<s>[INST] When does $\lfloor (n-1)x \rfloor + \lfloor x \rfloor = \lfloor nx \rfloor$? I am trying to find the conditions under which $\lfloor(n-1)x\rfloor + \lfloor x \rfloor = \lfloor nx \rfloor$. The trivial case is whenever $x \in \mathbb{Z}$. If $n = 2$, then $x - \lfloor x \rfloor \lt \frac{1}{2}$.
As $n$ increases however, the pattern becomes more complicated. For example, for $n = 3$, if $x - \lfloor x \rfloor \lt \frac{1}{3}$ then $\lfloor 3x \rfloor = 3\lfloor x \rfloor$, and $\lfloor 2x \rfloor = 2\lfloor x \rfloor$ so the identity is satisfied. However, if $ \frac{1}{2} \lt x - \lfloor x \rfloor \lt \frac{2}{3}$, then the identity is also satisfied.
I tried seeing whether Hermite's identity can help but I can't see an obvious way of applying it.
 [/INST] Let $m=\lfloor x\rfloor$ and $\alpha=x-m$. Then
$$\lfloor (n-1)x \rfloor + \lfloor x \rfloor =\lfloor(n-1)m+(n-1)\alpha\rfloor+m=nm+\lfloor(n-1)\alpha\rfloor\;,$$
and $$\lfloor nx \rfloor=\lfloor nm+n\alpha\rfloor=nm+\lfloor n\alpha\rfloor\;,$$
so $\lfloor (n-1)x \rfloor + \lfloor x \rfloor = \lfloor nx \rfloor$ iff $\lfloor(n-1)\alpha\rfloor=\lfloor n\alpha\rfloor$, i.e., iff there is an integer $k$ such that $$k\le(n-1)\alpha<n\alpha<k+1\;.\tag{1}$$
$(1)$ is equivalent to $$\frac{k}{n-1}\le\alpha<\frac{k+1}n\;,$$
so we want the non-empty intervals of the form $$\left[\frac{k}{n-1},\frac{k+1}n\right),\quad k=0,\dots,n-1\;.$$
Now $\frac{k}{n-1}<\frac{k+1}n$ iff $kn<(k+1)(n-1)=kn+n-k-1$ iff $n>k+1$, so
$$\lfloor (n-1)x \rfloor + \lfloor x \rfloor = \lfloor nx \rfloor\quad\text{iff}\quad x-\lfloor x\rfloor\in\bigcup_{k=0}^{n-2}\left[\frac{k}{n-1},\frac{k+1}n\right)\;.$$
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