# 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.

• 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.
– dxiv
Nov 18, 2021 at 19:38
• 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 Nov 18, 2021 at 19:41
• but .. (I am not much inside quantum ph.) are boxes with irrational sides physically admissible ? shouldn't they be quantitized ? Nov 18, 2021 at 20:01
• @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. Nov 18, 2021 at 20:15

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 $$(X,Y,Z)=(1,1,-1)$$.