number theory - Elementary proof that $4$ never divides $n^2 - 3$

I would like to see a proof that for all integers $n$, $4$ never divides $n^2 - 3$. I have searched around and found some things about quadratic reciprocity, but I don't know anything about that. I am wondering if there is a more elementary proof.

For example, I managed to show that $4$ never divides $x^2 - 2$ by saying that if $4$ does divide $x^2 - 2$, then $x^2 - 2$ is even. And then $x^2$ is even, which means that $x$ is even. So $x = 2m$ for some integer $m$, and so $x^2 - 2 = 4m^2 - 2$ is not divisible by $4$. So I would like to see a similar proof that $4$ doesn't divide $n^2 -3$.

Answer

$n $ is odd $\implies n=2k+1\implies n^2-3=2(2k^2+2k-1)$ where $2k^2+2k-1$ is odd and hence can't have $2$ as a factor.

In order for $4$ to divide $n^2-3$ it should have $4=2.2$ as a factor but note that $2$ appears as a factor only once if $n$ is odd.

$n$ is even $\implies n^2-3=4k^2-3$ which is odd