Mathematical Reflections 2014, Issue 2 - Problem U295

Problem:
Let $a$ be a real number such that $(\lfloor na \rfloor)_{n \geq 1}$ is an arithmetic sequence. Prove that $a$ is an integer.

Proposed by Mihai Piticari.

Solution:
Let $a_n=\lfloor na \rfloor$ and $b_n=n|a|$. By hypothesis, there exists $d \in \mathbb{R}$ such that $a_{n+1}-a_n=d$. Observe that $d \in \mathbb{Z}$ since it is the difference of two integers. Now, $(b_n)_{n \geq 1}$ is positive, increasing and unbounded and $$\lim_{n \to \infty} \dfrac{a_{n+1}-a_n}{b_{n+1}-b_n}=\dfrac{d}{|a|}.$$ By the Stolz-Cesaro Theorem, we obtain $\displaystyle \lim_{n \to \infty} \dfrac{a_n}{b_n}=\dfrac{d}{|a|}$, which gives $$\lim_{n \to \infty} \dfrac{\lfloor na \rfloor}{n}=d.$$ Moreover, since $na-1 < \lfloor na \rfloor \leq na$, we have $$\lim_{n \to \infty} \dfrac{na-1}{n} \leq \lim_{n \to \infty} \dfrac{\lfloor na \rfloor}{n} \leq \lim_{n \to \infty} \dfrac{na}{n},$$ and by the Squeeze Theorem, we obtain $\displaystyle \lim_{n \to \infty} \dfrac{\lfloor na \rfloor}{n}=a$. By uniqueness of the limit, we obtain $a=d$ and the conclusion follows.