Numbers of the form $\frac{xyz}{x+y+z}$

Question

It is easy to show that every natural number $n$ can be written as $n = \frac{xy}{x+y}$ with $x,y\in \mathbb{N}$ by setting $x = y = 2n$. Now I experimented a little bit with numbers of the form $n=\frac{xyz}{x+y+z}$ and it seems that every natural number $n$ can be written this way with $x,y,z \in \mathbb{N}$. Now I am asking myself if there is some formula which I am missing which would explain this, or if there is some counterexample. Notice that by setting $x=n, y=2n , z = 3n$ we get $n^2=\frac{xyz}{x+y+z}$, hence there are infinitely many numbers $n^2$ expressible in this way. If you happen to know a formula (maybe for some special case) then feel free to add it.

Edit: One last formula for $n\equiv 1 \mod(2)$: $x=3,y=\frac{n+3}{2},z=n$.

Answer 1

Chose $x = 2n$, $y = 2n+1$, $z = 1$

Then $$\frac{xyz}{x+y+z} = \frac{2n(2n+1)}{4n+2} = n$$

Answer 2

Also another formula which is based on what Aryabhata wrote: Let $n=a\cdot b$ and set $x = 2b,y = 2b+a, z = a$ Then

$\frac{xyz}{x+y+z} = \frac{2b(2b+a)a}{2b+2b+a+a} = \frac{2n(2b+a)}{4b+2a}=n$

This shows, that there are at least as many solutions as there are divisors of $n$.

Answer 3

$$\frac{n(n+2)2}{n+(n+2)+2}=n $$

Answer 4

Also another solution: $x = n+1, y = (n+1)^2-1, z=1$ Then $\frac{xyz}{x+y+z} = \frac{(n+1)((n+1)^2-1)}{(n+1)(n+2)} = \frac{n(n+2)}{n+2}=n$

Answer 5

Yet another solution: $x = 1, y = n+2, z = n(n+3)/2$