A *saturated linear hypergraph* is a [hypergraph](https://en.wikipedia.org/wiki/Hypergraph) $H=(V,E)$ such that 


1. $|e|\geq 2$ for all $e\in E$,
2. $|e\_1\cap e\_2| = 1$ for all $e\_1, e\_2\in E$ with $e\_1\neq e\_2$, and
3. $|\{e\in E:v\in e\}| = 2.$


Let $E$ be the set of $n\in\mathbb{N}$ such that it is impossible to have a saturated linear hypergraph on $\{1,\ldots, n\}$. (For instance, $4\in E$.) Is $E$ infinite?



Cardinalities of saturated linear hypergraphs
This is similar to an answer I gave before.


For each point $v$, let $L\_v$ denote the edges containing $v$. Then we know that each set $L\_v$ has cardinality 2, and they're distinct.


Moreover, if the edges are $e\_1, e\_2, \ldots , e\_m$, then since every pair of edges intersect, we'll need that the sets $L\_v$ are in bijection with the two-element subsets of $m$.


Thus, a configuration as you desire exists iff $n$ is of the form ${m \choose 2}$ (and $m \geq 3$ so that each edge will have at least 2 points).


So yes, the set $E$ is infinite, and it's equal to $\{1\}$ union the set of integers that aren't triangle numbers.


