Question

Show that cube of any position integer will be in the form of $8m$ or $8m+1$ or $8m+3$ or $8m+5$ or $8m+7$,where $m$ is a whole number.

Answer 1

As per Euclid's Division Lemma

If $a$ and $b$ are $2$ positive integers, then

$a=bq+r$

Where $0\le r\le b$

Let $b=8$,

Therefore,

$a=8q+r$

$r=0,1,3,5$

Case I:- $r=0$

Therefore,

$a=8q$

Cubing both sides, we get

${\left(a\right)}^3={\left(8q\right)}^3$

$a^3=512q^3$

$\Rightarrow a^3=8\times \left(64q^3\right)$

Here $m=64q^3$

Case II:- $r=1$

Therefore,

$a=8q+1$

Cubing both sides, we get

${\left(a\right)}^3={(8q+1)}^3$

$a^3={\left(8q\right)}^3+{\left(1\right)}^3+3{\left(8q\right)}^2\left(1\right)+3\left(8q\right){\left(1\right)}^2$

$a^3=512q^3+1+192q^2+24q$

$\Rightarrow a^3=8(64q^3+24q^2+3q)+1$

Here $m=(64q^3+24q^2+3q)$

Case III:- $r=3$

Therefore,

$a=8q+3$

Cubing both sides, we get

${\left(a\right)}^3={(8q+3)}^3$

$a^3={\left(8q\right)}^3+{\left(3\right)}^3+3{\left(8q\right)}^2\left(3\right)+3\left(8q\right){\left(3\right)}^2$

$a^3=512q^3+27+576q^2+216q$

$\Rightarrow a^3=8(64q^3+72q^2+27q+3)+3$

Here $m=(64q^3+72q^2+27q+3)$

Case IV:- $r=5$

Therefore,

$a=8q+5$

Cubing both sides, we get

${\left(a\right)}^3={(8q+5)}^3$

$a^3={\left(8q\right)}^3+{\left(5\right)}^3+3{\left(8q\right)}^2\left(5\right)+3\left(8q\right){\left(5\right)}^2$

$a^3=512q^3+125+960q^2+600q$

$\Rightarrow a^3=8(64q^3+120q^2+75q+15)+5$

Here $m=(64q^3+120q^2+75q+15)$

Case V:- $r=7$

Therefore,

$a=8q+7$

Cubing both sides, we get

${\left(a\right)}^3={(8q+7)}^3$

$a^3={\left(8q\right)}^3+{\left(7\right)}^3+3{\left(8q\right)}^2\left(7\right)+3\left(8q\right){\left(7\right)}^2$

$a^3=512q^3+343+1344q^2+1176q$

$\Rightarrow a^3=8(64q^3+168q^2+147q+42)+7$

Here $m=(64q^3+168q^2+147q+42)$

Thus cube of any positive number can be expressed as $8m$ or $8m+1$ or $8m+3$ or $8m+5$ or $8m+7$.

Hence proved.