Find $x;y;z\in \mathbb{Z}^+$ such that : $\left\{\begin{matrix} (xy+1)\vdots z & \\ (xz+1)\vdots y & \\ (yz+1)\vdots x & \end{matrix}\right.$

Question

Find $x;y;z\in \mathbb{Z}^+$ such that : $\left\{\begin{matrix} (xy+1)\vdots z & \\ (xz+1)\vdots y & \\ (yz+1)\vdots x & \end{matrix}\right.$

Thanks :)

I have tried that :

$\left\{\begin{matrix} (xy+1)\vdots z & \\ (xz+1)\vdots y & \\ (yz+1)\vdots x & \end{matrix}\right.$ $\Rightarrow \frac{(xy+1)(yz+1)(zx+1)}{xyz}=k(k\in \mathbb{Z})$ $\Rightarrow(x^2y^2z^2+x^2yz+xyz^2+xy^2z+xy+yz+zx+1)\vdots xyz$ $\Rightarrow (xy+yz+zx+1)\vdots xyz$

Then I don't know how to do next !?

What does $\vdots$ mean ?

Examples : $4\vdots 2$; $6\vdots 3$;$202\vdots 101$,...

Answer 1

This is a typical problem. See here for a similar variant, with a similar approach.

WLOG, $x \geq y \geq z$. Show that $ 0 \leq xy + yz + zx + 1 < 4xy$. Hence, $z= 1, 2, 3$.

If $z=1$, then $ xy + x+y + 1 \leq 4 xy$.
Solve for $x+y+1 = 2xy$ in the standard way of factorization / Simon's Favourite Factorization Trick. We have $2 = xy -x -y + 1 = (x-1)(y-1)$. Hence, this has solution $x=3, y = 2$ (since $x\geq y$). Solve $x+y+ 1 = xy$, and $x+y+1 = 3xy$ as above.

If $z=2$ and $z=3$, use the same approach.