Let me start with a hint that hopefully helps :)

In the problem we want $\lfloor \sqrt{n} \rfloor$ to be a factor of $n$. Let's try to take advantage of the fact that floors are always an integer.

If $k = \lfloor \sqrt{n} \rfloor$, then note that $k \leq \sqrt{n} < k+1$ (think about how we round down for the floor). Hence, $k^2 \leq n < k^2 + 2k + 1$. From here, try to consider how $k$ can be a factor of $n$. The advantage we have now is that we know both $k$ and $n$ are integers.

Hope this helps! If necessary, try to look at a few values of $k$ and try to find a pattern.