Proof by induction that $\frac13(n^3-n)$ is an integer. Am I on the correct path?

Question

I have been trying to prove $$\frac{n^3-n}{3}=k\in \mathbb N $$ I have tried the following calculations, have however difficulties in the final step.

Could you help me out?

Here are my calculations:

Answer 1

$\frac {n^3-n}3+n^2+n=\frac {n^3+3n^2+3n-n}3=\frac {n^3+3n^2+3n+1-n-1}3=\frac {(n+1)^3-(n+1)}3$

Answer 2

I suspect the $k$ in $k \in \mathbb N$ is confusing you

You have shown that $\dfrac{(n+1)^3-(n+1)}{3}=\dfrac{n^3-n}{3} +n^2+n$ and you know that

• $\dfrac{n^3-n}{3}$ is an element of $\mathbb N$ (by hypothesis)
• $n^2+n$ is an element of $\mathbb N$ (multiplication and addition of integers)
• $\dfrac{n^3-n}{3} +n^2+n$ is an element of $\mathbb N$ (addition of integers)

so $\dfrac{(n+1)^3-(n+1)}{3}$ is an element of $\mathbb N$, quod erat demonstrandum

Answer 3

Hint: Note that

$$\frac{n^3-n}{3} = \frac{n(n^2-1)}{3} = \frac{n(n+1)(n-1)}{3}$$

From here, you have a product of three consecutive integers in the numerator. What can you conclude from that?

Answer 4

Hypothesis : $3|(n^3-n)$.

Step $n+1$:

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

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

$(n^3-n)+ 3(n^3+n)$ ;

The second term is divisible by 3, so is the first term by hypothesis.

Answer 5

For the induction, you, in fact, reach your goal since $\frac{n^3-n}{3}$ is an integer by the inductive hypothesis.

If you do not to prove it with induction, there are other ways to do so.

• Consider Fermat Little Theorem. This theorem says that $n^p-n$ is always divisible by $p$ where $p$ is a prime. You just reach your goal in one step!
• You may also notice that $n^3-n = n(n^2-1) = (n-1)n(n+1)$. Now, one of these three consecutive integers must be divisible by $3$. This also follows from a theorem saying that, any $k$ consecutive integers is divisible by $k!$. According to this theorem, this number is divisible by $6$ too!