## plitter 3 years ago 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.

1. amistre64

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

2. amistre64

the proof might be induction tho

3. Mr.Math

Yes, induction is the best way.

4. plitter

Hmmm, give me a sec and I will try.

5. plitter

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?

6. plitter

I mean $$k=l-1$$

7. Mr.Math

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 $$\forall n\ge 0$$.

8. Mr.Math

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

9. plitter

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.

10. Mr.Math

$(k+1)^3-(k+1)=(k+1)((k+1)^2-1)=(k+1)(k^2+2k).$ 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.