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.
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.
Thanks for the feedback!