As the title says, I am trying to show that if a graph $G$ is such that $v(G)\geq4$ and the number of edges it has is 3 less than twice the number of vertices, i.e. that $$ e(G) = 2v(G) - 3, $$ then the graph has at least 2 cycles of equal length.
Clearly in order to be a cycle we need at least 3 vertices, so the possible cycle lengths are $$ 3,4,5, ... , v(G). $$ Now the minimum spanning tree $T$ is a tree and therefore its number of edges is $e(T) = v(G) - 1$, and so by adding edges to get to $G$ we find that there are $v(G)-2$ edges to add. And since this was a tree each new edge will create a cycle.
So we can assume that each additional edge creates a cycle with a distinct number of vertices, using all $v(G)-2$ possibilities.
Now I obviously should find a contradiction somewhere, but I'm not sure exactly how. Any help?