Question

Use Euclid’s Division Lemma, prove that for any positive integer n, ${{n}^{3}}-n$ is divisible by 6.

Answer 1

Hint:According to Euclid’s Division Lemma, if two positive integers “a” and “b” are there then there exists two unique whole numbers “q” and “r” such that $a=bq+r$ where $0\le r\le b$ so here in this question “a” is ${{n}^{3}}-n$ and “b” is equal to 6 and let us assume that any positive integer n can be in the form of 6m, 6m + 1, 6m + 2, 6m + 3, 6m + 4, 6m + 5 then substituting these integers one by one in n3 – n and apply the relation of Euclid’s Division Lemma i.e. $a=bq+r$ where $0\le r\le b$

Complete step-by-step answer:
Euclid’s Division Lemma states that if two positive integers “a” and “b” are there then there exists two whole numbers “q” and “r” such that these integers are following this relation $a=bq+r$ where $0\le r\le b$ .
As we have to show that for any positive integer n, ${{n}^{3}}-n$ is divisible by 6 so using Euclid’s Division Lemma that we have discussed above in this problem ${{n}^{3}}-n$ is “a” and 6 is “b” so substituting these values of “a” and “b” in the Euclid’s Division Lemma relation we get,
$\begin{aligned} & a=bq+r \\\ & \Rightarrow {{n}^{3}}-n=6q+r \\\ \end{aligned}$
Let us assume that any positive integer n can be in the form of 6m, 6m + 1, 6m + 2, 6m + 3, 6m + 4, 6m + 5 where m is any positive integer.
Now, substituting these positive integers one by one in ${{n}^{3}}-n$ and then see whether the resulting expression is in the form of 6q + r or not.
Putting $n=6m$ in ${{n}^{3}}-n$ we get,
$\begin{aligned} & {{\left( 6m \right)}^{3}}-6m \\\ & =6m\left( 36{{m}^{2}}-1 \right) \\\ \end{aligned}$
As you can see that the resulting expression is in the form of 6q + r where q is equal to $m\left( 36{{m}^{2}}-1 \right)$ and r = 0 so according to Euclid’s Division Lemma we can say that ${{n}^{3}}-n$ is divisible by 6.
Putting $n=6m+1$ in ${{n}^{3}}-n$ we get,
$\begin{aligned} & {{\left( 6m+1 \right)}^{3}}-\left( 6m+1 \right) \\\ & =\left( 6m+1 \right)\left( {{\left( 6m+1 \right)}^{2}}-1 \right) \\\ & =\left( 6m+1 \right)\left( 6m\left( 6m+1 \right) \right) \\\ & =6m{{\left( 6m+1 \right)}^{2}} \\\ \end{aligned}$
As you can see that the resulting expression is in the form of 6q + r where q is equal to $m{{\left( 6m+1 \right)}^{2}}$ and r = 0 so according to Euclid’s Division Lemma we can say that ${{n}^{3}}-n$ is divisible by 6.
Putting $n=6m+2$ in ${{n}^{3}}-n$ we get,
$\begin{aligned} & {{\left( 6m+2 \right)}^{3}}-\left( 6m+2 \right) \\\ & =\left( 6m+2 \right)\left( {{\left( 6m+2 \right)}^{2}}-1 \right) \\\ & =\left( 6m+2 \right)\left( \left( 6m+3 \right)\left( 6m+1 \right) \right) \\\ & =2\left( 3m+1 \right)\left( 3\left( 2m+1 \right)\left( 6m+1 \right) \right) \\\ & =6\left( 3m+1 \right)\left( \left( 2m+1 \right)\left( 6m+1 \right) \right) \\\ \end{aligned}$
As you can see that the resulting expression is in the form of 6q + r where q is equal to$\left( 3m+1 \right)\left( 2m+1 \right)\left( 6m+1 \right)$ and r = 0 so according to Euclid’s Division Lemma we can say that ${{n}^{3}}-n$ is divisible by 6.
Putting $n=6m+3$ in ${{n}^{3}}-n$ we get,
$\begin{aligned} & {{\left( 6m+3 \right)}^{3}}-\left( 6m+3 \right) \\\ & =\left( 6m+3 \right)\left( {{\left( 6m+3 \right)}^{2}}-1 \right) \\\ & =\left( 6m+3 \right)\left( \left( 6m+4 \right)\left( 6m+2 \right) \right) \\\ & =6\left( 2m+1 \right)\left( 3m+2 \right)\left( 6m+2 \right) \\\ \end{aligned}$
As you can see that the resulting expression is in the form of 6q + r where q is equal to $\left( 2m+1 \right)\left( 3m+2 \right)\left( 6m+2 \right)$ and r = 0 so according to Euclid’s Division Lemma we can say that ${{n}^{3}}-n$ is divisible by 6.
Putting $n=6m+4$ in ${{n}^{3}}-n$ we get,
$\begin{aligned} & {{\left( 6m+4 \right)}^{3}}-\left( 6m+4 \right) \\\ & =\left( 6m+4 \right)\left( {{\left( 6m+4 \right)}^{2}}-1 \right) \\\ & =\left( 6m+4 \right)\left( \left( 6m+5 \right)\left( 6m+3 \right) \right) \\\ & =6\left( 3m+2 \right)\left( 6m+5 \right)\left( 2m+1 \right) \\\ \end{aligned}$
As you can see that the resulting expression is in the form of 6q + r where q is equal to $\left( 3m+2 \right)\left( 6m+5 \right)\left( 2m+1 \right)$ and r = 0 so according to Euclid’s Division Lemma we can say that ${{n}^{3}}-n$ is divisible by 6.
Putting $n=6m+5$ in ${{n}^{3}}-n$ we get,
$\begin{aligned} & {{\left( 6m+5 \right)}^{3}}-\left( 6m+5 \right) \\\ & =\left( 6m+5 \right)\left( {{\left( 6m+5 \right)}^{2}}-1 \right) \\\ & =\left( 6m+4 \right)\left( \left( 6m+6 \right)\left( 6m+4 \right) \right) \\\ & =6\left( 6m+4 \right)\left( m+1 \right)\left( 6m+4 \right) \\\ \end{aligned}$
As you can see that the resulting expression is in the form of 6q + r where q is equal to $\left( 6m+4 \right)\left( m+1 \right)\left( 6m+4 \right)$ and r = 0 so according to Euclid’s Division Lemma we can say that ${{n}^{3}}-n$ is divisible by 6.
We have shown above that all the form of positive integers i.e. 6m, 6m + 1, 6m + 2, 6m + 3, 6m + 4, 6m + 5 when plugged into ${{n}^{3}}-n$ gives the value of r is equal to 0 which according to Euclid’s Division Lemma have shown that ${{n}^{3}}-n$ is divisible by 6.
Hence, we have proved that any positive integer n, ${{n}^{3}}-n$ is divisible by 6.

Note: We have taken the positive integers in the form of 6m, 6m + 1, 6m + 2, 6m + 3, 6m + 4, and 6m + 5 so that after substituting these integers in ${{n}^{3}}-n$ the answer is in multiples of 6 and we can easily show that the resulting answer is divisible by 6.