Proof by mathematical induction in Z Is it possible to proof the following by mathematical induction? If yes, how?
$a\in \mathbb{Z} \Rightarrow 3$  | $(a^3-a)$
I should say no, because in my schoolcarrier they always said that mathematical induction is only possible in $\mathbb{N}$. But I never asked some questions why it is only possible in $\mathbb{N}$...

Since $\mathbb{Z}$ is countable as $\mathbb{N}$ we can extend induction over $\mathbb{Z}$.

BASE CASE:

$$a=1 \implies 3|0$$

INDUCTIVE STEP 1 "UPWARD"

assume: $3|a^3-a$
$$(a+1)^3-(a+1)=a^3+3a^2+3a+1-a-1=a^3-a+3a^2+3a\equiv0\pmod 3$$
thus
$$3|(a+1)^3-(a+1)$$

INDUCTIVE STEP 2 "DOWNWARD"

assume: $3|a^3-a$
$$(a-1)^3-(a-1)=a^3-3a^2+3a-1-a+1=a^3-a-3a^2+3a\equiv0\pmod 3$$
thus
$$3|(a-1)^3-(a-1)$$

Thus: $$\forall a\in \mathbb{Z} \Rightarrow 3|a^3-a$$

