Question

For any positive integer n, prove that n³ – n is divisible by 6.

Answer 1

Euclid's division lemma states that for given positive integers a and b, there exists unique integers q and r satisfying $a=bq+r,0\le r<b$
Applying Euclid's division lemma om n and 6, we have
$n=6q+r,0\le r<6$
Therefore, n can have six values, i.e.
$$\begin{gathered} n=6q \\ n=6q+1 \\ n=6q+2 \\ n=6q+3 \\ n=6q+4 \\ n=6q+5 \end{gathered}$$
Case I: When $n=6q$
$$\begin{gathered} n^3=(6q{)}^3 \\ {n}^3-n=(6q{)}^3-6q \\ =6q(36q^2-1) \\ =6m\left[\mathrm{where}m=q(36q^2-1)\right] \end{gathered}$$
Hence, $\forall n=6q,n^3-n$ is divisible by 6

Case II:
When $n=6q+1$
$$\begin{gathered} n^3=(6q+1{)}^3 \\ {n}^3-n=(6q+1{)}^3-(6q+1) \\ =(6q+1)\left[(6q+1{)}^2-1\right] \\ =(6q+1)\left[36q^2+1+12q-1\right] \\ =(6q+1)\left[36q^2+12q\right] \\ =\left[216q^3+72q^2+36q^2+12q\right] \\ =6\left[36q^3+18q^2+2q\right] \\ =6m(\mathrm{where}m=36q^2+18q+2q) \\ Hence,\forall n=6q+1,n^3-n\mathrm{is}\mathrm{divisible}\mathrm{by}6 \end{gathered}$$

Case III: When $n=6q+2$
$$\begin{gathered} n^3=(6q+2{)}^3 \\ {n}^3-n=(6q+2{)}^3-(6q+2) \\ =(6q+2)\left[(6q+2{)}^2-1\right] \\ =(6q+2)\left[36q^2+4+24q-1\right] \\ =(6q+2)\left[36q^2+24q+3\right] \\ =\left[216q^3+144q^2+18q+72q^2+48q+6\right] \\ =\left[216q^3+216q^2+66q+6\right] \\ =6\left[36q^3+36q^2+11q+1\right] \\ =6m(\mathrm{where}m=36q^3+36q^2+11q+1) \\ Hence,\forall n=6q+1,n^3-n\mathrm{is}\mathrm{divisible}\mathrm{by}6 \end{gathered}$$

Case IV: When $n=6q+3$
$$\begin{gathered} n^3=(6q+3{)}^3 \\ {n}^3-n=(6q+3{)}^3-(6q+3) \\ =(6q+3)\left[(6q+3{)}^2-1\right] \\ =(6q+3)\left[36q^2+9+36q-1\right] \\ =(6q+3)\left[36q^2+36q+8\right] \\ =\left[216q^3+216q^2+48q+108q^2+108q+24\right] \\ =\left[216q^3+324q^2+156q+24\right] \\ =6\left[36q^3+54q^2+26q+4\right] \\ =6m \end{gathered}$$
Hence, $\forall n=6q+3,n^3-n$ is divisible by 6.

Case V: When $n=6q+4$
$$\begin{gathered} n^3=(6q+4{)}^3 \\ {n}^3-n=(6q+4{)}^3-(6q+4) \\ =(6q+4)\left[(6q+4{)}^2-1\right] \\ =(6q+4)\left[36q^2+16+48q-1\right] \\ =(6q+4)\left[36q^2+48q+15\right] \\ =\left[216q^3+288q^2+90q+144q^2+192q+60\right] \\ =\left[216q^3+432q^2+282q+60\right] \\ =6\left[36q^3+72q^2+47q+10\right] \\ =6m(\mathrm{where}m=36q^3+72q^2+47q+10) \end{gathered}$$
Hence, $\forall n=6q+4,n^3-n$ is divisible by 6.

Case VI: When $n=6q+5$
$$\begin{gathered} n^3=(6q+5{)}^3 \\ {n}^3-n=(6q+5{)}^3-(6q+5) \\ =(6q+5)\left[(6q+5{)}^2-1\right] \\ =(6q+5)\left[36q^2+25+60q-1\right] \\ =(6q+5)\left[36q^2+60q+24\right] \\ =\left[216q^3+360q^2+144q+180q^2+300q+120\right] \\ =\left[216q^3+540q^2+444q+120\right] \\ =6\left[36q^3+90q^2+74q+120\right] \\ =6m(\mathrm{where}m=36q^3+90q^2+74q+120) \end{gathered}$$
Hence, $\forall n=6q+5,n^3-n$ is divisible by 6.