Proof: How to prove $n$ is odd if $n^2 + 3$ is even New to the whole proof thing.
Trying to figure out that, for all integers $n$, if $n^2 + 3$ is even, then $n$ is odd.
Thank you for the help.

The notation "$m\mid n$" means "$m$ divides $n$," or "$n$ is divisible by $m$." In contrast, $m\not\mid n$ means "$m$ does NOT divide $n$," or put differently, "$n$ is not divisible by $m.$"

Assume $n^2 + 3$ is even.
$(1)$ $n^2 + 3$ is even means, by definition,  that $2$ divides $n^2 + 3$: 
$$2\mid (n^2 + 3) \iff n^2 + 3 = 2k\;\text{ for some}\; k \in \mathbb Z.$$
That is, $n^2 + 3 $ is some integer multiple of of $2$.
Hence $$n^2 = 2k - 3 = 2(k-2) + 1 \implies 2\not\mid 2(k - 2) + 1,\;\implies 2\not\mid n^2$$ since clearly, $\;2 \mid 2(k-2),\;$ but $\;2 \not\mid 1$. In words, $2$ does not divide $n^2$. So $n^2$ is not even, and hence it must be the case that $n^2$ is odd. From $(2)$, it will follow that $n$ is therefore odd.
$(2)$ $n^2$ odd $\iff n$ is odd.


*

*$ \implies$ We prove this by proving the contrapositive:
Assume $n$ is not odd (assume $n$ is even), so $n = 2k$ for some integer $k$. So then $n^2 = (2k)^2 = 4k^2 = 2(2k^2)$, and so $2$ divides $n^2$, and hence $n^2$  is even, ie. if $n$ is not odd, then $n^2$ is not odd, which gives us, contrapositively: $n^2$ is odd $\implies n$ is odd. N.B. (That's really all we need for your proof, but let's go ahead and show that the implication $(2)$ is bi-directional.)

*$\Longleftarrow$ Assume $n$ is odd. Then $n = 2k + 1 $ for some integer $k$. And so $$n^2 = (2k+1)^2 = 4k^2 + 4k + 1 = 2(2k^2 + 2k) + 1$$ Now $2 \not\mid [2(2k^2 + 2k) + 1]$, so  $n^2$ cannot be even: That is, if $n$ is odd, then $n^2$ is odd. 
