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How can I show that $n(n^2-1)$ is divisible by 24, if $n$ is an odd integer greater than $2$?

I am thinking that since odd numbers have the form of $2n-1$ in which if it is to be more than $2$, it will be $2n-1+1 = 2n+1$. So would it be correct to use this and try solving through induction?

Omari Celestine
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  • Easier to do directly. Hint: $24=2^3\times 3$. It is clear that your expression is divisible by $3$ (why?) and only slightly harder to show that it is divisible by $8$. – lulu Nov 07 '17 at 11:16
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    The "greater than $2$" part is unnecessary: $1(1^2-1)=0,$ and $0$ is divisible by $24,$ and $(-n)((-n)^2-1)=-(n(n^2-1)),$ so we can change $n$ to $-n$ without changing whether the result is divisible by $24.$ – David K Nov 07 '17 at 11:18
  • @DavidK The "greater than 2" part is part of the problem so I don't believe I can say that is is unnecessary. – Omari Celestine Nov 07 '17 at 11:20
  • @lulu I understand part of what you are demonstrating but I am not sure how to use that in solving the problem. Do you think you can demonstrate it by posting it as an answer to the question? – Omari Celestine Nov 07 '17 at 11:22
  • If you were to ignore the "greater than $2$" and decide to do a proof that only showed divisibility by $24$ when $n > 10,$ then indeed you would not have solved the problem, because you were required to show it for $n=3,5,7,9$ and you would not have done so. But if you prove divisibility for every odd integer $n,$ then surely that also is a proof that is valid for every odd integer greater than $2.$ – David K Nov 07 '17 at 11:25

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Induction on odd numbers is easier when expressed as $P(n_0)$ is true, $P(n) \implies P(n+2)$.

Let $f(n)=n(n^2-1)$. Then $f(1)=24$ and $f(n+2)-f(n)=6 (n + 1)^2$. If $n$ is odd, then $n+1$ is even and so $(n+1)^2$ is a multiple of $4$. Therefore, $f(n+2)-f(n)$ is a multiple of $24$ and so is $f(n+2)$ by induction.

For fun, here is a different approach using finite differences. Write $n=2t+1$. Then $$ n(n^2-1) = 8 t^3 + 12 t^2 + 4 t = 24 \binom{t}{1} + 72 \binom{t}{2} + 48 \binom{t}{3} $$ clearly a multiple of $24$.

lhf
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The product $p(p+1)$ of two consecutive numbers is divisible by $2$

because either $p$ or $p+1$ is even.

The product $p(p+1)(p+2)$ of three consecutive numbers is divisible by $3$

because either $p,\ p+1$ or $p+2$ is a multiple of $3$.


$f(n)=n(n^2-1)=(n-1)n(n+1)$ is then divisible by $3$.

Also for $n=2p+1$ odd then $(n-1)(n+1)=(2p)(2p+2)=4p(p+1)$ is divisible by $4\times 2$

So $f(n)$ is divisible by $8\times 3=24$.

zwim
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Easier with congruences:

  • $n^3-n\equiv 0\mod 3$ for all $n$ (that's Little Fermat's theorem),
  • If $n$ is odd, $n\equiv \pm 1,\pm3\mod 8$, so $\;n^2\equiv 1\mod8$,
  • last, use the Chinese remainder theorem.
Bernard
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Let $n=2m+1$.

Among $2m(2m+1)(2m+2)$, one of the extreme factors is a multiple of $4$ and the other is even, and one of the three factors is a multiple of $3$.

Check with

$$2\cdot3\cdot4,\\4\cdot5\cdot6=4\cdot5\cdot2.3,\\6\cdot7\cdot8=2.3\cdot7\cdot2.4,\\8\cdot10\cdot12=2.4\cdot10\cdot2.2.3,\\\cdots$$

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Because $N=n(n+1)(n-1)$ is a product of three consecutive numbers it is a multiple of both $2$ and $3$ so a multiple of $6$. Besides $N=2n+1$ gives $$N=(2n+1)(2n+2)(2n=4(2n+1)(n+1)n$$ and $N$ is also a multiple of $4$ so $N=24t$ for some integer $t$.(Note that $(n+1)n$ is even).

Ataulfo
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