4. Find all triples of natural numbers $x, y$ and $z$ such that $(x+1)(y+1)(z+1)=3 x y z$
Solution. Since the left side of the equation is divisible by 3, without loss of generality, let $\mathrm{x}=3 \mathrm{k}-1$, where $k \in N$, and we obtain the following estimate:

$3 k(y+1)(z+1)=3(3 k-1) y z \geq 3(3 k-k) y z$ or $(y+1)(z+1) \geq 2 y z$. Further, we have
$y z+y+z+1 \geq 2 y z \Leftrightarrow y+z+1 \geq y z \Leftrightarrow 2 \geq(y-1)(z-1)$. The last inequality holds independently of $x=3 \mathrm{k}-1$ and is symmetric with respect to the variables y and z.

Consider 2 cases with respect to the variable y.

1) Let $y-1>0$, then from $2 \geq(y-1)(z-1)$ it follows that either $z=3$ and $y=2(z=2$ and $y=3$ ) , or $z=y=2$. If $z=3$ and $y=2$ ( $z=2$ and $y=3$ ) then $x=2$. Taking into account the symmetry of the original equation with respect to the permutation of variables, we get $(x, y, z) \in\{(2,2,3),(2,3,2),(3,2,2)\}$

2) Now let $y=1$, then from the equation $3 k(y+1)(z+1)=3(3 k-1) y z$ we get

$2 k(z+1)=(3 k-1) z$ or $z=\frac{2 k}{k-1}=2+\frac{2}{k-1}$. The last equation in natural numbers has the following solutions $\mathrm{k}=2, \mathrm{z}=4$ and $\mathrm{k}=3, \mathrm{z}=3$.

$(x, y, z) \in\{(5,1,4),(5,4,1),(4,1,5),(4,5,1),(1,4,5),(1,5,4),(8,1,3),(8,3,1),(3,1,8),(3,8,1),(1,3,8),(1,8,3)\}$

Answer: $(x, y, z) \in\{(2,2,3),(2,3,2),(3,2,2),(5,1,4),(5,4,1),(4,1,5),(4,5,1),(1,4,5),(1,5,4),(8,1,3),(8,3,1),(3,1,8),(3,8,1),(1,3,8),(1,8,3)\}$