Proving $n^4 + 4 n^2 + 11$ is $16k$

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.

• $n=2k$:

$$n^4+4n^2+11\\=(n^2-1)(n^2+5)+16\\=(4k^2-1)(4k^2+5)+16\\=16k'^4+16k''^2+11\\=16k+11$$

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 $16k$.

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.

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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)).