Answer to Question #349162 in Discrete Mathematics for Nopi

Question

Prove by mathematical induction: 6 divides n^3 - n for n>=2

Answer 1

Let $P(n)$ be the proposition that for the  positive integer $n\ge 2: n^3-n$ is divisible by 6.

Basis Step:

$P(2)$is true, because $2^3-2=6$ is divisible by 6.

Inductive Step:

For the inductive hypothesis we assume that $P(k)$ holds for an arbitrary

positive integer $k.$ That is, we assume that $k^3-k$ is divisible by 6

$$k^3-k=6m, m\in \Z, m\ge 1$$

Under this assumption, it must be shown that $P(k + 1)$ is true, namely, that

$(k+1)^3-(k+1)$ is divisible by 6, is also true.

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

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

If $k$ is even, then $3k(k+1)$ is divisible by 6.

If $k$ is odd, then $k+1$ is even, and $3k(k+1)$ is divisible by 6.

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

This last equation shows that $P(k + 1)$ is true under the assumption that $P(k)$ is true. This completes the inductive step.

We have completed the basis step and the inductive step, so by mathematical induction we know that $P(n)$ is true for all integers $n\ge 2.$ That is, we have proved that for the  positive integer $n\ge 2: n^3-n$ is divisible by 6.