Question

Show that the square of any integer is either of the form $4q$ or $4q+1$ for some integer $q.$

Answer 1

By Euclid's division algorithm , $a=bq+r$ where $a,b,q,r$ are non-negative integers and $0\le r<b$.

On putting $b=4$ we get

$a=4q+r$ ....(i)

When $r=0,a=4q$

Squaring both sides, we get

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

$=4\left(4q^2\right)$

$=4m$ is perfect square for $m=4q^2$

When $r=1,a=4q+1$

Squaring both sides, we get

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

$=(4q{)}^2+(1{)}^2+2\left(4q\right)$

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

$=4m+1$ is perfect square for $m=4q^2+2q$

When $r=2,a=4q+2$

Squaring both sides, we get

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

$=(4q{)}^2+(2{)}^2+2\left(4q\right)\left(2\right)$

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

$=4m$ is perfect square for $m=4q^2+4q+1$

When $r=3,a=4q+3$

Squaring both sides, we get

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

$=(4q{)}^2+(3{)}^2+2\left(4q\right)\left(3\right)$

$=16q^2+9+24q$

$=16q^2+24q+8+1$

$=4(4q^{2}6q+2)+1$

$=4m+1$ is perfect square for some value of $m$.

Hence, the square on any integer is either of the form $4q$ or $(4q+1)$ for some integer $q$.