Induction in reverse So the statement that I have to prove is as follows $P(n)=n^3 -n$ is divisible by $3$.
Now I have to prove this with backwards induction for all negative numbers but I've already done the same thing but with "straight" induction for all positive numbers.
Do I need to start from the case $P(n+1)$ and go from there to $P(1)$ or $P(-1)$?
Likewise: 
$\begin{align}
P(n+1) & = (n+1)^3 - (n+1) \\
& = n^3 + 3\cdot n^2 + 3 \cdot n + 1 - n + 1      \\
& = 3n(n+1) + n^3 - n 
\end{align}$
$\text{I.H.: So let's suppose the above statement holds for any}$ $n \in \mathbb { Z_0^{-}} $ 
Then I don't know anymore how to continue. 
For the 2nd part I need to prove that the case of the negative numbers directly follows out of the positive case.
I don't know how to get this part because I'm not sure of the previous case.
Please help me out.
 A: To show that it holds for all negative numbers by backwards induction (your part b)), show that it holds for $n = 0$ (or maybe start at $n=-1$), and then show that anytime it holds for some $n$, it also holds for $n-1$
For part d): $n^3-n = n(n^2-1) = n(n+1)(n-1) = (n-1)n(n+1)$. Since these are $3$ consecutive integers, one of them will be divisible by $3$
A: hint
$$P (n-1)=(n-1)^3-(n-1) $$
$$=(n-1)\Bigl ((n-1)^2-1\Bigr) $$
$$=(n-1)(n-1+1)(n-1-1)$$
$$=(n-2)(n-1)n $$
=product of three consecutive integers.
One of them is a multiple of $3$.
A: if $n<0$ we have $n^3−n=-(\vert{n}\vert^3-\vert{n}\vert)$, $\vert{n}\vert=1,2, 3...$. 
A: If you prove a base case that $P(n)$ for $n= k$ 
And you prove positive induction step that $P(n)\implies P(n+1)$ then via induction you have proven this for all $n \ge k$.
And if you prove negative induction step that $P(n) \implies P(n-1)$ then via induction you have proven this for all $n \le k$.
And if you prove both positive and negative induction then you have proven it for all $n \ge k$ AND all $n \le k$ or in other words for all integers.
The neat thing about this is you can choose any base value you like. (But you do have to choose one.)
Ex:
Let $P(n) = 3|n^3 - n$
Base case:  Let $n = 13$ then $13^3 - 13 = 2184=3*728$.
Dual Induction step.
Assume $P(n)$ so $3|n^3 - n$  and $n^3 - n = 3k$ for some integer $k$.
$(n\pm 1)^3 - (n\pm 1) = (n^3 \pm 3n^2+ 3n \pm 1) - (n\pm 1) = n^3 \pm 3n^2 +2n$
$= n^3 - n + 3n \pm 3n^2 = 3k + 3n \pm 3n^2= 3(k+n \pm n^2)$.
So $3|(n\pm 1)^3 - (n\pm 1)$ so $P(n\pm 1)$.
So by induction $P(n)$ holds for all $n \ge 13$ and $P(n)$ holds for all $n \le 13$.
So $P(n)$ holds for all integers.
A: Hint $\ $ You've shown $\,3\mid P(n\!+\!1)-P(n)\,$ thus $\,3\mid P(n\!+\!1)\iff 3\mid P(n),\,$ and this bidirectional inference yields the sought induction step in both directions.
Remark $ $ Generally if $\,p(n)\,$ is true $\color{#c00}\Leftarrow\!\color{#0a0}\Rightarrow p(n\!+\!1)$ is true, $ $ then $p(n)$ is true for all integers $n$ $\iff p(n)$ is true for some integer $a$, since $p(n)\color{#0a0}\Rightarrow p(n\!+\!1)\,$ allows us to ascend truth of $p(a)$ to all integers $>a\,$ by normal induction, $ $ and $\,p(n)\color{#c00}\Leftarrow p(n\!+\!1)\,$ allows us to descend truth of $p(a)$ to all integers $< a\,$ by reverse induction. $ $ Above is the case $\,p(n)\, :=\, 3\mid n^3-n\,$ and e.g. $\,a=0.$
