Proving $n^4 + 4 n^2 + 11$ is $16k\,$ for odd $n$ 
if $n$ is an odd integer, prove that $n^4 + 4 n^2 + 11$  is of the form
$16 k$.

And I went something like:
$$\begin{align*}
n^4 +4 n^2 +11
&= n^4 + 4 n^2 + 16 -5 \\
&= ( n^4 +4 n^2 -5) + 16 \\
&= ( n^2 +5 ) ( n^2-1) +16
\end{align*}$$
So, now we have to prove that the product of  $( n^2 +5 )$ and $( n^2-1)$ is a multiple of 16.
But, how can we do this?
If anybody has any idea of how I can improve my solution, please share it here.
Edit updated to include the necessary hypothesis that $n$ is odd.

*

*$n=2k$:


$$n^4+4n^2+11\\=(n^2-1)(n^2+5)+16\\=(4k^2-1)(4k^2+5)+16\\=16k'^4+16k''^2+11\\=16k+11$$
Which is not $16k$.


*

*$n=2k+1$:


$$n^4+4n^2+11\\=(n^2-1)(n^2+5)+16\\=(4k^2+4k)(4k^2+4k+6)+16\\=8\underbrace{k(k+1)}
_{2k}(2k^2+2k+3)+16$$
Which is $16k$.
