I don't know if this is considered a research question but I stumbled upon collecting formulae 1 for $n=\frac{xyz}{x+y+z}$ and the question is:

Does the diophantine equation $xyz=n(x+y+z)$ for given $n\in \mathbb{N}$ have only finitely many solutions $x,y,z\in \mathbb{N}$ or is there a $n$ such that it has infinitely many solutions?


1 Answer 1


Ruling out the solution $(0,0,0)$, we have $xyz \neq 0$. WLOG $x \geq y \geq z$, then $n(x+y+z) \leq 3n x$ and so we must have $yz \leq 3n$, in particular $\max(y,z) \leq 3n$. So there are finitely many possible $(y,z)$ and finally $x$ is determined by $(y,z)$.

  • $\begingroup$ Thanks! That was too easy, I should have thought of this myself! $\endgroup$
    – user276611
    Aug 2, 2017 at 13:46

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .