Mathematical Reflections 2014, Issue 1 - Problem U289

Problem:
Let $a \geq 1$ be such that $\left(\left\lfloor a^n \right\rfloor \right)^{\frac{1}{n}} \in \mathbb{Z}$ for all sufficiently large integers $n$. Prove that $a \in \mathbb{Z}$.

Proposed by Mihai Piticari.

Solution:
Suppose that $\left(\lfloor a^n \rfloor \right)^{\frac{1}{n}}=m_n$ for all $n \geq n_0$, where $n_0 \in \mathbb{Z}^+$ and $m_n \in \mathbb{Z}$. Hence, $\lfloor a^n \rfloor=m_n^n$, so for all $n \geq n_0$ we have $a^n-1 < m_n^n \leq a^n$, i.e. $$\sqrt[n]{a^n-1} < m_n \leq a \qquad \forall n \geq n_0.$$
By the Squeeze Theorem, since $\displaystyle \lim_{n \to \infty} \sqrt[n]{a^n-1}=\lim_{n \to \infty} a=a$, we have that $\displaystyle \lim_{n \to \infty} m_n=a$. Since $\{m_n\}_{n \geq n_0}$ is a convergent sequence of integers, therefore must be constant for all $n \geq n_1 \geq n_0$. Indeed, $\displaystyle \lim_{n \to \infty} m_n=a$ implies that there exists $n_1 \in \mathbb{Z}^+$ such that for all $n \geq n_1$ we have $a-1/2<m_n<a+1/2$. Since $(a-1/2,a+1/2)$ has length $1$, it follows that there is only $m_n$ in this interval. This means that $m_n$ is constant for all $n \geq n_1$ and so $m_n=a$ for all $n \geq n_1$, which gives $a \in \mathbb{Z}$.