For $\forall m \in \mathbb{N}, m \ge 3$ there are m elements of S in arithmetic progression.

Question

Let $S=\{[n\pi], n \in \mathbb{N}\}$. Prove $\forall m \in \mathbb{N}, m \ge 3$ there are $m$ elements of $S$ in arithmetic progression.

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I don't know how to prove it, but I have the feeling the approximation of $\pi$ by rationals is helpful here. I mean: $\forall \epsilon > 0$ there is $\frac m n$ so that $|\frac m n - \pi| \lt e$

Answer 1

It is a simple consequence of Szemeredi's theorem. The set of numbers of the form $[n\pi]$ has a positive ($\geq\frac{1}{4}$) upper density in $\mathbb{N}$, hence such a set contains arbitrarily long APs.

On the other hand, if we take a convergent $\frac{p_m}{q_m}$ of the continued fraction of $\pi$, we have: $$\left|\pi-\frac{p_m}{q_m}\right|\leq\frac{1}{q_m^2}$$ hence it is enough to consider $n=q_m,2q_m,\ldots \left\lfloor\frac{q_m}{2}\right\rfloor q_m$ to have an AP with length $\left\lfloor\frac{q_m}{2}\right\rfloor$.

Answer 2

Suppose $x$ has a sequence of approximations $p/q$ with $o(1/q)$ error and $q \to \infty$.

Then for fixed $k$ and large enough $q$, the integers $[qx]$, $[2qx]$, $[3qx], \dots [kqx]$ form an arithmetic progression, since the difference between any two of their consecutive differences is an integer bounded by $q \times o(1/q)$.

Any real number has the necessary quality of approximations. Take $x= \pi$.