If a graph with $n$ vertices and $n$ edges there must a cycle? How to prove this question? If a graph with $n$ vertices and $n$ edges it must contain a cycle?
 A: Assume that $G$ contains no cycles. Then every connected component of $G$ is a tree.
Claim The number of edges in a tree on $n$ vertices is $n-1$.
Proof is by induction. The claim is obvious for $n=1$. Assume that it holds for trees on $n$ vertices. Take a tree on $n+1$ vertices. It's an easy exercise (look at a longest path in $G$) to show that a tree has at least one terminal vertex (i.e. with degree $1$). Removing this terminal vertex along with its edge, we get a tree on $n$ vertices, and induction takes us home.
Hence the number of edges in a graph without cycles is $n-k$, where $k$ is the number of connected components.
A: Here's is an approach which does not use induction:
Let $G$ be a graph with $n$ vertices and $n$ edges.
Keep removing vertices of degree $1$ from $G$ until no such removal is possible, and let $G'$ denote the resulting graph. Note that in each removal, we're removing exactly $1$ vertex and $1$ edge, so $G'$ cannot be empty, otherwise before the last removal we'd have a graph with $1$ vertex and $1$ edge, and $G'$ has the same number of vertices and edges.
Therefore the minimum degree in $G'$ is at least $2$, which implies that $G'$ has a cycle.
A: Here's another approach:
Let $P = a_1, a_2, \dots, a_n$ be a simple path of maximal length in G. 
Since the degree of each vertex is greater than two, there exists a vertex $b$ such that ${b,a_1}$ is an edge. 
Now, there can be two cases:


*

*$b$ is on the path P: In this case, the sub-path from $a_1$ to $b$ and the edge from $b$ to $a_1$ forms a cycle.

*$b$ is not on the path P: In this case, the path $b,a_1,a_2, \dots, a_n$ is a path of greater length than $P$, thus contradicting our assumption.
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