I would like to show that the following statement is true by the principle of mathematical induction (I must only use induction, not other theorems to justify my answer)
If $n$ is odd natural number, then $n^3-n$ is divisible by 24.
Base Case: For $n=1, n^3-n = 1-1 = 0$ which is divisible by $24$.
Induction hypothesis: Assume that that statement is true for $n=2k-1, k∈N$. This means that $(2k-1)^3 - (2k-1)$ is divisible by 24 and hence $(2k-1)^3 - (2k-1) = 24p, p∈N$.
$(2k-1) [(2k-1)^2-1] = 24p$
$(2k-1)[(2k-1-1)(2k-1+1)] = 24p$
$(2k-1)[(2k-2)(2k)] = 24p$
Inductive step: Show that the statement is true for $n=2k+1, k∈N$.
$n^3-n = (2k+1)^3-(2k+1)$
$ = (2k+1)[(2k+1)^2-1]$
$=24p+24k^2$ (Induction hypothesis)
Both expressions are divisible by $24$, hence the expression is divisible by 24.
We have shown that the statement is true for $n=2k+1$. Therefore, by induction, statement is true for all odd natural numbers n.
Please give me your suggestions. Thanks!