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Question

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

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Solution

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 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. $n=6q\phantom{\rule{0ex}{0ex}}n=6q+1\phantom{\rule{0ex}{0ex}}n=6q+2\phantom{\rule{0ex}{0ex}}n=6q+3\phantom{\rule{0ex}{0ex}}n=6q+4\phantom{\rule{0ex}{0ex}}n=6q+5$ Case I: When $n=6q$ ${n}^{3}=\left(6q{\right)}^{3}\phantom{\rule{0ex}{0ex}}{n}^{3}-n=\left(6q{\right)}^{3}-6q\phantom{\rule{0ex}{0ex}}=6q\left(36{q}^{2}-1\right)\phantom{\rule{0ex}{0ex}}=6m\left[\mathrm{where}m=q\left(36{q}^{2}-1\right)\right]$ Hence, $\forall n=6q,{n}^{3}-n$ is divisible by 6 Case II: When $n=6q+1$ ${n}^{3}=\left(6q+1{\right)}^{3}\phantom{\rule{0ex}{0ex}}{n}^{3}-n=\left(6q+1{\right)}^{3}-\left(6q+1\right)\phantom{\rule{0ex}{0ex}}=\left(6q+1\right)\left[\left(6q+1{\right)}^{2}-1\right]\phantom{\rule{0ex}{0ex}}=\left(6q+1\right)\left[36{q}^{2}+1+12q-1\right]\phantom{\rule{0ex}{0ex}}=\left(6q+1\right)\left[36{q}^{2}+12q\right]\phantom{\rule{0ex}{0ex}}=\left[216{q}^{3}+72{q}^{2}+36{q}^{2}+12q\right]\phantom{\rule{0ex}{0ex}}=6\left[36{q}^{3}+18{q}^{2}+2q\right]\phantom{\rule{0ex}{0ex}}=6m\left(\mathrm{where}m=36{q}^{2}+18q+2q\right)\phantom{\rule{0ex}{0ex}}Hence,\forall n=6q+1,{n}^{3}-n\mathrm{is}\mathrm{divisible}\mathrm{by}6$ Case III: When $n=6q+2$ ${n}^{3}=\left(6q+2{\right)}^{3}\phantom{\rule{0ex}{0ex}}{n}^{3}-n=\left(6q+2{\right)}^{3}-\left(6q+2\right)\phantom{\rule{0ex}{0ex}}=\left(6q+2\right)\left[\left(6q+2{\right)}^{2}-1\right]\phantom{\rule{0ex}{0ex}}=\left(6q+2\right)\left[36{q}^{2}+4+24q-1\right]\phantom{\rule{0ex}{0ex}}=\left(6q+2\right)\left[36{q}^{2}+24q+3\right]\phantom{\rule{0ex}{0ex}}=\left[216{q}^{3}+144{q}^{2}+18q+72{q}^{2}+48q+6\right]\phantom{\rule{0ex}{0ex}}=\left[216{q}^{3}+216{q}^{2}+66q+6\right]\phantom{\rule{0ex}{0ex}}=6\left[36{q}^{3}+36{q}^{2}+11q+1\right]\phantom{\rule{0ex}{0ex}}=6m\left(\mathrm{where}m=36{q}^{3}+36{q}^{2}+11q+1\right)\phantom{\rule{0ex}{0ex}}Hence,\forall n=6q+1,{n}^{3}-n\mathrm{is}\mathrm{divisible}\mathrm{by}6$ Case IV: When $n=6q+3$ ${n}^{3}=\left(6q+3{\right)}^{3}\phantom{\rule{0ex}{0ex}}{n}^{3}-n=\left(6q+3{\right)}^{3}-\left(6q+3\right)\phantom{\rule{0ex}{0ex}}=\left(6q+3\right)\left[\left(6q+3{\right)}^{2}-1\right]\phantom{\rule{0ex}{0ex}}=\left(6q+3\right)\left[36{q}^{2}+9+36q-1\right]\phantom{\rule{0ex}{0ex}}=\left(6q+3\right)\left[36{q}^{2}+36q+8\right]\phantom{\rule{0ex}{0ex}}=\left[216{q}^{3}+216{q}^{2}+48q+108{q}^{2}+108q+24\right]\phantom{\rule{0ex}{0ex}}=\left[216{q}^{3}+324{q}^{2}+156q+24\right]\phantom{\rule{0ex}{0ex}}=6\left[36{q}^{3}+54{q}^{2}+26q+4\right]\phantom{\rule{0ex}{0ex}}=6m$ Hence, $\forall n=6q+3,{n}^{3}-n$ is divisible by 6. Case V: When $n=6q+4$ ${n}^{3}=\left(6q+4{\right)}^{3}\phantom{\rule{0ex}{0ex}}{n}^{3}-n=\left(6q+4{\right)}^{3}-\left(6q+4\right)\phantom{\rule{0ex}{0ex}}=\left(6q+4\right)\left[\left(6q+4{\right)}^{2}-1\right]\phantom{\rule{0ex}{0ex}}=\left(6q+4\right)\left[36{q}^{2}+16+48q-1\right]\phantom{\rule{0ex}{0ex}}=\left(6q+4\right)\left[36{q}^{2}+48q+15\right]\phantom{\rule{0ex}{0ex}}=\left[216{q}^{3}+288{q}^{2}+90q+144{q}^{2}+192q+60\right]\phantom{\rule{0ex}{0ex}}=\left[216{q}^{3}+432{q}^{2}+282q+60\right]\phantom{\rule{0ex}{0ex}}=6\left[36{q}^{3}+72{q}^{2}+47q+10\right]\phantom{\rule{0ex}{0ex}}=6m\left(\mathrm{where}m=36{q}^{3}+72{q}^{2}+47q+10\right)$ Hence, $\forall n=6q+4,{n}^{3}-n$ is divisible by 6. Case VI: When $n=6q+5$ ${n}^{3}=\left(6q+5{\right)}^{3}\phantom{\rule{0ex}{0ex}}{n}^{3}-n=\left(6q+5{\right)}^{3}-\left(6q+5\right)\phantom{\rule{0ex}{0ex}}=\left(6q+5\right)\left[\left(6q+5{\right)}^{2}-1\right]\phantom{\rule{0ex}{0ex}}=\left(6q+5\right)\left[36{q}^{2}+25+60q-1\right]\phantom{\rule{0ex}{0ex}}=\left(6q+5\right)\left[36{q}^{2}+60q+24\right]\phantom{\rule{0ex}{0ex}}=\left[216{q}^{3}+360{q}^{2}+144q+180{q}^{2}+300q+120\right]\phantom{\rule{0ex}{0ex}}=\left[216{q}^{3}+540{q}^{2}+444q+120\right]\phantom{\rule{0ex}{0ex}}=6\left[36{q}^{3}+90{q}^{2}+74q+120\right]\phantom{\rule{0ex}{0ex}}=6m\left(\mathrm{where}m=36{q}^{3}+90{q}^{2}+74q+120\right)$ Hence, $\forall n=6q+5,{n}^{3}-n$ is divisible by 6.

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