$(1+1/x)(1+1/y)(1+1/z) = 3$ Find all possible integer values of $x$, $y$, $z$ given all of them are positive integers.

Question

Find all possible integer values of $x$, $y$, $z$ given all of them are positive integers and $$(1+1/x)(1+1/y)(1+1/z) = 3.$$

I know $(x+1)(y+1)(z+1) = 3xyz$ which is no big deal. I can't move forward now.

morever it is given $x$ is less than or equal to $y$ and $y$ is less than or equal to $z$

Answer 1

Hint: Suppose that $x, y, z \ge 3$. Then $$1 + \frac 1 x \le \frac{4}{3}$$ and likewise for the other two. Then

$$\left(1 + \frac 1 x\right)\left(1 + \frac 1 y\right)\left(1 + \frac 1 z\right) \le \frac{64}{27} < 3$$

So one of the numbers has to be pretty small; now consider cases with $x = 1$ and $x = 2$.

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Something else that you know: $3$ is a divisor of the right side, so it's a divisor of the left side. Since $3$ is prime, $3$ has to divide one of the individual terms.

Answer 2

Let $x$ be the smallest of the three.

Case 1: $x=1$.

Then $$ y=\frac{2\,z+2}{z-2} = 2 + \frac{6}{z-2}$$ So $z-2$ should divide $6$ So the possibilities are $z=3,4,5,8$ If we want $x \le y \le z$, we can eliminate two of them.

case 2: $x=2$

Then $$ y = \frac{z+1}{z-1} = 1 + \frac{2}{z-1}$$ So $z-1$ should divide 2, so z=2, or 3.

case 3:$x > 2$.

In this case show that $y < x$ and we can stop.