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.
My proof:
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 $(2k-1)^3 - (2k-1) = 24p, p∈N$.
$(2k-1) [(2k-1)^2-1] = 24p$
$(2k-1)[(2k-1-1)(2k-1+1)] = 24p$
$(2k-1)[(2k-2)(2k)] = 24p$
$8k^3-12k^2+4k=24p$
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]$
$=(2k+1)[(2k+1-1)(2k+1+1)]$
$=(2k+1)[(2k)(2k+2)]$
$=8k^3+12k^2+4k$
$=(8k^3-12k^2+4k)+24k^2$
$=24p+24k^2$ (Induction hypothesis)
$=24(p+k^2)$
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!
That's the right idea. Essentially you have verified the following equation
$$ \color{}{(2k+1)^3 -(2k+1)}\, =\ \color{#0a0}{(2k-1)^3 - (2k-1)}\, +\, 24k^2$$
Therefore $\ 24\mid \color{#0a0}{\rm green}\,\Rightarrow\,24\mid\rm RHS\,\Rightarrow\,24\mid LHS,\,$ which yields the inductive step.
Remark $\ $ A slicker way to prove that equality is to note that $\,f(k) = \rm RHS-LHS$ is at most quadratic (cubic terms cancel), so to verify that it is zero it suffices to show that it has $3$ roots, e.g. verify $\,0 = f(1) = f(2) = f(3).\,$ But this looks like the base of an inductive proof! Indeed, if you study the calculus of finite differences and/or telescopy you will learn even nicer ways to handle such inductive proofs. You can find some examples in my posts on telescopy.