Any number from $k^2$ to $k(k +2)$ shall have square root between $k$ and $k+1$ as $k(k+2) +1$ shall have square root $k + 1$
So the floor function of any number $k^2$ to $k(k+2)$ shall be $k$. For $k$ to divide the number with in the range it has to be of the form $k^2$ or$ k(k+1)$ or $k(k+2)$
So for any $k $ $k^2 , k(k+1),k(k+2)$ satisfy the criteria
