If $a$ and $b$ are integers then $a^2 -4b-2$ does not equal zero. I was just wondering if the rough outline of this proof was logically rigorous. 
If a and b are integers then $a^2 -4b-2$ does not equal zero. 
Proof: 
Suppose for the sake of contradiction that $a^2-4b-2=0$.
Then we have $a^2 = 2(2b+1)$. $a^2$ is even which implies that $a$ is even. 
If $a$ is even then $a = 2k$ for some integer $k$. 
$(2k)^2 = 2(2b+1)$ 
$2(2k^2) = 2(2b+1)$
$2k^2 = 2b+1$
Let $w = 2k^2$ where since$ k^2 $is an integer, $w$ is even. 
Let $v = 2b+1$ where since $b$ is an integer, $v$ is odd.
Then we have, $w = v$, which is a contradiction. 
QED
Thanks for the feedback!

I think the best way to think about this is to work modulo 4:
$0^2 \equiv 0$ mod 4
$1^2 \equiv 1$ mod 4
$2^2 \equiv 0$ mod 4
$3^2 \equiv 1$ mod 4
Clearly $4b + 2 \equiv 2$ mod 4. So indeed it is not possible that $a^2$ = 4b + 2. Similarly you can show $a^2$ - 4b - 3 is not equal to zero either.
