Mathematical Reflections 2013, Issue 5 - Problem J278

Problem:
Find all positive integers $n$ for which $$\{\sqrt[3]{n}\} \leq \dfrac{1}{n},$$ where $\{x\}$ denotes the fractional part of $x$.

Proposed by Ivan Borsenco.

Solution:
Clearly, every perfect cube satisfies the condition. Now, let $m \in \mathbb{Z}^+$ such that $m^3<n<(m+1)^3$, i.e. $n=m^3+k$ with $1 \leq k \leq (m+1)^3-1$. Then, $$\{\sqrt[3]{m^3+k}\} \geq \{\sqrt[3]{m^3+1}\}=\sqrt[3]{m^3+1}-m>\dfrac{1}{m^3+1},$$ for all $m>2$. Therefore, it's enough to find the integers which satisfy the condition in $(1,8) \cup (8,27)$. An easy check shows that the required integers are $n=2,9$. In conclusion, $n=2,9$ or $n=m^3$ for some $m \in \mathbb{Z}^+$.