Question

Use Euclid's Division Lemma, prove that for any positive integer $n,n^3-n$ is divisible by $6$.

Answer 1

Any positive integer is of the form $6m,6m+1,6m+2,6m+3,6m+4,6m+5$ for some positive integer $n$.

When $n=6m$,

$n^3-n={\left(6m\right)}^3-6m=216m^3-6m$

$=6m(36m^2-1)=6q$ where $q=m(36m^2-1)$

$n^3-n$ is divisible by $6$

When $n=6m+1$

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

$=(6m+1)\left(6m\right)(6m+2)$

$=6m(6m+1)(6m+2)$

$6q$ where $q=m(6m+1)(6m+2)$

When $n=6m+2$

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

$=(6m+1)(36m^2+30m+6)$

$=6m(36m^2+30m+6)+1(36m^2+30m+6)$

$=6\left[m(36m^2+30m+6)\right]+6(6m^2+5m+1)$

$=6p+6q,$ where $p=m(36m^2+30m+6)$ and $q=6m^2+5m+1$

$n^3-n$ is divisible by $6$

When $n=6m+3$

$n^3-n={(6m+3)}^2-(6m+3)$

$=(6m+3)[{(6m+3)}^2-1]$

$=6m({(6m+3)}^2-1)+3({(6m+3)}^2-1)$

$=6\left[m({(6m+3)}^2-1)\right]+3[36m^2+36m+8]$

$=6p+3q,$

where $p=m({(6m+3)}^2-1)$ and $q=36m^2+36m+8$

$n^3-n$ is divisible by $6$

When $n=6m+4$

$n^3-n={(6m+4)}^3-(6m+4)$

$=(6m+4)[{(6m+4)}^2-1]$

$=6m[{(6m+4)}^2-1]+4[{(6m+4)}^2-1]$

$=6m[{(6m+4)}^2-1]+4[36m^2+48m+15]$

$=6m[{(6m+4)}^2-1]+12[12m^2+16m+5]$

$=6m[{(6m+4)}^2-1]+6[24m^2+32m+10]$

$=6p+6q$, where $p=m[{(6m+4)}^2-1]$ and $q=24m^2+32m+10$

$n^3-n$ is divisible by $6$

When $n=6m+5$

$n^3-n={(6m+5)}^3-(6m+5)$

$=6m[{(6m+5)}^2-1]+5[{(6m+5)}^2-1]$

$=6m[{(6m+5)}^2-1]+5[36m^2+60m+24]$

$=6p+30q$

$=6(p+5q)$

$=6(p+5q)$ where $p=m[{(6m+5)}^2-1]$ and $q=6m^2+10m+5$

$n^3-n$ is divisible by $6$

Hence, $n^3-n$ is divisible by $6$ for any positive integer $n$