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 $n^2 -3$.
 A: $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
A: Here’s a different way. Consider:
$$ n^2-3=(n-1)(n+1) - 2$$
$$n^2-3=(n-2)(n+2) +1$$
Now if $n$ is odd then $(n-1)(n+1)$ is a multiple of 4 so $(n-1)(n+1)-2$ must not be.
If $n$ is even then $(n-2)(n+2)$ must be a multiple of 4 so $(n-2)(n+2)+1$ cannot be a multiple of 4
A: 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 $n^2-3$. 
A: You might also want to brute force this by doing a table of remainders modulo 4. 
For $ n \equiv 0 (4)$ and $n \equiv 2(4)$, $n^2 \equiv 0(4)$ so $n^2 -3 \equiv 1(4) $
For $n \equiv 1(4)$ and $n \equiv 3(4)$, $n^2 \equiv 1(4)$ so $n^2 -3 \equiv 2(4)$
A: If $n^2-3$ is divided by $4$ then $n$ should be odd.
Let $n=2k-1$.
Thus, $$(2k-1)^2-3$$ is divided by $4$ or
$$4(k^2-k)-2$$ is divided by $4$, which is wrong.
A: If $n$ is even, then $n^2-3$ is odd and therefore $4\nmid n^2-3$.
If $n$ is odd, $n=4k\pm1$ for some $k$, and therefore $n^2-3=16k^2\pm8k-2=4\times(4k^2\pm2k)-2$, which is not a multiple of $4$.
