# 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 $\lceil
\sqrt{n}\rceil$
divides n?

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

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