<s>[INST] Prove that $n^4-n^2$ is divisible by $8$ if $n$ is an odd positive integer. 
Prove that $n^4-n^2$ is divisible by $8$ if $n$ is an odd positive integer.

I'm supposed to use proof by induction, but I failed at it miserably. So far I have this:
$$(n^4) - (n^2) = (n^2)((n^2)-1) = n(n-1)n(n+1)$$
Let $n = 2k + 1$ because its always odd. Then
$$\begin{align}
&(2k+1)(2k)(2k+1)(2k+2)
\\\implies\quad &(2k+1)(2k)(2k+1)2(k+1)
\\\implies\quad &(2k+1)2(k)(2k+1)2(k+1)
\\\implies\quad &4(2k+1)(k)(2k+1)(k+1)
\\\implies\quad &4((2k+1)^2)(k)(k+1)
\\\implies\quad &4((2k+1)^2)((k^2)+k)
\end{align}$$
Now $(2k+1)^2$ is odd and $k^2 + k$ is always even because
$$\begin{cases}
\mathrm{odd}^2 + \mathrm{odd} = \mathrm{even} \\
\mathrm{even}^2 + \mathrm{even} = \mathrm{even}
\end{cases}$$
So that means I can take out a $2$ and write $((k^2)+k) = 2X$.
$$\begin{align}
&4(2X)((2k+1)^2)
\\\implies\quad &8X((2k+1)^2)
\end{align}$$
and because there is an $8$ I can say its always divisible by $8$, no matter what odd integer $k$ is? How can this be done using proof by induction?
 [/INST] You can do it more simply.  Your statement that if $n$ is odd you can express it as $2k+1$ is correct.  Then write $n^4-n^2=(2k+1)(2k)(2k+1)(2k+2)=(2k+1)^2(2k)(2k+2)$  Now if $k$ is odd, $2k+2$ has two factors of $2$-it is divisible by $4$.  If $k$ is even, $2k$ is divisible by $4$.  Either way, $2k(2k+2)$ is divisible by $8$.  Another approach is just to check the odd residues $\bmod 8$  Induction was well covered by Zelzy
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