Question

Use Euclid division Lemma to show that the cube of any positive integer is either of the form $9m,9m+1$ or, $9m+8$ for some integer $m$

Answer 1

According to Euclid’s Division Lemma if we have two positive integers $a$ and $b$, then there exist unique integers $q$ and $r$ which satisfies the condition $a=bq+r$ where $0\le r<b.$

In the given question, Let ${}^{\prime }{a}^{\prime }$ be any positive integer and $b=3.$

By Euclid'd Division lemma, $a=3q+r,$ $0\le r<3,$ [$\because b=3$ ], where $q\ge 0$

and $r=0,1,2$

so, $a=3q$ or $3q+1$ or $3q+2$

Then, ${}^{\prime }{a}^{\prime }$ is of the form $3q$ or, $3q+1$ or, $3q+2.$ So, we have the following cases:

Case $I$ When $a=3q$
In this case, we know

$a^3=(3q{)}^3$

$=27q^3$

$=9\left(3q^3\right)$

$=9m$, where $m=3q^3$

Case $II$, When $a=3q+1$
In this case, we have
$a^3=(3q+1{)}^3$
$\Rightarrow a^3=27q^3+27q^2+9q+1$
$\Rightarrow a^3=9q(3q^2+3q+1)+1$
$\Rightarrow a^3=9m+1$, where $m=q(3q^2+3q+1)$

Case $III$, When $a=3q+2$
In this case, we have
$a^3=(3q+2{)}^3$
$\Rightarrow a^3=27q^3+54q^2+36q+8$
$\Rightarrow a^3=9q(3q^2+6q+4)+8$

$a^3=9m+8$, where $m=q(3q^2+6q+4)$

Hence $a^3$ is either of the form $9m$ or, $9m+1$ or, $9m+8$.