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,
Best Answer
$(n+1)^3-(n+1)=n^3+3n^2+3n+1-(n+1)=n^3-n+6\frac{n(n+1)}2$ $=n^3-n+6k$ 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 $6$
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$(2,3)=6$