Every 2-connected graph has a cycle of at least length 5 Is it true that every 2-connected simple graph of at least 10 vertices has at least one cycle of length 5 or more?  I know that any two vertices lie on a common cycle and I am trying to use this by assuming all cycles are of length 3 and 4 but then constructing one that is length 5.  Any thoughts?
 A: What I would try:
Pick a vertex $v$... it is on some cycle, $C_{v}$--for fun, make $C_{v}$ the longest cycle including $v$.  How long is the cycle $C_{v}$?  If it is at most $4$ then there are 6 vertices (at least) not on $C_{v}$.  Pick one of them and call it $w$.  Can you find a cycle through $w$?  Sure, call it $C_{w}$.  If $C_{v}$ and $C_{w}$ are disjoint, it should be easy to connect them up and make a larger cycle.  If they intersect, use the fact that $w$ is not on $C_{v}$ but build a larger cycle by working $w$ into $C_v$.  
I'm not certain that this works (I haven't carefully went through the details), but something like it should be possible.  If the proof doesn't work (because the statement is false), it should help you build a counter-example.
A: I know this question is 6 years old, but I want to answer it for the sake of anyone who stumbles on this page.
The answer is no. The graph $K_{2,t}$ is $2$-connected and has arbitrarily many vertices, but its only cycle length is $4$.
