Question

Use division algorithm to show that the square any positive integer is of the form or $3p$ or $3p+1$.

Answer 1

Let ${}^{\prime }{a}^{\prime }$ be any positive integer and $b=3$,
Then, by division algorithm $a=3q+r$,
for some integer $q⩾0$ and $0⩽r⩽3$.
So, $a=3q,3q+1,3q+2$.
Then, the square of positive integer,

$a=3q$
$a^2=(3q{)}^2$
$=9q^2$
$=3\left(3q^2\right)$
$=3p$
(Where $p=3q^2$)

$a=3q+1$

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

$=9q^2+6q+1$

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

$=3p+1$

(Where $p=3q^2+2q$)

$a=3q+2$

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

$=9q^2+12q+4$

$=3(3q^2+4q+1)+1$

$=3p+1$

(Where $p=3q^2+4q+1$)

Since $p$ is some positive integer.
$\therefore$ The square of any positive integer is of the form $3p$ or $3p+1$.