TITLE: Showing there is a node in the graph with one and only one edge
QUESTION [6 upvotes]: We have an undirected simple graph with $n$ vertices where for every pair of vertices $v_1,v_2$, if $d(v_1)=d(v_2)$ then the set of neighbours of $v_1$ is disjoint from the set of neighbours of $v_2$. Assuming the graph contains at least one edge, prove that there is a vertex of degree exactly $1$ in the graph.
For example the following graph has vertices of degree exactly $1$:

While this problem concerns a graph, I feel like there is a way to apply pigeonhole theory to prove this.  Is this possible?

REPLY [6 votes]: Let $v$ be the vertex of the greatest degree and let $\operatorname{deg}(v)=k>0$.
Let $N(v)$ be neighbors of vertex $v$.
Then the degrees of all vertices from $N(v)$ are pairwise distinct,
and if there are no vertices of degree 1 among them,
then there must be a vertex of degree $k+1$ or more.
This contradicts the choice of vertex $v$.