$f(x) = x^3 - x$ then $f(n)$ is multiple of 3

Question

If $f(x) = x^3 - x$ then $f(n)$ is multiple of 3 for all integer $n$.

First I tried $$f(n) = n^3-n=n(n+1)(n-1)\qquad\forall n\ .$$

When $x$ is an integer then at least one factor on the right is even, and exactly one factor on the right is divisible by $3$. It follows that for any $n\in{\mathbb Z}$ the right hand side is divisible by $6$, and so is the left hand side. That is to say: $n^3=n$ mod $6$ for all integers $n$.

Is that correct? or there are another simple solution for this? Thanx.

Answer 1

If $3\mid n$ then the result obviously follows. Otherwise, by FL(ittle)T, $n^3 \equiv n \pmod 3$ so we are done.

Answer 2

After you get $f(n)=n(n+1)(n-1)$, meaning $f(n)$ has factors $n$, $n+1$, $n-1$. Now you want to show $3$ divides one of them.

If $3\mid n$ then you get what you want.

If $3\nmid n$, then $n\equiv 1$ or $n\equiv 2 \pmod 3$. Meaning $3\mid n-1$ or $3\mid n+1$