Find all positive integers

Question

Find all of non-null positive integers such that:

$$(x-\frac{1}{yxz})\cdot (y-\frac{1}{yxz})\cdot (z-\frac{1}{xyz}) \in \mathbb{N}$$

Answer 1

a) Extract the product, then $(x-\frac{1}{y})\cdot (y-\frac{1}{z})\cdot (z-\frac{1}{x}) \in \mathbb{N}$ if and only if $A=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}-\frac{1}{xyz} \in \mathbb{N}$. Since $x,y,z \geq 1$ then $0<A<3$.

Thus $A=1$ or $A=2$.

Case 1: $A=1$ then we have $xy+yz+zx=xyz+1$. WOLG assume that $x\geq y\geq z\geq 1$. Then $3xy\geq xyz +1 >xyz$, which implies $3>z$. Since $z$ is positive integer, $z\in \{1,2\}$.

If $z=1$ then $xy+x+y=xy+1$, which is impossible with $x,y\geq 1$.

If $z=2$ then $xy+2x+2y=2xy+1$, equivalent to $(x-2)(y-2)=3$, then since $x\geq y \geq z=2$ and they are integer, we conclude that $x=3,y=5$.

So $(x,y,z)=(5,3,2)$ and its permutations.

Case 2: $A=2$ then $xy+yz+zx=2xyz+1$. Same argument leads to $z=1$. We have $xy+x+y=2xy+1$, equivalent to $(x-1)(y-1)=0$, Since $x\geq y \geq 1$, we conclude $y=1$. and $x$ is any positive integer.

So $(x,y,z)=(a,1,1)$ with any integer $a \geq 1$ and its permutations.

b) $(x+\frac{1}{y\cdot z})\cdot (y+\frac{1}{x\cdot z})\cdot (z+\frac{1}{x \cdot y}) \in \mathbb{N}$ if and only if $(xyz)^2 \mid (xyz+1)^3$, which implies $xyz \mid (xyz+1)^3$, and then $xyz \mid 1$.

Since $x,y,z$ are positive integer, only possible case is $x=y=z=1$.