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.
 A: 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 $4|(n^2+3)$
A: If we divide $n^2$ by $4$ we can have either $0$ or, $1$ as remainder. Now, note that, if we can show that $$4|2(n^4-3)-(n^2+3)\text{ or }4|2(n^4-3)+(n^2+3)$$we are done. For odd $n$, $4|(2n^4-n^2-9)=2n^4-n^2-1$ and for even $n$ we have $4|(2n^4+n^2-3) $. Hence, $4$ divides $2(n^4+3)\big($which means $2$ divides $(n^4+3)\big)$, iff $4$ divides $(n^2+3)$. 
A: 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 $2|n^4 -3$.
.....
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 $\Leftarrow$.
