Prove by induction that $n^3-n$ is divisible by $24$ for all odd positive integers

Question

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?

Answer 1

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 $n = 2k+1$.

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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,
$8k^3 − 12k^2 + 4k = 24m$.

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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 $n = 2k+1$.

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

Answer 2

It's fine the way it is, but at this point, note that $k$ is odd, so $k=2l+1$ for some integer $l$. Insert that into your expression.

Of course, you could also start the induction step by saying "Assume $(2k-1)^3-(2k-1)=24x$. Now we look at $(2k+1)^3-(2k+1)$", but in my opinion, that's just more arithmetical hassle for no real gain. Doing it your way we don't have to expand two cubic expressions, just one quadratic, and the substitution to $24x$ is simpler, thus not as prone to minor mistakes.