Distinct integer solutions to linear equations involving irrational numbers Let $\alpha$ and $\beta$ be irrational numbers such that $\alpha/\beta$ is also irrational.  Can we prove that there are no possible integer solutions for
$$
x+ \alpha y + \beta z = x' + \alpha y' + \beta z'
$$
with $x \neq x'$, $y \neq y'$, and $z \neq z'$?
It is obvious that no solutions exist with only one of the pairs "mismatched" (i.e., $x \neq x'$, $y = y'$, $z = z'$).  And it is fairly trivial to prove that there exist no solutions with two "mismatched" pairs (for example, $x \neq x'$ and $y \neq y'$ but $z = z'$);  the proof proceeds by contradiction.  But I cannot see how to straightforwardly extend the proof to show that no solutions exist where all three values are distinct.
For context, this is related to a question over on Physics.SE concerning the degeneracies of the energy levels in a 3-D box.  I realized in writing up my answer there that I didn't have a pithy proof for the statement I made about boxes whose side lengths are irrational multiples of each other, though I would be surprised if it turned out to be false.

Note that your question is equivalent to asking whether there are integral solutions to
$$X+\alpha Y+\beta Z=0,$$
by taking $X=x-x'$, $Y=y-y'$ and $Z=z-z'$.
Simple counterexamples have been given in the comments, for example $\beta=\alpha+1$ for any irrational $\alpha$, with $(X,Y,Z)=(1,1,-1)$.
