$\triangleright$ For $r=2$, we get $5(p+1)(q+2)=8 p q$, from which it follows that $p=5$ or $q=5$. For $p=5$, we do not get any solutions in the domain of prime numbers, while for $q=5$, we get the third solution to the given equation, which is the triple $(7,5,2)$.

$\triangleright$ For $r=3$, we get $(p+1)(q+2)=2 p q$, from which $q=2+4 /(p-1)$, which is an integer only for prime numbers $p \in\{2,3,5\}$. Among the corresponding values $q \in\{6,4,3\}$, there is only one prime number, for which we get the solution $(p, q, r)=(5,3,3)$, which we already know.

$\triangleright$ For $r=5$, we get $2(p+1)(q+2)=5 p q$, from which it follows that $p=2$ or $q=2$. For $p=2$, we get the already known solution $(2,3,5)$, while for $q=2$, we get $p=4$.