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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.

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  • $\begingroup$ The assumptions are not enough, consider for example $1+\sqrt{8} = 1 + 2 \sqrt{2}$ or $1+\sqrt{2}=\sqrt{5 + 2 \sqrt{2}}$. This would only hold true if $\alpha,\beta$ are rationally independent. $\endgroup$
    – dxiv
    Nov 18 '21 at 19:38
  • $\begingroup$ you should add the condition that also $m \alpha +n \beta$ should be irrational. In fact if $\alpha =pi , \beta = 1-\pi$ then you can find integer solutions $\endgroup$
    – G Cab
    Nov 18 '21 at 19:41
  • $\begingroup$ but .. (I am not much inside quantum ph.) are boxes with irrational sides physically admissible ? shouldn't they be quantitized ? $\endgroup$
    – G Cab
    Nov 18 '21 at 20:01
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    $\begingroup$ @dxiv: There's the concept I was looking for. And it looks like if $\alpha$, $\beta$ and 1 are rationally independent, then the proof follows by definition of rational independence. $\endgroup$ Nov 18 '21 at 20:15
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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 $(X,Y,Z)=(1,1,-1)$.

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