Finding all positive integer solutions for $x+y=xyz-1$ How do I manually solve $x+y=xyz-1$ assuming that $x, y$ and $z$ are positive integers? I was able to guess all possible solutions, but I do not know how to show that these are the only ones:
$x=1, y=1, z=3$
$x=1, y=2, z=2$
$x=2, y=1, z=2$
$x=2, y=3, z=1$
$x=3, y=2, z=1$  
Any hints would be appreciated.

Alternative solution:
$x\mid y+1$ and $y\mid x+1$, so $x-1\le y\le x+1$. Simply check cases:


*

*If $x-1=y$, then $y\mid x+1$ gives $x-1\mid x+1$, i.e. $x-1\mid 2$, i.e. $x\in\{3,2\}$, so $y\mid x+1\in\{4,3\}$, so $y\in\{1,2,3,4\}$.

*If $x=y$, then $x\mid 1, y\mid 1$.

*If $x+1=y$, then $x\mid y+1=x+2$, so $x\mid 2$, so $(x,y)\in\{(1,2),(2,3)\}$.
It's left to check $11$ cases (knowing $(x,y)$ find $z$).
