<s>[INST] 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.
 [/INST] Hint $\rm\,\ n\,$ odd $\rm\:\:\! \Rightarrow\ 2\:|\:\color{#90f}{n^2 + 5}$
and, furthermore $\ \:\!\rm 8\:|\:\color{#0a0}{n^2}-\color{#c00}1,\ $ by $\ \rm mod\ 8\!:\ \color{#0a0}{odd^2} \equiv \{\pm 1,\pm 3\}^2\equiv \color{#c00}1$
multiplying  $\rm \Rightarrow  2\cdot 8\:\!\:\!|\:\!(n^2\!-1)\:\!(\color{#90f}{n^2\!+\!5}).\,\ \small QED$
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