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.

The claim is false, for example
$$n=2\Longrightarrow n^4+4n^2+11=16+16+11=43$$
which is not a multiple of 16. Check your expression.
Now, if $\,n=2k+1\,$ is odd, then the claim is true, since then
$$n^4+4n^2+11=8k(k+1)(2k^2+2k+3)+16$$
and since $\,8k(k+1)=0\pmod {16}\,$ no matter what parity $\,k\,$ has, we're done.
