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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?

  • 1
    I know what you meant, but it's confusing when in both of steps (2) and (3) you use $6x$. It would be clearer if you had said in (2) $n^3-n=6x$ for some integer $x$ and in (3) you had said $(n+1)^3-(n+1)=6y$ for some integer $y$.2012-10-11

3 Answers 3