For $n\ge4$, let G be a simple n-vertex graph with at least $2n - 3$ edges. Prove that G has two cycles of equal length. (West's Introduction to Graph Theory Q 2.1.42)
I am trying to prove the above claim. So far I have reasoned as follows:
G must be connected because $2n-3\ge n-1$ for $n\ge4$, and any simple n-vertex graph with at least $n-1$ edges must be connected.
Let T be a spanning tree of G. T must have $n-1$ edges. Therefore T has at least $2n-3 - (n-1) = n-2$ fewer edges than G. That is, $|e(G) \setminus e(T)| \ge n-2$.
Each one of these 'extra' edges adds exactly cycle to G, so G has at least $n-1$ cycles, each with length 3 to n.
Because there are at least n-2 cycles in G, it suffices to show that $n-2$ of these edges cannot all have distinct sizes. (I'm unsure about this line... is this a correct statement?)
From here I've lost my way, but I think I am on the right track. For what it's worth, this question appears right after the section that introduces trees, spanning trees, and their properties etc.