# Maximum number of edges in “square” hypergraph

For any set $$X$$ and any cardinal $$\kappa$$, let $$[X]^\kappa$$ denote the subsets of $$X$$ having cardinality $$\kappa$$.

A linear hypergraph is a hypergraph such that for all $$e\neq e_1 \in E$$ we have $$|e\cap e_1|\leq 1$$. For any positive integer $$n$$ let $$[n] = \{1,\ldots,n\}$$. We say that $$([n^2], E)$$ is a square hypergraph if it is linear and every element of $$E$$ has $$n$$ elements.

For any integer $$n>1$$ let $$m(n)$$ denote the maximum cardinality of $$E$$ where $$([n^2], E)$$ is a square hypergraph. For instance we have $$m(2) = 6$$. What is the value of $$\lim (\sup)_{n\geq 2}\frac{m(n)}{n^2}\;?$$ (Note: I added $$(\sup)$$ in case the limit doesn't exist.)

In a linear hypergraph, any pair of vertices is contained in at most one hyperedge. Since any hyperedge contains $$n$$ vertices, it contains $${n \choose 2}$$ pairs. Double counting gives $$m(n) \leq \frac{n^2 \choose 2}{n \choose 2} = n (n+1).$$
This bound is sharp when $$n$$ is a prime power (take an affine plane of order $$n$$), so $$\limsup \frac {m(n)}{n^2} = 1.$$