TITLE: If given the girth and the minimum degree of a simple graph $G$, can we give a lower bound on the number of vertices it has?
QUESTION [6 upvotes]: I'm trying to prove that every simple graph $G$ of girth $g(G)=5$ (length of smallest cycle), and minimum degree $\delta$, has at least $\delta^2 + 1$ vertices. 
I tried using induction on $\delta$ without any results, and also tried to apply the pigeonhole principle, to no avail.
Help?

REPLY [8 votes]: If you start from any vertex, it is connected to at least $\delta$ vertices.  Each of those is connected to at least $\delta-1$ others.  These are all distinct or you would have a 4 (or less)-cycle.  $1+\delta+\delta(\delta-1)=\delta^2+1$.  The pentagon and Petersen graph follow exactly this, then connect appropriate ones of the second tier of vertices.