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As the title suggests: how many edges (as a function of $n$) can a bipartite graph with $n$ nodes on each side have if it does not have $K_{4, 4}$ as a subgraph? Is this known?

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    I guess this is incorrect question. It could be possible to get lower bound , but about upper bound it's neccesary to have more information about number of vertices in both part. – 2017-01-13
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    For example : consider graph which contain $K_{4,4}$, then add some vertices in both sizes and delete edges , which contains in $K_{4,4}$. We could add infty many vertices , and also infty many edges to them to get this graph be bipartite and connected. – 2017-01-13
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    Thanks, you're right. I specifically meant balanced bipartite graphs, i.e. $K_{n, n}$, and then would be looking for an answer as a function of $n$. I edited the question to reflect this. – 2017-01-13

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Quite surprisingly (at least to me) the general case seems to be an open problem, the Zarankiewicz problem. However, some estimates are due to Kővári–Sós–Turán, and maybe the specific $K_{4,4}$ case is entirely solved. I am afraid I am not able to be more helpful than this (i.e. very little).

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    Thanks! For posterity: this link gives an upper bound of the form $O(n^{7/4})$ edges. It's open whether or not this is tight, but it is conjectured to be tight (at least up to the hidden constant factor). – 2017-01-13
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    Unconditionally, the best lower bound seems to be of the form $\Omega(n^{5/3})$ (this graph even lacks $K_{3, 3}$). – 2017-01-13
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    @GMB: the upper bound $n^{7/4}$ might be sharp from a topological argument, like the one Lovasz used to prove Kneser's conjecture. We may see every vertex of the original bipartite graph as a point in the upper or lower part of a hypersphere. We want that every time we choose 4 points in the upper hemisphere and 4 points in the lower hemisphere, at least a segment joining them is not an arc in the original graph. For such a purpose, it is enough to join every point on the hypersphere with every point in the opposite hemisphere that is far enough from the antipodal point. – 2017-01-13
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    By the Lusternik-Schnirelmann theorem, if four closed set cover $S^3$, at least one of them contains a couple of antipodal points. So the above construction *should be* optimal and the bound $n^{7/4}$ *should be* sharp. – 2017-01-13