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.

If $2|k=>16|n^4$  and $4|n^2=>16|(n^4+4n^2)=>n^4+4n^2+11≡11\pmod{16}$
Else 
$n$ is odd$=2k+1$(say), $n^2=(2k+1)^2=8\cdot\frac{k(k+1)}{2}+1≡1\pmod{8}=>8|(n^2-1)$  
(i)So, $n^4+4n^2+11=(n^2-1)^2+6(n^2-1)+16≡0\pmod{16}$ if  $n$ is odd.
(ii)When  $n$ is odd, $2|(n^2+1)$ and $8|(n^2-1)$(already proved) $=>2\cdot8|(n^2-1)\cdot(n^2+1)=>16|(n^4-1) $
(iii)When  $n$ is odd, $n^2≡1\pmod{8}=1+8m$(say),
So, $n^4=(n^2)^2=(1+8m)^2=1+16m+64m^2≡1\pmod{16}$
So  using (ii) or (iii), $n^4≡1\pmod{16}$  and $n^2≡1\pmod{8}=>4n^2≡4\pmod{32}$ if  $n$ is odd,
So, $n^4+4n^2+11≡1+4+11\pmod{16}≡0\pmod{16}$ if  $n$ is odd.

Alternatively, using Carmichael Function, $\lambda(16)=\frac{\phi(16)}{2}=4$  and $\lambda(8)=\frac{\phi(8)}{2}=2$ 
So, $n^4≡1\pmod{16}$  and $n^2≡1\pmod{8}=>4n^2≡4\pmod{32}$ if $(16,n)=1$ i.e.,  $n$ is odd,
So, $n^4+4n^2+11≡0\pmod{16}$ if  $n$ is odd(like (ii)).
