Among any $5$ vertices there is a vertex connected to $4$ others. What is the minimum number of edges in such a graph with 40 vertices? 
In an undirected graph with $40$ vertices among any $5$ vertices there is a vertex connected to $4$ other vertices. Tell the minimum number of edges a graph satisfying this property can have.

I got stuck with this problem. I tried applying some sort of induction or trying to solve for numbers less than $5$ and then generalize but I cannot really come up with anything.
Could you provide some hints please?
 A: The result depends on what "isolated" means in the complementary question, corresponding to two interpretations of "connected" in the original question.

In an undirected graph with $40$ vertices, among any $5$ vertices there is an isolated vertex. What is the maximum number of edges under this condition?

If vertices are "connected" by individual edges alone, "isolated" means "degree $0$ in the induced subgraph" and the following argument applies. Suppose that five vertices $v_1,\dots,v_5$ in the complementary graph are linked by the edges $v_1v_2,v_2v_3,v_4v_5$. None of the vertices is isolated, so to maximise the edge count in the complement, all the edges have to be non-adjacent, i.e. the complement consists of $20$ disjoint edges, and the original graph must have at least $\frac{40×39}2-20=760$ edges. (That is, the minimal admissible graph is $K_{40}$ minus a perfect matching.)
If vertices are "connected" in the sense of connected components, the complement cannot have any connected component of five or more vertices, and the number of edges is maximised by ten disjoint copies of $K_4$, having $60$ edges in total; the original graph has at least $\frac{40×39}2-60=720$ edges.
