# Thread: The ceiling of the square root of n problem.

1. ## The ceiling of the square root of n problem.

How can one find all integers n for which $\displaystyle \lceil \sqrt{n}\rceil$ divides n?

2. Let $\displaystyle n \in \mathbb{N}$, and let $\displaystyle \lceil \sqrt{n} \rceil = a$. Clearly, $\displaystyle (a-1)^2 < n \le a^2$. Furthermore, the only numbers between $\displaystyle (a-1)^2$ and $\displaystyle a^2$ which $\displaystyle a$ divides are $\displaystyle a^2$ and $\displaystyle a^2-a$ ($\displaystyle a^2-2a$ is already outside the interval, as $\displaystyle a^2-2a<a^2-2a+1=(a-1)^2$). Thus, $\displaystyle \lceil \sqrt{n} \rceil \mid n \Leftrightarrow n=a^2-a\;\:\text{or}\:\;n=a^2,\:\;1<a \in \mathbb{N}$

The last step is not fully explained, and I'll let you complete the gap.