As homework, I have to prove that
$\forall n \in \mathbb{N}: n^3-n$ is divisible by 6
I used induction
1) basis: $A(0): 0^3-0 = 6x$ , $x \in \mathbb{N}_0$ // the 6x states that the result is a multiple of 6, right?
2) requirement: $A(n):n^3-n=6x$
3) statement: $A(n+1): (n+1)^3-(n+1)=6x$
4) step: $n^3-n+(n+1)^3-n=6x+(n+1)^3-n$
So when I resolve that I do get the equation: $n^3-n=6x$ so the statement is true for $\forall n \in \mathbb{N}$
Did I do something wrong or is it that simple?
Best Answer
No, your argument is not quite right (or at least not clear to me). You must show that if $A(n)$ is true, then $A(n+1)$ follows. $A(n)$ here is the statement "$n^3-n$ is divisible by $6$". Assuming $A(n)$ is true, then $$(n+1)^3-(n+1)=n^3+3n^2+3n+1-n-1=(n^3-n)+3n(n+1)$$ is divisible by $6$ because (1) $n^3-n$ is a multiple of $6$ by assumption, and (2) $3n(n+1)$ is divisible by $6$ because one of $n$ or $n+1$ must be even (this is related to what Alex was pointing out). Therefore $A(n)$ implies $A(n+1)$ and, if $A(n)$ is true for some value of $n$, then all higher integer values of $n$ follow. You correctly showed that $A(0)$ is true.