Show that for each integer $n\geq 2$. there is exactly one bipartite graph order $n$ size $\lfloor n^2/4 \rfloor$
I know that there is no bipartite order $n$ can have size bigger than $\lfloor n^2/4 \rfloor$. But I'm not sure I know how to use this fact to prove this theorem
HINT: If $n$ is even, then the complete bipartite graph that arises from distributing the vertices equally between the partite sets has size $$ \frac{n}{2}\cdot\frac{n}{2}=\frac{n^2}{4}=\left\lfloor\frac{n^2}{4}\right\rfloor. $$
If $n$ is odd, then the complete bipartite graph that arises from distributing the vertices as equally as possible between the partite sets has size $$ \frac{n-1}{2}\cdot\frac{n+1}{2}=\frac{n^2-1}{4}=\left\lfloor\frac{n^2}{4}\right\rfloor. $$
HINT 2: Given the vertex distribution above, suppose we move $k$ vertices from one partite set to the other. What is the size of the resulting complete bipartite graph then? Can it ever equal $\lfloor n^2/4\rfloor$?