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

 A: 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 
A: 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?
A: $\frac {n^3-n}3+n^2+n=\frac {n^3+3n^2+3n-n}3=\frac {n^3+3n^2+3n+1-n-1}3=\frac {(n+1)^3-(n+1)}3$
A: 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. 
A: 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!

