A simple graph $G$ with $n$ vertices has at least $n+4$ edges. Prove that it has two cycles having no common edge.
- It is enought to prove the statement if graph has exactly $n+4$ edges.
- It is clearly if some vertex has degree $1$ then we can remove that vertex and we can proccede with induction. So we can assume that all vertices have degree at least $2$.
- Also we can assume that graph is connected, else it has at least two component and since all vertex has degree at least $2$ we have a cycle in each and thus a conclusion.
- By Handshake lemma we can see that at least $n-8$ vertices (if $n\geq 8$) has degee exactly $2$, so we have at most $8$ vertices with degree at least $3$.
Also, we see that $n\geq 5$ since ${n\choose 2} \geq n+4$.
That is all I can find.
