<s>[INST] 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 $n^2 -3$.
 [/INST] It is obviously that if $n$ is even that $n^2-3$ is odd and so it is not divisible even by $2$. 
Now suppose $n$ is odd. Among 4 consecutive integer exactly one is divisible by 4. So among $$n^2-3,\;\;\;\;n^2-2,\;\;\;\;n^2-1,\;\;\;\;n^2$$ exactly one is divisible by $4$. Since $n^2-1 = (n-1)(n+1)$ we see that $4|n^2-1$. So $4$ doesn't divide $n^2-3$. 
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