nicetry007 wrote: Suppose $ n = k^{2}$. Then $ \;\sqrt{n}\; = k$ and $ k \;| \;k^{2}$. Suppose $ k^{2}\;< \;n \;< \;(k+1)^{2}$ and $ n = k^{2}+c$. Then $ \lfloor \sqrt{n}\rfloor = k$. $ k\; |\; n\; \Leftrightarrow \;k\; |\; k^{2}+c \;\Leftrightarrow\; k\;|\; c \;\Leftrightarrow\; c = k$ or $ 2k$. Therefore, whenever $ n = k^{2}, \; k^{2}+k$ or $ k^{2}+2k\;$, $ \;\lfloor\sqrt{n}\rfloor\; |\; n$.

Write $n = k^2 + q$ with $k = \lfloor \sqrt{n}\rfloor$; $k|n \iff q\in\{0, k, 2k\}$. Thus, $n$ is of the form $k^2$, $k(k+1)$, or $k(k+2)$. Clearly all such numbers work.