Prove by induction that $n^3-n$ is divisible by $24$ for all odd positive integers Q: Prove by induction that $n^3-n$ is divisible by 24 for all odd positive integers
After proving the first part for n=1
Assume true for some positive integer $n=k$
ie $k^3-k=24x$ where x is an integer
Prove true for $n=k+2$
ie $(k+2)^3-(k+2)=24y$ where y is an integer
$=k^3+6k^2+11k+6$
$=24x+12k+6k^2+6$
But how do I get this in the form $24y$? Am i supposed to use $2k+1$ instead?

We can see that when $n = 1 -----> n (n^2 − 1) = 1 (1 − 1) = 1 (0) = 0$ 
$n = 3 -----> n (n^2 − 1) = 3 (9 − 1) = 3 (8) = 24$ 
We can see that this is true for $n = 1$ and $3$. 
Assume statement is true for $n = 2k−1$. Then we must show it is true for $n = 2k+1$.

Then for $n = 2k-1 -----> (2k−1) ((2k−1)^2 − 1) = 24m$, for some integer m, giving us, 
$(2k−1) ((2k−1)^2- 1) = 24m$ 
$\implies (2k−1) ((2k−1−1)(2k−1+1)) = 24m$ 
$\implies (2k−1) (2k−2) (2k) = 24m$. Thus,
$8k^3 − 12k^2 + 4k = 24m$. 

When $n = 2k+1:$
\begin{align} 
n (n^2 + 1) = (2k+1) ((2k+1)^2 − 1)
\\& = (2k+1) ((2k+1−1)(2k+1+1)) 
\\& = (2k+1) (2k) (2k+2) 
\\& = 8k^3 + 12k^2 + 4k 
\\& = (8k^3 − 12k^2 + 4k) + 24k^2 
\\& = 24m + 24k^2 
\\& = 24 (m + k^2)  \text{divisible by 24} 
\end{align}
Statement true for $n = 2k−1 \Rightarrow$ statement true for $n = 2k+1$.

Therefore, by principles of mathematical induction, statement is true for all odd positive integers $n$. Hope it helps.
