How do I prove that $n^3-n$ is always divisible with 6? I sort of see the solution since $n(n-1)(n+1)$ will always have a part that is a 2 and a part that is 3. But I can't formulate it mathematically.

Question

amistre64 · Answer

3(2n) is always divisible by 6 ... not that it helps

amistre64 · Answer

the proof might be induction tho

mr.math · Answer

Yes, induction is the best way.

anonymous · Answer

Hmmm, give me a sec and I will try.

anonymous · Answer

First $f(n) = \frac{n^3-n}{6}$ 
$$f(1) = \frac{0}{6} = 0$$ so valid, then $f(k)$ and $f(k+1)$ 
$$f(k+1) = \frac{(k+1)^3-(k+1)}{6}$$ which boils down to 
$$f(k+1) = \frac{k(k+1)(k+2)}{6}$$ which basically is 
$$f(k+1) = \frac{l^3-l}{6} $$ where $k=l+1$, is that good enough?

anonymous · Answer

I mean $k=l-1$

mr.math · Answer

I'll prove it for $n\ge 0$. Let $P(n)$ be the statement that $n^3-n$ is divisible by $6$. $P(0)$ is obviously true as you said. Now assume that $P(k)$ is true, that is $k^3-k$ is divisible by $6$ then for P(k+1):
$$(k+1)^3-(k+1)=(k+1)(k^2+2k)=k^3+3k^2+2k=k^3-k+3k(k+1).$$
By the induction hypothesis $k^3-k$ is divisible by $6$ and it's obvious that $3k(k+1)$ is also divisible by $6$ since either $k$ or $k+1$ is divisible by $2$. Therefore $P(k)$ implies $P(k+1)$ and thus $P(n)$ is true $\forall n\ge 0$.

mr.math · Answer

You can easily show that it's also true for $n<0$ since [call $f(n)=n^3-n$ ] $f(-n)=(-n)^3+n=-(n^3-n).$

anonymous · Answer

I'm not sure about the $(k+1)^3-(k+1) = (k+1)(k^2+2k)$ of your equations, but the rest is right and with less work :) but I hope you approve of my method as well.

mr.math · Answer

$$(k+1)^3-(k+1)=(k+1)((k+1)^2-1)=(k+1)(k^2+2k).$$
As for your method, you didn't state clearly what is the induction hypothesis and you didn't show that f(k) implies f(k+1). That's at least what I think.