Question

Show that cube of any positive integer is of the form $4m,4m+1$ or $4m+3$ , for some integer $m$ .

Answer 1

By euclids division Lemma

$a=bq+r$

Here $a$ is any positive integer & $b=4$

& $0\le r<4r=0,1,2,3$

$a=4q+r.....\left(1\right)$

Put $r=0$ in equation $\left(1\right)$

$a=4q$

Cubing on both sides

$(a{)}^3=(4q{)}^3$

$(a{)}^3=64q^3=4(16q^3)$

$a^3=4m$ where $m=16q^3$

Put $=1$ in equation $\left(1\right)$

$a=4q+1$ cubing both side

$a^3=(4q+1{)}^3$

we know

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

so apply

$a^3=(4q{)}^3+(1{)}^3+3(4q{)}^2\times 1+3(4q)\times (1{)}^2$

$a^3=64q^3+1+3\times 16q^2+12q$

$a^3=64q^3+1+48q^2+12q$

$a^3=4(16q^3+12q^3+3q)+1$

$a^3=4m+1$

where $m=16q^3+12q^2+3q$

again put $r=2$ in equation $\left(1\right)$

$a=4q+2$

cube both sides

$a^3=(4q+2{)}^3$

$a^3=(4q{)}^3+(2{)}^3+3\times (4q{)}^2\times 2+3\times 4q(2)$

$a^3=64q^3+8+3\times 16q^2\times 2+3\times 4q\times 4$

$a^3=64q^3+8+96q^2+48q$

$a^3=4(16q^3+2+24q^3+12q)$

$a^3=4m$

$m=16q^3+2+24q^3+12q$

Put $r=3$ in equation $\left(1\right)$

$a=4q+3$

again cube

$a^3=(4q+3{)}^3$

$a^3=(64q^3+27+144q^2+108q)+3$

$a^3=(64q^3+24+144q^2+108q)+3$

$a^3=4m+3$

when $m=16q^3+6+36q^2+27$

hence $a^3$ is of the for

$4m,4m+1,4m+3$