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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 $A$.

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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 $k>\frac{1+a-n_0\alpha}{\alpha-\frac d{n_1-n_0}}$.

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Good, but I was wondering should it be $0<\alpha-\frac d{n_1-n_0}<\frac{1+a-n_0\alpha} {k(n_1-n_0)}$? – fmat Nov 30 '12 at 9:17

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,
$\large A<\alpha N<A+1$
$\large A+kD<\alpha(N+M_k)<A+kD+1$
$\Large\forall k\ \ \ \frac{kD-1}\alpha<M_k<\frac{kD+1}\alpha$
which implies
$\Large\forall k\ \ \{\frac{kD+1}\alpha\}<\frac 2\alpha$
that contradicts with uniform distribution of $\large\ \{\beta k\}\ $ for irrational $\large\ \beta=\frac D\alpha$

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