## Task 14/82

Determine all triples $(x ; y ; z)$ of integers $x ; y ; z$ that satisfy the equation:

$$
\frac{x y}{z}+\frac{y z}{x}+\frac{x z}{y}=3
$$
From the given equation, it follows by multiplication

$$
\frac{(x y)^{2}+(y z)^{2}+(z x)^{2}}{3}=x y z
$$

Since the left side of the equation is certainly positive, it follows that $x y z>0$ and thus (due to the required integrality) even $x y z \geq 1$. By the well-known relationship between the arithmetic and geometric means, we further have

$$
x y z=\frac{(x y)^{2}+(y z)^{2}+(z x)^{2}}{3}=\sqrt[3]{(x y)^{2} \cdot(y z)^{2} \cdot(z x)^{2}}
$$

Since $x y z>0$, it follows that $1 \geq \sqrt[3]{x y z}$, and thus also $1 \geq x y z$. Therefore, $x y z=1$. Due to the integrality, we then have $|x|=|y|=|z|=1$, and the following solutions arise:

$$
\left(x_{1} ; y_{1} ; z_{1}\right)=(1 ; 1 ; 1) \quad ; \quad\left(x_{2} ; y_{2} ; z_{2}\right)=(1 ;-1 ;-1)
$$

$$
\left(x_{3} ; y_{3} ; z_{3}\right)=(-1 ; 1 ;-1) \quad ; \quad\left(x_{4} ; y_{4} ; z_{4}\right)=(-1 ;-1 ; 1)
$$

(from $x y z>0$ it follows that either none or exactly two of the three values $x ; y ; z$ are negative).