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I want to proove that 2 and 3 divide $x^3 - x, x \in \mathbb{N}$ and I'm stuck at the inductive step, here's where I'm at:

For all $x \in \mathbb{N}$, let $P(x)$ be the proposition:
2 and 3 divide $x^3 - x$

Basic step:
the first term in $\mathbb{N}$ is $0$, then: $\frac{0^3 - 0}{2} = 0$ et $\frac{0^3 - 0}{3} = 0$, thus $P(0)$ is true.

Inductive step:
For the inductive hypothesis, we assume that $P(k)$ is true for an arbitrary nonnegative integer k bigger than 0. That is, we assume that 2 and 3 divide $k^3 - k$

To carry out the inductive step using this assumption, we must show that when we assume that $P(k)$ is true, then $P(k + 1)$ is also true. That is, we must show that 2 and 3 divide $(k+1)^3 - (k+1)$

Is the next step here is that we need to prove that $\frac{(k+1)^3-(k+1)}{2}$ and $\frac{(k+1)^3-(k+1)}{3}$ are integers? thus 2 and 3 divide $(k+1)^3 - (k+1)$?