Question

Show that the square on any odd integer is of the form $(4q+1)$ 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 $r=0,a=4q$ which is even (as it is divisible by $2$)

When $r=1,a=4q+1$ which is odd (as it is not divisible by $2$)

Squaring the odd number $(4q+1)$ , 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 $r=2,a=4q+2$ [From(i)]

$\Rightarrow a=2(2q+1)$ which is even (as it is divisible by $2$)

When $r=3,a=4q+3=4q+2+1$

$=2[2q+1]+1$ which is odd (as it is not divisible by $2$)

Squaring the odd number $(4q+3)$ , we get

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

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

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

$=4[4q^{2}6q+2]+1$

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

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