$(1+1/x)(1+1/y)(1+1/z) = 3$ Find all possible integer values of $x$, $y$, $z$ given all of them are positive integers. 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 $z$
 A: 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 $x = 2$.

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.
A: 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.
A: This is a variation on Fermat's theorem.  The number of solutions is nite.  Will write a more General equation:
$$(x+r)(y+r)(z+a)=3xyz$$
If  $z,a$ - Will ask themselves.  Then the solution can be written;
$$r=ps(2z-a)$$
$$x=((p+3s)z+ap)s$$
$$y=(z+a)(p+s)p$$
$$***$$
$$x=((z+a)p+s)s$$
$$y=(z+a)(3zp+s)p$$
$p,s$ - integers asked us.  You must consider that you can solve then reduce by common divisor.
