Question

Use Euclid's division lemma to show that the cube of any positive integer is of the form $9m,9m+1$ or $9m+8$.

Answer 1

Let $a$ be a positive integer and $b=3$.

By Euclid's division lemma, we get $a=3q+r,0\le r<3$ , where $q$ is a positive integer.

If $r=0$ we get $a=3q$

Cubing $a=3q$ on both sides

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

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

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

If $r=1$ we get $a=3q+1$

Cubing $a=3q+1$ on both sides

$a^3=(3q+1{)}^3$

$=(3q{)}^3+3(3q^2\left)\right(1)+3(3q\left)\right(1{)}^2+(1{)}^3$

$=27q^3+27q^2+9q+1$

$=9(3q^3+3q^2+q)+1$

$a^3=9m+1$ where $m=3q^3+3q^2+q$

If, $r=2$ we get $a=3q+2$

Cubing $a=3q+2$ on both sides

$a^3=(3q+2{)}^3$

$=(3q{)}^3+3(3q{)}^2\left(2\right)+3\left(3q\right)(2{)}^3$

$=27q^3+54q^2+36q+8$

$=9(3q^3+6q^2+4q)+8$

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

Hence cube of any positive integer is of the form $9mor9m+1or9m+8$