Prove that $16$ divides $n^{4} + 4 n^{2} + 11$ if n is an odd integer

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Solution

Proving that $16$ divides $n^{4} + 4 n^{2} + 11$ where $n$ is any odd integer
$= n^{4} + 4 n^{2} + 11$
$= (n^{2} - 1) (n^{2} + 5) + 16$
Put $n = 2 k + 1$ where $k$ is any integer
$= (4 k^{2} + 4 k) (4 k^{2} + 4 k + 6) + 16$
$= 8 k (k + 1) (2 k^{2} + 2 k + 3) + 16$
$= 16 (\frac{k (k + 1)}{2} (2 k^{2} + 2 k + 3)) + 16$ $[\frac{k (k + 1}{2} i s a n i n t e g e r]$
$= 16 x + 16$
$9 +$ is divisible by $16$

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