Showing $n^3 - n$ is divisible by $6$

Question

How would you show that $n^3-n$ is divisible by $6$, when $n=k+1$ ?

Answer 1

Let for $n=k$, the number $n^3-n$ be divisible by $6$ then we have $$k^3-k=6\lambda$$ Now, substituting $n=k+1$, $$(k+1)^3-(k+1)$$$$=(k+1)(k^2+2k+1-1)$$$$=k^3+3k^2+2k$$ $$=(k^3-k)+3k^3+3k$$ substituting $k^3-k=6\lambda$ $$=6\lambda+3k(k+1)$$ since, $\color{red}{\text{product of two consecutive integers is divisible by}\ 2}$ so we can let $k(k+1)=2m$ where, $m$ is an integer. Hence, we get $$6\lambda+3k(k+1)=6\lambda+3(2m)=6(\lambda+m)$$ since $(\lambda+m)$ is an integer hence the number $6(\lambda+m)$ is divisible by $6$

Thus, the number $n^3-n$ is divisible by $6$ for $n=k+1$

Answer 2

HINT: use that $$n^3-n=(n-1)n(n+1)$$ without induction

Answer 3

Using induction:

\begin{align} (n+1)^3-(n+1) & = n^3+3n^2+3n+1-n-1 \\ & = n^3+3n^2+2n \\ & = (n^3-n)+3(n^2+n) \\ & = (n^3-n)+3n(n+1) \\ & = (n^3-n)+6k(2k \pm 1) \\ \end{align}

for $k = \lceil n/2 \rceil$.

Answer 4

$n^3-n=n(n^2-1)=n(n-1)(n+1)$ it is obvious that $2\vert n(n-1)(n+1)$ (three consecutive integers, atleast one of them is even), easily you can check $3\vert n(n-1)(n+1)$ too.