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


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


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.

$n=6 q$

$n=6 q+1$

$n=6 q+2$

$n=6 q+3$

$n=6 q+4$

$n=6 q+5$

Case I: When $n=6 q$

$n^{3}=(6 q)^{3}$

$n^{3}-n=(6 q)^{3}-6 q$

$=6 q\left(36 q^{2}-1\right)$

$=6 m\left[\right.$ 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 $n=6 q+1$

$n^{3}=(6 q+1)^{3}$

$n^{3}-n=(6 q+1)^{3}-(6 q+1)$

$=(6 q+1)\left[(6 q+1)^{2}-1\right]$

$=(6 q+1)\left[36 q^{2}+1+12 q-1\right]$

$=(6 q+1)\left[36 q^{2}+12 q\right]$

$=\left[216 q^{3}+72 q^{2}+36 q^{2}+12 q\right]$

$=6\left[36 q^{3}+18 q^{2}+2 q\right]$

$=6 m\left(\right.$ 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 $n=6 q+2$

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

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

$=(6 q+2)\left[(6 q+2)^{2}-1\right]$

$=(6 q+2)\left[36 q^{2}+4+24 q-1\right]$

$=(6 q+2)\left[36 q^{2}+24 q+3\right]$

$=\left[216 q^{3}+144 q^{2}+18 q+72 q^{2}+48 q+6\right]$

$=\left[216 q^{3}+216 q^{2}+66 q+6\right]$

$=6\left[36 q^{3}+36 q^{2}+11 q+1\right]$

$=6 m\left(\right.$ 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 $n=6 q+3$

$n^{3}=(6 q+3)^{3}$

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

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

$=(6 q+3)\left[36 q^{2}+9+36 q-1\right]$

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

$=\left[216 q^{3}+216 q^{2}+48 q+108 q^{2}+108 q+24\right]$

$=\left[216 q^{3}+324 q^{2}+156 q+24\right]$

$=6\left[36 q^{3}+54 q^{2}+26 q+4\right]$

$=6 m$

Hence, $\forall n=6 q+3, n^{3}-n$ is divisible by 6

Case V: When $n=6 q+4$

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

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

$=(6 q+4)\left[(6 q+4)^{2}-1\right]$

$=(6 q+4)\left[36 q^{2}+16+48 q-1\right]$

$=(6 q+4)\left[36 q^{2}+48 q+15\right]$

$=\left[216 q^{3}+288 q^{2}+90 q+144 q^{2}+192 q+60\right]$

$=\left[216 q^{3}+432 q^{2}+282 q+60\right]$

$=6\left[36 q^{3}+72 q^{2}+47 q+10\right]$

$=6 m\left(\right.$ 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 $n=6 q+5$

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

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

$=(6 q+5)\left[(6 q+5)^{2}-1\right]$

$=(6 q+5)\left[36 q^{2}+25+60 q-1\right]$

$=(6 q+5)\left[36 q^{2}+60 q+24\right]$

$=\left[216 q^{3}+360 q^{2}+144 q+180 q^{2}+300 q+120\right]$

$=\left[216 q^{3}+540 q^{2}+444 q+120\right]$

$=6\left[36 q^{3}+90 q^{2}+74 q+120\right]$

$=6 m\left(\right.$ 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 .



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