# Maximum number of unit squares that can be fit into side length $\sqrt{24}$ square

I was inspired by this question Algorithm to find the largest number of identical squares that fit into a square with a specified area (ex: 972), where OP and we surmised that the solution had to use unit tiles (since the identical squares needed to have integer side lengths)

One question posed toward my solution of $$\displaystyle \lfloor \sqrt{A}\rfloor^2$$ for the max number of unit tiles that could fit inside a square of area $$A$$ was:

Is there some way to fit in more than that amount?

So I thought (for example purposes) there would be ample room to experiment with if $$A=n^2-1$$ for some integer $$n$$.

I don't know how to begin, and I personally don't think it's possible, since there is neither enough horizontal space for more than $$\displaystyle \lfloor \sqrt{A}\rfloor$$ tiles in any column (# of tiles touching some $$x=x_1$$) nor enough vertical space for more than $$\displaystyle \lfloor \sqrt{A}\rfloor$$ tiles in any row (# of tiles touching some $$y=y_1$$).

Wouldn't this necessarily bound the number of tiles to $$\lfloor \sqrt{A}\rfloor^2$$?

Packing problems are hard. There are often irregular configurations that allow denser packing. The Wikipedia page gives an example of packing five unit squares in a square of side $$2+\frac 1{\sqrt 2} \approx 2.707$$ The formula $$\lfloor \sqrt A \rfloor^2$$ would say you can only put four in. That formula assumes that all the packing squares are aligned with the outer square.