Prove that an Eulerian graph $G$ has even size iff $G$ has an even number of vertices $V$ which $\deg(v) \equiv 2\pmod 4$

Question

Prove that an Eulerian graph $G$ has even size iff $G$ has an even number of vertices $V$ which $\deg(v) \equiv 2 \pmod 4$.

Let $m=2k$ because $G$ hase even size. So by the first theorem of graph theory $\sum \deg(v) =4k$ for $v \in G$. Since $G$ is Eulerian, every vertex in $G$ is even, so it either $\deg (v) \equiv 0\pmod 4$ or $\deg(v) \equiv 2\pmod 4$.

But I can't see why $G$ has to have an even number of vertices $V$ which $\deg(v) \equiv 2\pmod 4$ ?

Answer 1

Given a graph, let $m$ be the number of edges. if $G$ is eulerian all nodes have even degree, so let $2a_i$ be the degree of node $i$.
By the first theorem of graph theory, you have that $m=\sum a_i$.
But $a_i$ is even if and only if the grade of node $i$ is a multiple of 4, whereas it is odd iff the grade is $deg(i)\equiv 2(mod 4)$.
So The parity of $m$ is decided by the parity of the nodes with $deg(i)\equiv 2(mod 4)$.

More precisely, $m$ has the same parity of the nodes with $deg(i)\equiv 2(mod 4)$.

Answer 2

In order to be more explicit, following the answer and notation from @Exodd, every degree of a vertex $v_i$ can be written as $deg(v_i)=2a_i$. If $a_i$ is even then $a_i=2a_i'$ and so $deg(v_i)=2\ (2a_i')=4a_i'$; hence $deg(v_i)\equiv\ 0\ (mod\ 4)$. On the other hand, if $a_i$ is odd, then $a_i=2a_i'+1$ and so $deg(v_i)=2\ (2a_i'+1)=4a_i'+2$; hence $deg(v_i)\equiv 2\ (mod4)$. We have to prove that the number of nodes in which the corresponding $a_i$ is odd, is even.

Proof: Let's suppose that there is an odd number of nodes in which the corresponding $a_i$ is odd. We can split the sum of degrees in the following way: $$ 2m=4k=\sum_{d_i\ even}deg(v_i)+\sum_{d_i\ odd}deg(v_i)$$ Since the sum of even numbers is even, $\sum_{d_i\ even}deg(v_i)$ is even, i.e. $\sum_{d_i\ even}deg(v_i)=2p$ for a $p\in \mathbb{N}$. On the other hand, the sum of an odd amount of odd numbers is odd, therefore $\sum_{d_i\ odd}deg(v_i)$ is odd, i. e. $\sum_{d_i\ odd}deg(v_i)=2q+1$ for a $q\in \mathbb{N}$. We have then $4k=2p+(2q+1)$ which is a contradiction because the left part of the equation is even and the right part is odd.