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

Question:

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

Solution:

Euclid's division lemma states that for given positive integers $a$ and $b$, there exists unique integers $q$ and $r$ satisfying $a=b q+r, 0 \leq r Applying Euclid's division lemma om and 6, we have$n=6 q+r, 0 \leq r<6$Therefore, n can have six values, i.e.$n=6 qn=6 q+1n=6 q+2n=6 q+3n=6 q+4n=6 q+5$Case I: When$n=6 qn^{3}=(6 q)^{3}n^{3}-n=(6 q)^{3}-6 q=6 q\left(36 q^{2}-1\right)=6 m\left[\right.$where$\left.m=q\left(36 q^{2}-1\right)\right]$Hence,$\forall n=6 q, n^{3}-n$is divisible by 6 Case II: When$n=6 q+1n^{3}=(6 q+1)^{3}n^{3}-n=(6 q+1)^{3}-(6 q+1)=(6 q+1)\left[(6 q+1)^{2}-1\right]=(6 q+1)\left[36 q^{2}+1+12 q-1\right]=(6 q+1)\left[36 q^{2}+12 q\right]=\left[216 q^{3}+72 q^{2}+36 q^{2}+12 q\right]=6\left[36 q^{3}+18 q^{2}+2 q\right]=6 m\left(\right.$where$\left.m=36 q^{2}+18 q+2 q\right)$Hence,$\forall n=6 q+1, n^{3}-n$is divisible by 6 Case III: When$n=6 q+2n^{3}=(6 q+2)^{3}n^{3}-n=(6 q+2)^{3}-(6 q+2)=(6 q+2)\left[(6 q+2)^{2}-1\right]=(6 q+2)\left[36 q^{2}+4+24 q-1\right]=(6 q+2)\left[36 q^{2}+24 q+3\right]=\left[216 q^{3}+144 q^{2}+18 q+72 q^{2}+48 q+6\right]=\left[216 q^{3}+216 q^{2}+66 q+6\right]=6\left[36 q^{3}+36 q^{2}+11 q+1\right]=6 m\left(\right.$where$\left.m=36 q^{3}+36 q^{2}+11 q+1\right)$Hence,$\forall n=6 q+1, n^{3}-n$is divisible by 6 Case IV: When$n=6 q+3n^{3}=(6 q+3)^{3}n^{3}-n=(6 q+3)^{3}-(6 q+3)=(6 q+3)\left[(6 q+3)^{2}-1\right]=(6 q+3)\left[36 q^{2}+9+36 q-1\right]=(6 q+3)\left[36 q^{2}+36 q+8\right]=\left[216 q^{3}+216 q^{2}+48 q+108 q^{2}+108 q+24\right]=\left[216 q^{3}+324 q^{2}+156 q+24\right]=6\left[36 q^{3}+54 q^{2}+26 q+4\right]=6 m$Hence,$\forall n=6 q+3, n^{3}-n$is divisible by 6 Case V: When$n=6 q+4n^{3}=(6 q+4)^{3}n^{3}-n=(6 q+4)^{3}-(6 q+4)=(6 q+4)\left[(6 q+4)^{2}-1\right]=(6 q+4)\left[36 q^{2}+16+48 q-1\right]=(6 q+4)\left[36 q^{2}+48 q+15\right]=\left[216 q^{3}+288 q^{2}+90 q+144 q^{2}+192 q+60\right]=\left[216 q^{3}+432 q^{2}+282 q+60\right]=6\left[36 q^{3}+72 q^{2}+47 q+10\right]=6 m\left(\right.$where$\left.m=36 q^{3}+72 q^{2}+47 q+10\right)$Hence,$\forall n=6 q+4, n^{3}-n$is divisible by 6 . Case VI: When$n=6 q+5n^{3}=(6 q+5)^{3}n^{3}-n=(6 q+5)^{3}-(6 q+5)=(6 q+5)\left[(6 q+5)^{2}-1\right]=(6 q+5)\left[36 q^{2}+25+60 q-1\right]=(6 q+5)\left[36 q^{2}+60 q+24\right]=\left[216 q^{3}+360 q^{2}+144 q+180 q^{2}+300 q+120\right]=\left[216 q^{3}+540 q^{2}+444 q+120\right]=6\left[36 q^{3}+90 q^{2}+74 q+120\right]=6 m\left(\right.$where$\left.m=36 q^{3}+90 q^{2}+74 q+120\right)$Hence,$\forall n=6 q+5, n^{3}-n\$ is divisible by 6 .