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.
 A: 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)).
A: 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.
A: *

*$n=2k$:


$$n^4+4n^2+11\\=(n^2-1)(n^2+5)+16\\=(4k^2-1)(4k^2+5)+16\\=16k'^4+16k''^2+11\\=16k+11$$
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 $16k$.
A: If $m$ is odd, then $m^2 \equiv 1$ (mod 8), since $(2a+1)^{2} = 4(a^{2}+a) +1$ and $a^2 +a$ is always even when $a$ is an integer. If $h$ is an integer congruent to $3$ (mod 8), then $h^{2}-9 = (h-3)(h+3)$ is divisible by $16$. Now when $n$ is odd, we have $n^{2}+2 \equiv 3$ (mod $8$), so $(n^{2}+2)^{2} \equiv 9$ (mod $16$), so $n^{4} + 4n^{2} + 11 \equiv 9 +7 \equiv 0$ (mod 16).
A: 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  $\rm \Rightarrow  2\cdot 8\:\!\:\!|\:\!(n^2\!-1)\:\!(\color{#90f}{n^2\!+\!5}).\,\ \small QED$
A: *

*$n=2k+1$:


$$n^4+4n^2+11\\=(n^2-1)(n^2+5)+16$$
Now, below is square of an odd integer hence it can be reperesented as (8a+1). 
$$n^2$$
Therefore $$8a(8p+6) + 16\\= 64ap + 48a+16\\=16(4ap + 3a + 1)$$
This proves that when n is an odd integer it will be divisible by 16
