Integer Part and Arithmetic Progressions For $A=\{\lfloor n \alpha \rfloor: n\in\Bbb Z \}$, where $\alpha$ irrational, $\alpha \gt 2$, we aim to show the following:


*

*There exists $m$ elements contained in $A$ that form an arithmetic progression, for any $m \gt 2$, $m \in\Bbb N$.

*There exists no infinite arithmetic progression in $A$.
 A: Find $n$ such that $\{n\alpha\}=n\alpha-\lfloor n\alpha\rfloor<\frac 1m$.
Then $0<kn\alpha-k\lfloor n\alpha\rfloor<1$ for $k=1,\ldots, m$, hence $\lfloor kn\alpha\rfloor =k\lfloor n\alpha\rfloor$.
Assume there is an infinite arithmetic progression, i.e. we have $a,d\in\mathbb Z$, $d\ne 0$ such that $a+k d\in A$ for all $k\in\mathbb N_0$.
Select $n_k$ such that $\lfloor n_k\alpha\rfloor=a+kd$.
Note that $\alpha>1$ implies that $n_k$ is uniquely determined.
Let $m_k=n_0+k(n_1-n_0)$.
Claim: $m_k=n_k$.
This is clear for $k=0$ and for $k=1$.
Assume we know already $m_i=n_i$ for all $i\le k$.
Then $$\begin{align}m_{k+1}\alpha &= (m_k+n_1-n_0)\alpha\\
&=n_k\alpha+n_1\alpha-n_0\alpha\\
&=(a+kd)+\{n_k\alpha\}+(a+d)+\{n_1\alpha\}-a-\{n_0\alpha\}\\
&=a+(k+1)d+\{n_k\alpha\}+\{n_1\alpha\}-\{n_0\alpha\},\end{align}$$
i.e. $-1<m_{k+1}\alpha-(a+(k+1)d)<2$.
Since also $0<n_{k+1}\alpha-(a+(k+1)d)<1$, we conclude
$$-2<(m_{k+1}-n_{k+1})\alpha<2$$
and with $\alpha>2$ we obtain $m_{k+1}=n_{k+1}$.
This proves the claim.
Thus we find that 
$$0<(n_0+k(n_1-n_0))\alpha-(a+kd)<1$$
for all $k\in\mathbb N$. But this implies
$$0<\alpha-\frac d{n_1-n_0}<\frac{1+a-n_0\alpha}k$$
which is absurd if we choose $k>\frac{1+a-n_0\alpha}{\alpha-\frac d{n_1-n_0}}$.
A: As it has already been mentioned by Hagen von Eitzen, we could obtain arbitrary long finite arithmetic progression $\large\ \left\lfloor kN_\varepsilon\alpha\right\rfloor\ $ selecting $\large\ N_\varepsilon\ $ such that $\large\ \left\{N_\varepsilon\alpha\right\}<\varepsilon\ $ for sufficiently small $\large\ \varepsilon$.  
Let infinite arithmetic progression exists:
$$\large\exists\ D,A,N\in\mathbb{N}\ \ \ \ \ A=\left\lfloor\alpha N\right\rfloor\ \ \ \ \forall\ k>0\ \ \ \ \exists\ M_k\in\mathbb{N}\ \ \ \ A+kD=\left\lfloor\alpha(N+M_k)\right\rfloor$$
So,
$\large A<\alpha N<A+1$
$\large A+kD<\alpha(N+M_k)<A+kD+1$
thus
$\Large\forall k\ \ \ \frac{kD-1}\alpha<M_k<\frac{kD+1}\alpha$
which implies
$\Large\forall k\ \ \{\frac{kD+1}\alpha\}<\frac 2\alpha$
that contradicts with uniform distribution of $\large\ \{\beta k\}\ $ for irrational $\large\ \beta=\frac D\alpha$
