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.

Proof:
Since the statement is biconditional we must prove the following two statements:
$(a)$ If $2|(n^4 − 3)$ then $4|(n^2 + 3)$
$(b)$ If $4|(n^2 + 3)$ then $2|(n^4 − 3)$
We will begin with statement $(b) ($since it should be the easiest to prove$)$
We will prove this statement directly. Assume that $4|(n^2 + 3)$. This implies that there is an integer $x$ such that
$n^2 + 3 = 4x$. Rearranging we get $n^2 = 4x − 3$. Now if we evaluate $n^4 − 3$ we get
$$n^4=(n^2)^2-3$$$$=(4x-3)^2-3$$$$=(16x^2-24x+9)-3$$$$2(8x^2-12x+3)$$
Notice that $8x^2 − 12x + 3$ is an integer. This implies that $2|(n^4 − 3)$
To prove statement (a) we could approach this directly or by contrapositive. Both directions could prove informative
so both will be presented here.
Direct: Assume that $2|(n^4 − 3)$. This means that there is some integer $y$ such that $n^4 − 3 = 2y$. We wish to prove something about $n^2$ so we will have to (somehow) reduce the power on $n$. To do this, notice that $n^4 = 2y+3 = 2(y+1)+1$.
Since $y + 1$ is an integer, we see that $n^4$ is odd.
If $n^4$ is odd then $n^2$ is odd. We see that $n$ must be odd. Therefore there is an integer $k$ such that $n = 2k + 1$.
Now we can write $$n^2+3=(2k+1)^2+3$$$$=4k^2+4k+1+3$$$$=4(k^2+k+1)$$
Since $k^2+k+1$ is an integer, we see that $4|(n^2+3)$
