Proof by induction that $n^3 - n$ is divisible by $6$

Question

Show using induction that $n^3-n$ is divisible by 6 $\forall n\ge1, \quad n \in \mathbb{N}$

First off i show that the basis step: $1^3-1=0, \quad \frac{0}{6}=0$

Now I factorised it and set it equal to a multiple of 6: $\mathbf{n(n+1)}(n-1)=6A$

Assuming the result is true for k terms, and trying for $k+1$ terms:

$\mathbf{k(k+1)}(k+2)=6B$

I'm stuck here, I realise that the bold terms are the same, but $k+2$ and $n-1$ are not. Could someone show me what do to next to solve this.

Also is it possible to prove this using modular arithmetic?

Thanks,

Answer 1

$(n+1)^3-(n+1)=n^3+3n^2+3n+1-(n+1)=n^3-n+6\frac{n(n+1)}2$ $=n^3-n+6k$ where $k=\frac{n(n+1)}2$ which is an integer as $2\mid n(n+1)$ for any integer $n$

$\implies (n+1)^3-(n+1)$ will be divisible by $6\iff 6\mid(n^3-n)$

Now,

for $n=1, n^3-n=1^3-1=0$ which is divisible by $6$

for $n=2, n^3-n=2^3-2=6$ which is divisible by $6$

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Alternatively, $n^3-n=n(n^2-1)=(n-1)n(n+1)$ which is a product of $3$ consecutive integers, hence is divisible by $3$ and by $2$.

Hence, $n^3-n$ is divisible by lcm$(2,3)=6$

Answer 2

Suppose as inductive hypothesis $\rm\: \color{#0A0}{f(k)} = (k\!-\!1)k(k\!+\!1)\,=\,\color{#0A0}{6A}.\:$ Then $\rm\,6\mid f(0)\,$ and

$\rm\quad \color{#C00}{f(k\!+\!1)\!-\!f(k)}\, =\, \color{90f}{k(k\!+\!1)}(k\!+\!2\!-\!(k\!-\!1))\, =\, \color{#90f}{k(k\!+\!1)}(\color{}3)\, =\, \color{#C00}{6B},\:$ by $\color{#90f}{\rm\ 2\mid k\,\ or\,\:2\mid k\!+\!1}$

so $\rm\ f(k\!+\!1)\, =\, \color{#C00}{f(k\!+\!1)\!-\!f(k)} +\color{#0A0}{f(k)}\, =\, \underbrace{\color{#C00}{6B}\!+\!\color{#0A0}{6A}}_{\small \rm\textstyle 6(B\!+\!A)},\: $ so $\rm\ \color{#0a0}{6\mid f(k)}\,\Rightarrow\,6\mid f(k\!+\!1)$
Note $ $ Viewing the above proof telescopically makes $\,\color{darkorange}6\,$ divides $\rm \,f(n)\,$ crystal clear, viz.

$\qquad\displaystyle\rm n^3 - n=f(n)\!-\!f(0) = \sum_{k=0}^{n-1} \left[\color{#c00}{f(k\!+\!1)\!-\!f(k)}\right] = \color{darkorange}6\,\sum_{k=0}^{n-1} \frac{\color{#90f}{k(k+1)}}2 $