TITLE: If $n$ is an integer then $4$ does not divide $n^2 - 3$
QUESTION [9 upvotes]: Having some trouble proving this. Was going to attack with using the contrapositive of this statement but can't seem to show that $n$ isn't an integer.

REPLY [3 votes]: I think the easiest way is by simple algebraic manipulation. Given some integer $k$, if $n = 2k$ (meaning that $n$ is even), then $n^2 - 3 = 4k^2 - 3$; if $n = 2k + 1$ (meaning $n$ is odd), then $n^2 - 3 = 4k^2 + 4k - 2$. Notice that $4k^2$ and $4k^2 + 4k$ are both multiples of 4, while $4k^2 - 3$ and $4k^2 + 4k - 2$ are not.