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**Danneedshelp** So, if I further factore I have that $\displaystyle n^{3}-n=n(n^{2}-1)=n(n-1)(n+1)$. Now, if $\displaystyle 6|n(n-1)(n+1)$, then there exists a $\displaystyle K$ such that $\displaystyle \frac{n(n-1)(n+1)}{6}=k$. Now, for n=0 and n=1 we see that k=0, but 0 is a multiple of any number. Furthermore, for any n>1, at least one of $\displaystyle (n+1)$ or $\displaystyle (n-1)$ will be divisible by 3 whenever n is even. Similarly, if n>2 and odd, then n will at least be divisible by 3 and $\displaystyle (n+1)$ or $\displaystyle (n-1)$ will be divisible by 2. Since 6=2*3, we have show that $\displaystyle 6|n^{3}-n$, because $\displaystyle n^{3}-n=n(n+1)(n-1)$.

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