Let's find all integer roots of the following equation: $x+y+z=x y z$.
Since the roots $x, y, z$ are interchangeable, we can assume that $x \leq y \leq z$. If $z=0$, then $x=y=0$, because neither can be positive and $x+y=0$. For the rest, it is sufficient to look for solutions where $z>0$, because the others can be obtained from these by multiplying by -1, which only multiplies both sides of the equation by -1.

If $z>0$ and $x \leq y \leq z$, then $x y z \leq 3 z$ and thus $x y \leq 3$. If $x y=0$, then $x+y+z=0$ and one of $x, y$ must be 0, so $y=0, x=-z$. It is not possible for both $x$ and $y$ to be negative, because then $x+y+z>0$. Thus, two cases are possible: $x=1, y=2, z=x+y=3$; or $x=1, y=3$, but this already gives 2 for $z$, which arises from the previous one by interchange. Therefore, the complete solutions are $1, 2, 3$, and the numbers $-a, 0, a$ where $a$ is any integer, as well as any solutions obtained by arbitrary permutations of these or by multiplying them by -1.

Solved by: Csernók L., Gacsályi S., Gehér L., Gősy S., Jankó B., Kővári T., Vermes R.

Incomplete solutions: Czibere T. and Nagy F., Kossuth G., Kovács J., Párkány M., Perjes P., Tarnay Gy., Tarnóczi T., Ungár Veronika, Vata L., Vörös M.