An algebra problem by Aira Thalca

Putting $\xi = x$ , $\eta = 2y$ , $\zeta= 4z$ , we see that $\xi,\eta,\zeta$ are the roots of the cubic $0 \; = \; X^3 - 9X^2 + 26X - 24 \; = \; (X-2)(X-3)(X-4)$ and so $\{\xi,\eta,\zeta\} = \{2,3,4\}$ . No matter how we assign the values $2,3,4$ to $\xi,\eta,\zeta$ , $x=\xi$ will always be an integer. Provided that $\eta$ is even, $y$ will be an integer. The only way that $z$ will be an integer is if $\zeta=4$ .

Of the six ways of assigning the three values $2,3,4$ to $\xi,\eta,\zeta$ , either $y$ or $z$ will be an integer unless $\eta$ is odd and $\zeta \neq 4$ . The only way this can happen is if $\xi=4$ , $\eta=3$ and $\zeta=2$ . In all of the other $\boxed{5}$ arrangements, either $y$ or $z$ is an integer, and so at least two of $x,y,z$ are integers.