## Task 20/89

Determine all triples $(x ; y ; z)$ of natural numbers $x, y, z$ for which the equation $x+y+z+2=x y z$ holds.
Since the equation is symmetric in the variables, we first set $y=x+a, z=x+b$ with $a ; b \in N$. The equation then takes the form:

$$
x+x+a+x+b+2=x(x+a)(x+b), \quad 3 x+a+b+2=x^{3}+(a+b) x^{2}+a b x
$$

1. Let $x=1$. Then, after simplification, $a b=4$, so $a=1$ and $b=4$ or $a=b=2$ ( $a=4$ and $b=1$ only represents a swap of $y$ and $z$ and will be captured later), and we obtain the first solutions (which are confirmed by verification):

1.1. $x=1 ; y=2 ; z=5$ and 1.2. $x=1 ; y=3 ; z=3$.

2. Let $x=2$. Then, after simplification, $3(a+b)+2 a b=0$. This equation is only solvable for $a ; b \in N$ with $a=b=0$, and we obtain the (verified) third solution: $x=2 ; y=2 ; z=2$.
3. Let $x \geq 3$, so $x=3+c$ with $c \in N$. Then, after simplification, $0=16+f(a ; b ; c)$, where $f(a ; b ; c) \geq 0$. This equation is not solvable for any triplet $(a ; b ; c)$ with $a ; b ; c \in N$. Thus, there is no solution for $x \geq 3$.

Since $x=0$ clearly does not provide a solution, all triples (up to permutation) have been found. By swapping, a total of 10 solutions are obtained:

$(1 ; 2 ; 5),(1 ; 5 ; 2),(2 ; 1 ; 5),(2 ; 5 ; 1),(5 ; 1 ; 2),(5 ; 2 ; 1),(1 ; 3 ; 3),(3 ; 1 ; 3),(3 ; 3 ; 1),(2 ; 2 ; 2)$.

## Solution by Katrin Böhme:

It can be easily verified that none of the numbers $x, y, z$ can be zero. Furthermore, two of the numbers cannot simultaneously be equal to 1.

W.L.O.G., assume $x \leq y \leq z$. Then $x \geq 1$ and $y \geq 2$ and

$$
\begin{gathered}
x+y+z+2=x y z \quad \rightarrow \quad y+z+2=x(y z-1) \quad \rightarrow \quad \frac{y+z+2}{y z-1}=x \geq 1 \\
y+z+2 \geq y z-1 \quad \rightarrow \quad z+3 \geq y(z-1) \quad \rightarrow \quad \frac{z+3}{z-1} \geq y \geq 2 \\
z+3 \geq 2 z-2 \quad \rightarrow \quad 5 \geq z
\end{gathered}
$$

Thus, the solution set is restricted:

The largest of the three numbers $x, y, z$ is not greater than 5. Using (1) and (2), the solutions are

$$
(x=1, y=3, z=3), \quad(x=1, y=2, z=5), \quad(x=2, y=2, z=2)
$$

and the permutations thereof.

## Solution by Karsten Wolter:

1. $x=y=z=0$ is not a solution.
2. The assumption $x>2, y>2, z>2$ (so $x=2+\bar{x}, y=2+\bar{y}, z=2+\bar{z}$ with $\bar{x}, \bar{y}, \bar{z}>0$) leads to a contradiction.
3. From the symmetry of the equations with respect to $x, y$ and $z$, the solutions are also symmetric.

Thus, only the following two cases need to be investigated:

1. $z=1$ : It follows that $y=\frac{x+3}{x-1}$. Integer (natural!) solutions exist only for $x=2, y=5 ; x=3, y=$ $3 ; x=5, y=2$.
2. $z=2$ : It follows that $y=\frac{x+4}{2 x-1}$. Integer (natural!) solutions exist only for $x=1, y=5$; $x=2, y=2 ; x=5, y=1$.

From 3., permutations yield a total of 10 solutions.

Solution by Gerhard Fritzsche:

First, assume $x \leq y \leq z$. Clearly, $x \neq 0$, as the assumption $x=0$ leads to a contradiction. Similarly, the assumption $x=y=1$ leads to a contradiction.

Let $(x ; y ; z)$ be a solution to the equation. If one of the numbers $x, y, z$ is increased by 1, the left side increases by 1, while the right side increases by the product of the other two numbers, which is more than 1. Therefore, to find another solution starting from the solution $(x ; y ; z)$, at least one of the numbers must be increased and another must be decreased.

First, set $x=y=z$. Then $3 x+2=x^{3}$, and this equation has the solutions $x=2$ and $x=-1$ (double solution).

Thus, $(x ; y ; z)=(2 ; 2 ; 2)$ is a solution.

When transitioning to further solutions, at least one of the numbers must be less than 2. Suppose this is $x=1$.

We set $y=z$ and obtain the equation $2 y+3=y^{2}$ with the solutions $y=z=3$ and $y=z=-1$. Thus, $(x ; y ; z)=(1 ; 3 ; 3)$ is a solution.

If further solutions exist, $y>3$ must hold. Since $x=y=1$ is excluded, only $y=2$ remains. We get $z+5=2 z$, so $z=5$. This yields the third solution $(x ; y ; z)=(1 ; 2 ; 5)$.

The method is now exhausted, and no further solutions can exist. If we now drop the assumption $x \leq y \leq z$, we obtain a total of 10 solution triples by permutation.