First year college question on divisibility of integers I'm having a hard time with a practice question.

Given $n$ is an integer, prove $2$ divides $(n^4 -3)$ iff $4$ divides $(n^2 +3)$.

So I know since it's an iff statement, I have to show the implication going both ways. Let's start with the left side first.
There exists an integer $r$ such that $2 r  =  n ^ 4  - 3 $. 
Here, I'm thinking in my head how can I get the equation to look like the conclusion, that is $4a = (n^2 + 3)$ for an integer $a$. I see that we can play with the $3$ on both sides.
$$3 = n^4 - 2r$$
$$3 + n^2 = n^4 + n^2 - 2r$$
So here's where I'm stuck. How can I show I can factor out a $4$ out if this right side? Thanks.

Trick:  If $2|n^4 -3$ then $n$ is odd.  So let $n = 2r + 1$ then $n^2 +3 = (2r+1)^2 + 3 = 4r^2 + 4r + 1 + 3 = 4r^2 + 4r + 4 = 4(r^2 + r + 1)$ and $4|n^2 + 3$.
Trick:  If $4|n^2 + 3$ then $n$ is odd.  So if $n$ is odd then $n^4$ is odd and $n^4 -3$ is even.  So $2|n^4 -3$.
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In general to prove $4|n^2 +3$ we want to assume $n \equiv k \pmod 4$ and prove that $k^2 + 3\equiv 0\pmod 4$.
If $2|n^4 - 3$ then $k^4 \equiv 3\equiv 1 \pmod 2$ so $k \equiv 1 \pmod 2$ and $k \equiv 1,3\pmod 4$. And $k^2 + 3 \equiv 1,9 + 3 \equiv 4,12 \equiv 0 \mod 4$.  So that is $\implies$.
To prove $2|n^4 -3$ we want to assume $n \equiv k \pmod 2$ and prove that $k^4 -3 \equiv 0 \pmod 2$.
If $4|n^2 +3$ then $n^2\equiv -3 \equiv 1 \pmod 4$ so $n\equiv \pm 1 \pmod 4$ so $n\equiv 1\pmod 2$.  So $n^4 -3 \equiv 1^4 -3 \equiv 0 \mod 2$.  So that is $\Leftarrow$.
