Prove that if a graph of order $3n (n ≥ 1)$ has $n$ vertices of each of the degrees $n − 1$, $n$ and $n + 1$, then n is even.

Question

I'm currently working in the following graph theory excercise:

Prove that if a graph of order $3n (n ≥ 1)$ has $n$ vertices of each of the degrees $n − 1$, $n$ and $n + 1$, then n is even.

So by lemma ($m$ order of the graph):

$$\sum_{v \in V(G)} deg(v) = 2m$$

I'm doing:

$$(n-1)+n+(n+1) = 2(3n)$$

Reaching

$$3n = 2(3n)$$

I feel I'm missing something cause the conclusion is not so clear for me, any hint or help will be really appreciated.

Answer 1

No, you have $$n\cdot (n-1)+n\cdot n+n\cdot (n+1) = 2m$$

so $$3n^2 =2m \implies 2\mid 3n^2 \implies 2\mid n^2 \implies 2\mid n$$