Question

Prove that for any integer of the form $n^3-n$ is divisible by $3$.

Answer 1

$N^3-n=n(n^2-1)=n(n-1)(n+1)$ is divided by $3$ then possible remainder $0,1$ and $2[\because$i $P=ab+r$, then $0\le r<a$ by Euclid lemma]

$\therefore$ Let $n=3r,3r+1,3r+2$, where $r$ is an integer

Case 1 :- when $n=3r$

Then, $n^3-n$ is divisible by $3[\because n^3-n=n(n-1\left)\right(n+1)=3r(3r-1\left)\right(3r+1)$ , clearly shown it is divisible by $3]$

Case 2:- when $n=3r+1$

e.g., $n-1=3r+1-1=3r$

Then, $n^3-n=(3r+1)\left(3r\right)(3r+2)$, it is divisible by $3$

Case 3:- when $n=3r-1$

e.g., $n+1=3r-1+1=3r$

Then, $n^3-n=(3r-1)(3r-2)\left(3r\right)$, it is divisible by $3$