Proving a set of numbers has arithmetic progressions of arbitrary length, but none infinite For each real number $x$, let $[x]$ be the largest integer less than or equal to $x$. For example,
$$[5] = 5$$
$$[7.9] = 7,$$ and $$[−2.4] = −3.$$ 
An arithmetic progression of length $k$ is a sequence $a_1, a_2, \dots, a_k$ with the property that there exists a real number $b$ such that $a_{i+1} − a_i = b$ for
each $1 \leq i \leq k − 1.$
Let $\alpha > 2$ be a given irrational number. Let $S = \{[n · α] : n ∈ Z\}$, the set of all integers that are equal to $[n  \alpha]$ for some integer $n$.
(a) Prove that for any integer $m \geq 3$, there exist m distinct numbers contained in $S$ which form an arithmetic progression of length $m$.
(b) Prove that there exist no infinite arithmetic progressions contained in $S$.
I'm thinking that for (a), some application pigeonhole principle can be considered... but otherwise, I don't really know how to prove this. 
Some suggestions on a strategy would be awesome. Thanks.
 A: Outline: We want to construct an arithmetic sequence of length $n$ that is a subsequence of the sequence $(\lfloor i\alpha\rfloor)$.
For $i=1,2,3, \dots, 10n+1$, let $b_i$ be the fractional part of $i\alpha$, that is, $i\alpha-\lfloor i\alpha\rfloor$.  These fractional parts are all distinct, since $\alpha$ is irrational. By the Pigeonhole Principle, there exist $i$ and $j$ with $i\lt j$ such that $|b_j-b_i|\lt \frac{1}{10n}$.  Let $k=j-i$.  Then either (i) $b_k\gt 1-\frac{1}{10n}$ or (i) $b_k\lt \frac{1}{10n}$.
We look first at case (i). Consider the numbers $k\alpha$, $2k\alpha$, $3k\alpha$ and so on up to $nk\alpha$. The fractional parts of these are slowly decreasing.  Thus the numbers $\lfloor k\alpha\rfloor$,
$\lfloor 2k\alpha\rfloor$, $\lfloor 3k\alpha\rfloor$, and so on up to $\lfloor nk\alpha\rfloor$ form an arithmetic progression of length $n$. There is a lot of slack, we can continue well beyond $\lfloor nk\alpha\rfloor$.
Case (ii) is essentially the same, except that the fractional parts of the $ik\alpha$ slowly increase instead of decreasing.     
The argument that our sequence has no infinite arithmetic progression uses similar ideas.  
