<s>[INST] If $x+y$ and $y+z$ are even, prove $x+z$ is even I'm stuck on a practice problem for an upcoming midterm. Can someone tell me if I'm starting off on the right foot here? 
Q: $x,y,z \in \mathbb{Z}$. $x + y$ and $y + z$ are even. Prove that $x + z$ is even.
I thought it would be easier to prove using the contrapositive so I am proving that if $x+z$ is odd, then either $x+y$ or $y+z$ are odd.
By the definition of an odd number, $x+z = 2n + 1, n \in \mathbb{Z}$.
Case 1 : suppose x is even. $x = 2a, a \in\mathbb{Z}$. 
         Then $x+z = 2(a+b) + 1 = (2a) + (2b + 1)$, where $n = a+b, a,b \in\mathbb{Z}$.
         This means $z$ must be odd.
         The same logic holds for $x+y$, meaning that $y$ must also be odd.
         Therefore, $y = 2c + 1$.
         It follows that $y+z = 2b +1 + 2c + 1 = 2d$, where $d = b+c+1$.
Case 2 : suppose $x$ is odd. $x = 2a +1, a \in\mathbb{Z}$. 
         Then $x+z = 2(a+b) + 1 = (2a +1) + (2b)$, where $n = a+b, a,b\in\mathbb{Z}$.
         This means $z$ must be even.
         The same logic holds for $x+y$, meaning that $y$ must also be even.
         Therefore, $y = 2c$.
         It follows that $y+z = 2b + 2c = 2d$, where $d = b+c$.
I just don't think this proves that one of the two expressions is odd. Any help would be much appreciated!
 [/INST] Suppose $x+y$ and $y+z$ are even. Then $x+y=2a$ and $y+z=2b$ for some $a,b\in\Bbb Z$, so that 
$$\begin{align}
x+z&=(x+y)+(y+z)-2y\\
&=2a+2b-2y\\
&=2(a+b-y).
\end{align}$$
But $a+b-y\in \Bbb Z$. Thus $x+z$ is even.
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