Proving a Statement using Mathematical Induction I'm trying to prove that $6 \mid (n^3 - n)$ where $n$ is a nonnegative integer. I started off by proving the basic step with $P(6)=4$. The next step would be the induction. However I'm having a bit f trouble understanding and using the method. Could someone show me how I would prove the statement and explain to me how I would get the answer. Thanks in advance.

Hint $\ $ Let $\,f(n) = n^3-n\,$ The base case is clear $\,6\mid f(0) = 0.\ $ For the inductive step we need to prove that $\,6\mid f(n)\,\Rightarrow\, 6\mid f(n\!+\!1).\,$ For this it suffices to prove that $\,6\,\mid f(n\!+\!1)-f(n)=: g(n)\,$ since then $\,6\mid g(n),f(n)\,\Rightarrow\,6\mid g(n)+f(n) = f(n\!+\!1).\ $  But this is easy
$$ f(n\!+\!1) - f(n)\, =\, (n\!+\!2)\color{#0a0}{(n\!+\!1)n - (n+1)n}(n-1) \,=\, 3\color{#0a0}{n(n+1)}$$
which is divisible by $6$ since $\,n\,$ or $\,n\!+\!1\,$ is even (or, use the same method recursively: note that for $\,p(n) = n(n\!+\!1)\,$ we have $\,2\mid p(0)\,$ and $\,2\mid p(n\!+\!1)-p(n) = 2(n\!+\!1))\,$ 
Remark $\ $ This is a special case of a powerful general method known as telescopic induction. You can find many more examples in my posts on telescopy.
