<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 are almost done in your first step. You showed
$$
(n^4)−(n^2)=(n^2)((n^2)−1)=n(n−1)n(n+1).
$$
Now $n$ is odd hence $n-1$ and $n+1$ are even, since every second even number is divisible by $4$ either $4k_1 =n-1$ and $2k_2=n+1$ or $4k_1=n+1$ and $2k_2=n-1$ for some $k_1,k_2\in\mathbb{N}$, this is
$$
(n^4)−(n^2) = n^2(4k_1)(2k_2).
$$
But if you want to prove it by induction you can try this: Let $n$ be odd such that $8|(n^4)−(n^2)$, we will show it for $n+2$.
First note that $8|(n^4)−(n^2)=n^2(n−1)(n+1)$ implies that $8|=(n−1)(n+1)$, hence $\exists k\in\mathbb{N}$ such that $8k=(n-1)(n+1)$, and compute for $n+2$
$$
((n+1)^4)−((n+2)^2) =(n+2)^2(n+1)(n+3)= (n+2)^2(n+1)((n-1)+4)=(n+2)^2(8k+(n+1)4)
$$
since $n$ is odd $n+1$ is even and hence $8|4(n+1)$ and we are done.
</s>