Bipartite graph $G=(A,B)$ with $\delta(A)=3n/2$ and no $C_4$ has a matching which saturate each vertex in $A$. 
Say $G$ is a bipartite graph with bipartition $(A,B)$ and $G$ is $C_4$-free. Prove that if every vertex in $A$ has degree at least $\frac32 n$ and $|A|\leq n^2$, then $G$ has a matching which uses every vertex in $A$.


My proof:
Use the Hall marriage theorem. Take any $X\subseteq A$ and let $|X|=k$. Also let $Y=N(X)$ and $|Y|=l$. Let us prove that $l\geq k$. Assume that $l<k$.
Let $Y^*$ be a set of all unordered pairs $\{y_i,y_j\}$, $i\ne j$ of elements in $Y$, and connect a pair $\{y_i,y_j\}$ with $x\in X$ iff both $y_i$ and $y_j$ are adjacent with $x$ in $G$. Then the degree of each pair in this new bipartite graph $G^*$ (on vertex set $X \cup Y^*$) is at most $1$ (since there is no $C_4$ in $G$) and the degree of each $x\in X$ is at least $\displaystyle{{3n\over 2}\choose 2}$. So we have $$k\cdot {{3n\over 2}\choose 2}\leq {l\choose 2} .$$
Since we assume $l<k$ we have $${{3n\over 2}\choose 2}< {k-1\over 2}$$ so $${3n(3n-2)\over 4} < k-1 .$$ Since $k\leq n^2$ we have $$3n(3n-2) \leq  4n^2-8$$so
$$5n^2\leq 6n-8$$ which is obviously not true.
Edit: After Darij's confirmation that the proof is correct, I will award any solution with better bound than $\delta (G)=n\sqrt{2}$ (instead of $3n/2$).
 A: It has been shown for $C_4$-free subgraphs of $K_{n,n}$ that:

*

*Any such subgraph contains at most $n^{3/2} + 2n$ edges. (Kövári, Sós, Turán)

*For infinitely many $n$, there is such a subgraph with at least $n^{3/2} - n$ edges. (Füredi)

The first of these solves this problem, when $G$ has $|A| \le n^2$ and $\delta(G) > n+2$. Let $X \subseteq A$ with $|N(X)| < |X| = k$. We can add isolated vertices to $N(X)$ to get a $C_4$-free bipartite graph with $k$ vertices on each side. We know it has at most $k^{3/2} + 2k$ edges; on the other hand, it has more than $(n+2)k$ edges. Therefore $kn + 2k < k^{3/2} + 2k$, or $n^2 < k$, contradicting $|A| \le n^2$.
The second one of these happens to show that we cannot improve $\delta(G)$ significantly. Assuming $n$ is a prime power, take a finite field $F$ of order $n$ and let $A = B = F^2 - \{(0,0)\}$; put an edge between $(a_1, a_2) \in A$ and $(b_1, b_2) \in B$ if $a_1 b_1 + a_2 b_2 = 1$.  The system $$ \begin{cases}a_1 x + a_2 y = 1 \\ a_1'x + a_2'y = 1\end{cases}$$ has at most one solution, and so this graph is $C_4$-free; every vertex in $A$ has degree $n$, since $(a_1, a_2) \in A$ has neighbors $(t, \frac{1 - a_1 t}{a_2})$ for all $t \in F$.
Now delete one vertex in $B$, so that there is no matching that saturates $A$. In this graph $G'$, we still have $|A| \le n^2$, and every vertex in $A$ has degree at least $n-1$.
