My proof that $4$ either divides $n^2$ or $n^2 − 1$ I am attempting to prove the following statement

Prove that if n is any integer then 4 either divides n^2 or n^2 − 1

I have started with the case of n = 2k
Consider the case n = 2k

  n = 2k
  n^2 = 4k^2
⇒ n = 4k

∴ 4 divides n^2 as there is some integer k in which n = k 

Would this be considered as a correct proof for the first case here? Is there any additions I should make?
I also attempted case 2, where n = 2k+1, however I am less sure of the direction I have taken this and is incomplete, so some advice on this would also be appreciated.
Consider the case n = 2k+1

n       = 2k+1
n^2     = (2k+1)^2
n^2     = 4k^2 + 1
n^2 - 1 = 4k^2
Consider the case $n=2k$, then $$n^2=4k^2$$
$$\therefore \space 4|n^2 \space \text{that is if $n$ is even then $n^2$ is divisible by $4$}$$
For the case $n=2k+1$ we have $$n^2-1=(2k+1)^2-1=(2k)^2+2k+2k+1-1=4k^2+4k=4(k^2+k)$$
$$\therefore \space 4|(n^2-1) \space \text{that is if $n$ is odd then $n^2-1$ is divisible by $4$}$$
Since all integers are of the form $2k$ or $2k+1$ for some $k\in \mathbb Z$, we are done.
Here's another simple proof using parity. $[$Actually the same thing what @Ares did, but not taking the form of $2k$ or $(2k + 1)$$]$
Suppose $n$ is even, then $2|n$, which implies $4|n^2$, so $n$ has to be odd.
Now if $n$ is odd, then  note that $(n^2 - 1) = (n + 1)(n - 1)$ . Here in this case both $(n + 1)$ and $(n - 1)$ will be even as $n$ is odd.
So $2|(n+1)$ , $2|(n-1)$ which implies that :-  $$(2*2)|(n+1)(n-1)$$
$$\rightarrow 4|(n^2 - 1)$$
In other words, $4|n^2$ if $n$ is even, else $4|(n^2 - 1)$ if $n$ is odd.