Triplet Threat

Given that

$\begin{aligned} x + 2yz & = xyz \\ xyz - x - 2yz & = 0 \\ x(yz-1) -2(yz-1) & = 2 \\ (x-2)(yz-1) & = 2 \end{aligned}$

$\implies \begin{cases} (x-2)(yz-1) = 1 \cdot 2 & \implies x = 3 & \implies yz = 3 \implies \begin{cases} y = 1, z = 3 \\ y = 3, z = 1 \end{cases} \\ (x-2)(yz-1) = 2 \cdot 1 & \implies x = 4 & \implies yz = 2 \implies \begin{cases} y = 1, z = 2 \\ y = 2, z = 1 \end{cases} \end{cases}$

Therefore there are $\boxed 4$ ordered $(x,y,z)$ solutions: $(3,1,3), (3,3,1), (4,1,2), (4,2,1)$ .

We need to find positive integers $(x,y,z)$ such that $x+2yz=xyz$

The above equation is equivalent to $x=xyz-2yz=yz(x-2) \implies \frac{x}{(x-2)}=yz$ .

However, $yz$ is a positive integer and so is $x$ . This implies that $x-2 > 0$ , and since $x$ is an integer, $x \ge 3$ [A]

Since $a-2>0$ and $a>(a-2)$ , $\frac{a}{a-2} > 1$ . However, this quantity should be an integer ( $yz$ ).

This in turn would imply, $\frac{x}{(x-2)} \ge 2$ or $x \ge 2(x-2)$ or $4 \ge x$ . [B]

Combining [A] and [B] and considering the fact that $x$ is an integer, we get $x=3,4$ .

If $x=3$ , we have $yz = \frac{x}{x-2} = 3$ for which we get $(y,z)=(3,1), (1,3)$ or the triplets $(x,y,z) = (3,3,1), (3,1,3)$ .

Similarly, if $x=4$ , we have $yz = 2$ from which we get the triplets $(x,y,z)=(4,2,1),(4,1,2)$ .

Thus, giving $\boxed{4}$ solutions.

$x=yz(x-2)$ . Hence, $(x-2)|x=(x-2)+2$ . So, $(x-2)|2$ . Thus, $x-2\in (-1,\quad 1,\quad 2)$ . From that, $x\in (1,\quad 3,\quad 4)$ . For $x=1$ , we get $yz=-1$ , which is not possible since all are positive integers. For $x=3$ , we get $yz=3$ , hence 2 ordered pairs. Similarly, for $x=4$ , we get 2 ordered pairs.