Question

Prove that for any positive interger $n,n^3-n$ is divisible by $6$.

Answer 1

Let $f\left(n\right)=n^3-n$.

Now $f\left(1\right)=0$, divisible by $6$.

Again $f\left(2\right)=6$, divisible by $6$.

So the given statement is true for $n=1,2$.

Let us take the statement is true for some integer $k$.

$\therefore f\left(k\right)=k^3-k$ is divisible by $6$.

Let $f\left(k\right)=6k_1$ [$k_1$ being an integer].......(1).

Now

$f(k+1)$

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

$=k^3+3k^2+2k$

$=k^3-k+3k^2=3k$

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

$k(k+1)$ is the product of two consecutive integers, which is always divisible by $2$.

Then we can write $k(k+1)=2k_2$ [Where $k_2$ is some integer.].

Using this and (1) form (2) we get,

$f(k+1)=6k_1+6k_2=6(k_1+k_2)$

$k_1,k_2$ are integers then $k_1+k_2$ is also an integer.

So $f(k+1)$ is also divisible by $6$.

So the statement is true for $n=k+1$ if we assume it to be true for $n=k$.

By the principle of mathematical induction the statement is true $\forall n\in N$.