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.