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.

  • $\begingroup$ 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
  • 2
    $\begingroup$ @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

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.