I want to show the lower bound $\Omega(n^{3/2})$ for the Turan number $ex(n,C_4)$.
For a prime power $q$, a finite affine plane of order $q$ has $q^2$ points and $q^2 + q$ lines; each line contains $q$ points, and each point is on $q + 1$ lines. I think this is of interest since we can form a bipartite graph $G$ where the bipartitions are the set of lines and the set of points, and an edge connects the line $l$ with the point $p$ if $p \in l$. By the property of affine planes, this graph is $C_4$-free. Furthermore, $G$ has $q^2 + q$ vertices and $q^3 + q^2$ edges, so the order of the number edges is the same as the lower bound we are trying to prove.
Of course since $2q^2 + q$ is not necessarily equal to $n$, I’m thinking we can either choose $q$ such that $2q^2 + q \geq n$, then delete the extra vertices from $G$, or we can choose $q$ such that $2q^2 + q \leq n$, then add artificial vertices to get $n$ vertices (and possibly form a $C_4$-free component on those vertices to increase the number of edges). Here I’m not sure which of the two ways I should go, and what is the order of the alteration.
Say we go the first route. Then for each vertex deleted, we remove at most $q+1$ edges. So the total number of edges removed really just depends on the value $x$ such that $(2q^2 + q)-x = n$.
Having $q$ to be too small and then adding some vertices works. You do not even need to add any edges.
From Bertrand's postulate, we know that there exists a prime $q$ such that $$\tfrac12(n/2)^{1/2}<q<(n/2)^{1/2}.$$ For such a $q$, consider the incidence graph of the affine plane of order $q$, but where all vertical lines are removed (for simplicity). There are now $q^2$ points and $q^2$ lines, for $2q^2$ vertices total. Our restriction on $q$ implies $2q^2< n$, so we can add some dummy vertices to this graph to make it have exactly $n$ vertices.
Each point has $q$ lines through it, and each line has $q$ points on it, so this graph has exactly $q^3$ edges. Therefore, $$ \text{# of edges}=q^3>\left[\tfrac12(n/2)^{1/2}\right]^3=\Omega(n^{3/2}). $$