Given that $x$, $\frac{1}{x}$, $y$, $\frac{1}{y}$, $z$ and $\frac{1}{z}$ are all integers, how many distinct values of $x+ y+ z$ are possible?
I notice that if $x$ is an integer, then $\frac{1}{x}$ can only be an integer if $x$ is $\pm 1$. Similarly, $y$ and $z$ must also be $\pm 1$. So the possible values of $x$, $y$ and $z$ are $\pm 1$, and there are $2^3 = 8$ ways to assign them. For example, one way is $x = 1$, $y = 1$, $z = -1$. In this case, $x + y + z = 1 + 1 - 1 = 1$. Another way is $x = 1$, $y = -1$, $z = -1$. In this case, $x + y + z = 1 - 1 - 1 = -1$. I can see that the possible values of $x + y + z$ are $\pm 1$ and $\pm 3$. There are four distinct values in total.

