Prove the relation is an equivalence relation. Problem

Define the relation $R$ on the set of natural numbers as $(a,b) \in R
> \iff 2 \vert(a^2 + b) $. Prove that $R$ is an equivalence relation.


This is what I have so far.
Claim:
Define the relation $R$ on the set of natural numbers as $(a,b) \in R
> \iff 2 \mid(a^2 + b) $. The relation $R$ is an equivalence relation. 
Proof:
Part 1 (Reflixivity): 
Let $R = \{(a,b) \in \Bbb{N} \times \Bbb{N} \mid 2 \mid (a^2+b)\}$ be given and suppose that $b \in \Bbb{N}$. 
Then, for some integer $k$:
$$\require{enclose}
\enclose{downdiagonalstrike}{\begin{align}
2 \mid b^2 +b \iff 2k &= b^2 +b \\
 & = b^2 + b - (b^2 + b) + (b^2 + b) \\ 
 & = 2b^2 + 2b - (b^2 + b) \\ 
 & = -2b^2 - 2b + (b^2 + b)
\end{align}}$$
Therefore, 
$$\enclose{downdiagonalstrike}{\begin{align}
2 \mid b^2 +b \iff b^2 + b &= 2k -2b^2 - 2b \\
 & = 2(k - b^2 - b)\\  
 & 
\end{align}}$$
$\enclose{horizontalstrike}{\text{Thus, $2 \mid b^2 +b$ for some integer $(k - b^2 - b)$. Which implies that $R$ is Reflexive.}} $

EDIT: Thanks to some positive feed back I have been let known that this is not showing Reflexivity.  

Part 2 (Symmetry)
Let $R = \{(a,b) \in \Bbb{N} \times \Bbb{N} \mid 2 \mid (a^2+b)\}$ be given and suppose that, for any $a,b \in \Bbb{N}$, $ a\mathbf{R}b \leftrightarrow b\mathbf{R}a.$ 
Then, for some integer $k$: 
$$\enclose{updiagonalstrike}{\begin{align}
2 \mid a^2 + b  & \iff 2k = a^2 + b \\
 & \iff 2k + (a + b^2) = (a^2 + b) + (a + b^2)\\ 
 & \iff (a + b^2) = (a + b) + (a^2 + b^2) - 2k \\ 
 & 
\end{align}}$$
$\enclose{horizontalstrike}{\text{Since, the relation $R$ is proven to be Reflexive, let the integer $ m = a = b $ and let the integer $ n = a^2 = b^2 $. Then, }}$

EDIT: This is not a valid way to show Symmetry, since Part 1 (Reflexivity) has not been proven. 

$$\enclose{updiagonalstrike}{\begin{align}
a + b^2 = (a + b) + (a^2 + b^2) - 2k & \iff (a + b^2) = 2m + 2n - 2k \\
 & \iff (a + b^2) = 2(m + n -k)\\  
 & 
\end{align}}$$
$\enclose{horizontalstrike}{\text{Thus, $2 \mid b^2 + a$ which implies that $R$ is Symmetric since $ a\mathbf{R}b \leftrightarrow b\mathbf{R}a $.}}$
Part 3 (Transitivity)
Let $R = \{(a,b) \in \Bbb{N} \times \Bbb{N} \mid 2 \mid (a^2+b)\}$ be given and suppose that, for any $a,b,c \in \Bbb{N}$, $ a\mathbf{R}b \text{ and } b\mathbf{R}c.$ Then, let $k$ and $h$ be some integers: 
\begin{align}
2k = a^2 + b \text{ and } 2h = b^2 + c & \implies 2(k + h) = (a^2 + b) + (b^2 + c) \\
 & \\ 
 & \\ 
 & 
\end{align}

Comment: And, this is where I get stuck. I am struggling to find a way to show that $ (2 \mid a^2 + b) \land (2 \mid b^2 + c) \implies 2 \mid a^2 + c$. I've also tried eliminating $b$ like so, $2h = (2k - a^2)^2 + c$, but this doesn't seem to get me any where. I feel like I'm running in circles here. 

My Question
Can you argue that $R$ is transitive since it has already been shown that $2 \mid b^2 + b$? 
This would imply something like "$2(k - b^2 - b) + 2h = (a^2 + c) +(b^2 + b)$ is logically equivalent to $(2 \mid b^2 + b) \land (2 \mid a^2 + c)$." And, this simplifies to just $2 \mid a^2 + c$ by the inference rule of simplification [$(p \land q) \to p$]. Which ultimatily I believe gets me to my goal, but I'm not sure if it is two far of a leep to go from $2(k - b^2 - b) + 2h = (a^2 + c) +(b^2 + b) \implies (2 \mid b^2 + b) \land (2 \mid a^2 + c)$. 
I hope my question was specific enough. Otherwise, I would much appreciate some guidance on showing how this relation is transitive if anyone is feeling generous. Thanks! 

Your approach is too technical. 
Note that $$2| (a^2 + b)$$ if and only if both $a$ and $b$ are odd or both are even. 
Reflexivity: $a$ and $a^2$ are either both even or both odd.  
Symmetry: If both $a$ and $b$ are odd then both $b$ and $a$ are also odd. Similarly for the even case. 
Transitivity: If $a$ and $b$ are related and $b$ and $c$ are related then all three of them are odd or all three are even .
In any case a and c are both odd or both even.
