6. Find all natural numbers that can be represented in the form $\frac{x y+y z+z x}{x+y+z}$, where $x, y$ and $z$ are three distinct natural numbers. (D. Khramtsov)
Answer. All natural numbers except 1. Solution. Any natural number $n>1$ can be obtained by setting $x=1, y=n$ and $z=n^{2}: \frac{x y+y z+z x}{x+y+z}=\frac{n\left(1+n^{2}+n\right)}{1+n+n^{2}}=n$. We will show that the number 1 cannot be obtained. For this, it is sufficient to show that if the natural numbers $x, y$ and $z$ are distinct, then $x y+y z+z x>x+y+z$. Indeed, $x y \geq x, y z \geq y$ and $z x \geq z$, and these inequalities turn into equalities only when $x=y=z=1$. Remark. We will present another way to represent any number $n>1$. First, choose numbers $x>y$ such that $x+y=n+1$ (then $x \geq 2$). Note that

$$
\frac{x y+y z+z x}{x+y+z}=x+y-\frac{x^{2}+x y+y^{2}}{x+y+z}
$$

Thus, by setting $z=\left(x^{2}+x y+y^{2}\right)-(x+y)$, we obtain the desired triplet if we verify that $z$ is different from $x$ and $y$. Since $x \geq 2$, we have $x^{2}-x \geq x$ and $y^{2}-y \geq 0$, hence $z=\left(x^{2}-x\right)+\left(y^{2}-y\right)+x y \geq x+0+y>\max (x, y)$. This is what we needed to show.