Let $A_{n,m}$ be the maximum number of edges that a bipartite graph with $n,m$ vertices can have when it doesn't contain $4$-cycle. I have calculated some values:
$A_{2,2}=3$, $A_{3,3}=6$, $A_{4,4}=9$, $A_{4,5}=10$, $A_{5,5}=12$.
I am trying to find formula for $A_{n,m}$. Does anyone know it or a hint how to find?
Especially I need the value of $A_{6,6}$. I only know that $A_{6,6}\ge16$.
I will give simple proofs of the evaluations $A_{4,6}=12,\ A_{5,6}=14,\ A_{6,6}=16,\ A_{6,7}=18,\ $ and $A_{7,7}=21$.
1. Since the ratio $\dfrac{A_{m,n}}{mn}$ is a nonincreasing function of $m$ and $n$, we have the recursive upper bound $$A_{m,n+1}\le\lfloor\frac{n+1}nA_{m,n}\rfloor.$$
2. It's easy to see that $A_{1,n}=n$ and $A_{2,n}=n+1$ for all $n$.
3. $$A_{3,4}\le\lfloor\frac32A_{2,4}\rfloor=\lfloor\frac{15}2\rfloor=7$$ $$A_{3,5}\le\lfloor\frac54A_{3,4}\rfloor\le\frac{35}4\rfloor=8$$ $$A_{4,5}\le\lfloor\frac43A_{3,5}\rfloor\le\lfloor\frac{32}3\rfloor=10$$ $$A_{4,6}\le\lfloor\frac65A_{4,5}\rfloor\le\lfloor\frac{60}5\rfloor=12$$ $$A_{{5,5}}\le\lfloor\frac54A_{4,5}\rfloor\le\lfloor\frac{50}4\rfloor=12$$ $$A_{5,6}\le\lfloor\frac65A_{5,5}\rfloor\le\lfloor\frac{72}5\rfloor=14$$ $$A_{6,6}\le\lfloor\frac65A_{5,6}\rfloor\le\lfloor\frac{84}5\rfloor=16$$ $$A_{6,7}\le\lfloor\frac76A_{6,6}\rfloor\le\lfloor\frac{112}6\rfloor=18$$ $$A_{7,7}\le\lfloor\frac76A_{6,7}\rfloor\le\lfloor\frac{126}6\rfloor=21$$
4. Let $G$ be the bipartite graph with vertex sets $U,V$ where $|U|=6,|V|=7$, and:
The graph $G$ witnesses the inequality $A_{6,7}\ge18$; moreover, by deleting one, two, or three of the degree $2$ vertices of $G$, we get witnesses to $A_{6,6}\ge16,\ A_{5,6}=A_{6,5}\ge14$, and $A_{4,6}=A_{6,4}\ge12$. Combined with the previously obtained upper bounds, this verifies the exact evaluations $A_{4,6}=12,\ A_{5,6}=14,\ A_{6,6}=16,$ and $A_{6,7}=18$.
5. The point-line incidence graph of a finite geometry is a $C_4$-free bipartite graph. In this way we get the lower bounds $$A_{n^2,\ n^2+n}\ge n^3+n^2$$and $$A_{n^2+n+1,\ n^2+n+1}\ge(n^2+n+1)(n+1)$$ whenever a projective plane of order $n$ exists, e.g., whenever $n$ is a prime power.
Setting $n=2$ (the Fano plane) we get $A_{4,6}\ge12$ (again) and $A_{7,7}\ge21$ which, combined with the previously obtained upper bound, establishes that $A_{7,7}=21$.
Update. Thanks to @David for pointing out that "the graph G in 4 is in fact the incidence graph of the Fano plane with one line (or point) removed."
A slight simplification of bof's answer: sections 4 and 5 could be combined, since the graph $G$ in 4 is in fact the incidence graph of the Fano plane with one line (or point) removed.
