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This post is the analogous of that one (about $\sqrt{2}$) but with a much stronger expectation here.

We observed, and then this comment of Lucia proved, that for $\phi$ the golden ratio, if $n$ and $m=\lfloor n\phi\rfloor$ are odd then so is $\lfloor m\phi\rfloor$. Next EmilJeřábek3.0 provided a family of quadratic algebraic integers satisfying this property (also in comments, by extended Lucia's proof).

Let $L$ be the set of irrational positive numbers $\alpha$ satisfying the property that if $n$ and $m=⌊nα⌋$ are odd then so is $⌊mα⌋$.

Question: What is the set $L$, explicitly?

Remark: Then $L \cap S = \emptyset$, with $S$ defined in this post on the borderline Collatz-like problems.

Improve this question
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    $\begingroup$ Yes, by some easy algebra. Write $n\phi = m+\theta$, and then $m\phi = n\phi^2 -\theta\phi = n\phi + n - \theta \phi = m+n +\theta -\theta \phi$. This is a little less than the even number $m+n$. $\endgroup$
    – Lucia
    Feb 25, 2020 at 13:48
  • $\begingroup$ @EmilJeřábek3.0: Correct! It should be asked whether it is unique up to the addition of an even integer. $\endgroup$
    – Sebastien Palcoux
    Feb 25, 2020 at 16:49
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    $\begingroup$ It is far from unique. If I didn’t make a numerical mistake (which should only affect the side conditions anyway), Lucia’s argument applies to all numbers of the form $\frac12\bigl(b+\sqrt{b^2+4c}\bigr)$ where $b,c$ are integers such that $b>0$, $b^2+4c$ is a positive nonsquare, and either $b\ge c>0$ and $b\equiv c\pmod2$, or $c<0$ and $b\not\equiv c\pmod2$. $\endgroup$
    – Emil Jeřábek
    Feb 25, 2020 at 17:07
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    $\begingroup$ I don’t know. I just tried to push Lucia’s argument to its limit. The argument in fact gives that if $\alpha^2=b\alpha+c$, then $m\alpha=bm+cn+(b-\alpha)\theta$. The conditions I gave ensure that $|b-\alpha|<1$ and that it has the right sign w.r.t the parity of $bm+cn$ (i.e., $b+c$); thus, if correct, I think that this might give a full classification for the limited case when $\alpha$ is a quadratic algebraic integer. Whether there are examples of a different nature, I don’t know. $\endgroup$
    – Emil Jeřábek
    Feb 25, 2020 at 17:33
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    $\begingroup$ Actually, pretty much all such numbers are conterexamples. But to be completely explicit, just take $\alpha=\phi+2$, $n=1$, $m=\lfloor \alpha\rfloor=3$, then $\lfloor m\alpha\rfloor=10$. Here is what goes wrong: assume $\alpha'$ satisfies the property and $\alpha=\alpha'+2k$. Let $n$ and $m=\lfloor n\alpha\rfloor$ be odd. Then $m'=\lfloor n\alpha'\rfloor=m-2kn$ is odd, hence $\lfloor m'\alpha'\rfloor$ is odd (by assumption), hence $\lfloor m'\alpha\rfloor=\lfloor m'\alpha'\rfloor+2km'$ is odd, but there is no reason for $\lfloor m\alpha\rfloor=\lfloor m'\alpha+2k\alpha\rfloor$ to be odd. $\endgroup$
    – Emil Jeřábek
    Feb 26, 2020 at 14:27

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