Is there exist an Euler graph on 2014 vertices with 3007 edges?

Question

I have to prove or disprove the statement "there exist an Euler graph on 2014 vertices with 3007 edges"

here is my argument

let $G=(V,E)$ be a graph with $|V|=2014, |E|=3007$

let $v_i\in V (i=1,2\dots2014)$ and suppose $deg(v_i)=2m_i$ wehere $m_i\in \Bbb N$

by handshaking lemma

$\sum\limits_{i=1}^{2014}deg(v_i)=2\cdot3007$

$\sum\limits_{i=1}^{2014}2m_i=2\cdot3007$

$\sum\limits_{i=1}^{2014}m_i=3007$

now by taking some sequence of $m_i$ which has the property $\sum\limits_{i=1}^{2014}m_i=3007$ gives me a Euler graph.since degree of every vertex is even.

am I correct or wrong? if i am correct is there any better way to prove this?if i am wrong how should i prove?any hint?

Answer 1

Consider the "union" of the two cyclic graphs $$0 - 1 - 2 - 3 - \cdots - 2013 - 0 \quad\text{ and }\quad 0 - 2 - 4 - 6 - \cdots - 1984 - 0$$ It has $2014$ vertices and $(2013+1)+(\frac{1984}{2}+1) = 3007$ edges.

Start at vertex $0$, first walks around the first graph, next walks around second graph and finally back to $0$ give us an Euler circuit.

Answer 2

You're wrong or incomplete.

You have to show that such a sequence $\{m_i\}$ exists, i.e. the sequence $\{2 m_i\}$ is graphical. Not all sequences will give valid graphs.