I have these notes, but I'm confused on what is happening at the inductive step.


$\forall n \in \mathbb{N} 3 | (n^3-n) $

Inductive Step:

For $n \geq 0, show P(n) \Rightarrow P(n+1) is True$

Assume P(n) true i.e. $3 | (n^3-n) $


$(n+1)^3 - (n+1) $ = $n^3 + 3n^2 + 3n + 1 - (n+1)$
= $n^3 + 3n^2 + 2n $
= $n^3 - n + 3n^2 + 3n$

I understand what we're trying to prove but

  1. where does $(n+1)^3 - (n+1) $ come from?
  2. What is happening here $n^3 - n + 3n^2 + 3n$, how did we get there from the previous point? I assume $n^3 + 3n^2 + 2n $ is not obvious if it was divisible by 3, thus we assumed $3 | (n^3-n) $ is true... But how does step 2 become 3?

Thanks in advance


$P(n+1)$ is the statement that $(n+1)^3 - (n+1)$ is divisible by 3. Just fill in $n+1$ in $P(n)$. So you have to simplify this expression first.

The equality between 2 and 3 is just algebra (you write $2n$ from the second term as $3n - n$, which is true for all $n$).

Why would you do this? Because you want to use your assumption that $P(n)$ is true, which says that $n^3 - n$ is divisible by 3. The second term has no term $n^3 - n$ but it has an $n^3$ and then substracts $n$ to get the number $n^3 - n$, about which we know something (!), and add $n$ to another term to compensate and keep the equality.

Now note that the last line is a sum of multiples of 3, the first one, $n^3 - n$, by the inductive assumption.

  • $\begingroup$ thanks for your answer, it's really obvious now, I've +1 since it helped me get a better understanding :) $\endgroup$
    – boywonder
    Oct 30 '15 at 10:25
  1. $(n+1)^3-(n+1)$ comes from the expression $n^3-n$ when you replace $n$ with $n+1$. This is the induction step - to show that $(n+1)^3-(n+1)$ is divisable by $3$ if we assume that $n^3-n$ is.
  2. Here the following happens: $n^3-n$ is divisable by $3$ (by the inductive assumption), $3n^2$ is divisable, $3n$ is divisable, therefore, their sum $n^3-n+3n^2+3n$ is divisable too. (Note that $+2n=-n+3n$ here from 2 to 3.)

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