Question

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

Answer 1

Using Euclid division algorithm, we know that $a=bq+r,$ $0\le r\le b$ $----\left(1\right)$

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

Substitute $b=3$ in equation $\left(1\right)$

$a=3q+r$ where $0\le r\le 3,$ $r=0,1,2$

If $r=0,a=3q$

Cube the value, we get

$a^3=27q^3$

$a^3=9\left(3q^3\right)$, where m = $3q^3$ $---\left(2\right)$

If $r=1,a=3q+1$

Cube the value, we get

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

$a^3=(27q^3+27q^2+9q+1)$

$a^3=9(3q^3+3q^2+1)+1$, where m = $3q^3+3q^2+q$ $---\left(3\right)$

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

Cube the value, we get

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

$a^3=(27q^3+53q^2+36q+8)$

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

From equation $2,3$ and $4,$

The cube of any positive integer is of the form $9m,9m+1$ or $9m+8.$