Question
For any positive integer n, prove that $n^3 - n$ is divisible by $6.$
✓
Answer
Let $n=6 q$ or $6 q+1,6 q+2,6 q+3 \ldots 6 q+5$
If $n=6 q$, then
Then $n ^3- n =(6 q)^3-6 q=216 q^3-6 q$
$=6\left(36 q^3-q\right)$
Which is divisible by 6
If $n =6 q +1$, then
$n^3-n=(6 q+1)^3-(6 q+1)$
$=216 q^3+108 q^2+18 q+1-6 q-1$
$=216 q^3+108 q^2+12 q$
$=6\left(36 q^3+18 q^2+2 q\right)$
Which is also divisible by 6
$\text { If } n=6 q+2 \text {, then }$
$n^3-n=(6 q+2)^3-(6 q+2)$
$=216 q^3+216 q^2+72 q+8-6 q-2$
$=216 q^3+216 q^2+66 q+6$
$=6\left(36 q^3+36 q^2+11 q+1\right)$
Which is divisible by 6
Hence we can similarly, prove that $n ^2- n$ is divisible by $6$ for any positive integer n .Hence proved.
If $n=6 q$, then
Then $n ^3- n =(6 q)^3-6 q=216 q^3-6 q$
$=6\left(36 q^3-q\right)$
Which is divisible by 6
If $n =6 q +1$, then
$n^3-n=(6 q+1)^3-(6 q+1)$
$=216 q^3+108 q^2+18 q+1-6 q-1$
$=216 q^3+108 q^2+12 q$
$=6\left(36 q^3+18 q^2+2 q\right)$
Which is also divisible by 6
$\text { If } n=6 q+2 \text {, then }$
$n^3-n=(6 q+2)^3-(6 q+2)$
$=216 q^3+216 q^2+72 q+8-6 q-2$
$=216 q^3+216 q^2+66 q+6$
$=6\left(36 q^3+36 q^2+11 q+1\right)$
Which is divisible by 6
Hence we can similarly, prove that $n ^2- n$ is divisible by $6$ for any positive integer n .Hence proved.
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