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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.

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3(2n) is always divisible by 6 ... not that it helps
the proof might be induction tho
Yes, induction is the best way.

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Hmmm, give me a sec and I will try.
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?
I mean \(k=l-1\)
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 \(\forall n\ge 0\).
You can easily show that it's also true for \(n<0\) since [call \(f(n)=n^3-n\) ] \(f(-n)=(-n)^3+n=-(n^3-n).\)
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.
\[(k+1)^3-(k+1)=(k+1)((k+1)^2-1)=(k+1)(k^2+2k).\] 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.

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