Would someone be willing to verify this proof? (NB: There are other posts on this same problem, but my proof differs.)
If $a, b \in \mathbb{Z}$, then $a^2 - 4b - 3 \neq 0$
Suppose, for the sake of contradiction, that $a^2 - 4b - 3 = 0$. Then, $a^2 - 4b = 3$.
Case 1: Suppose $a$ is odd. Then, $a = 2n-1; n \in \mathbb{Z} \rightarrow a^2 = 4n^2 - 4n + 1$.
Then, $a^2 - 4b = 3 \rightarrow 4(n^2 - n - b) = 2 \rightarrow 2(n^2 - n - b) = 1$. This is a contradiction, as 1 is not even.
Case 2: Suppose $a$ is even. Then, $a = 2n; n \in \mathbb{Z} \rightarrow a^2 = 4n^2$.
Then, $a^2 - 4b = 3 \rightarrow 4(n^2 - b) = 3$, which is a contradiction, as 3 is not even.
It's perfect (well, maybe end with $2(2n^2 - 2b) = 3$ instead of $4(n^2 - b) = 3$, but that's just me nitpicking).
For language, maybe it is better to use $\implies$ instead of $\to$, as it may look like a converging sequence. (Another nitpick: write "$a = 2n$ for some $n \in \mathbb{Z}$" instead of "$a = 2n; n \in \mathbb{Z}$").