Question

Show that the square of any positive integer is of the form 4m or 4m + 1 for any integer m.

Answer 1

Applying Euclids division algorithm with $a,b,q$ and $r$ where $b=4$ (Theorem $1.1$)

$\Rightarrow a=bq+r,0\le r<b$

$\Rightarrow a=4q+r,0\le r<4$

i) When $r=0,a=4q$

Thus $a^2=16q^2=4\left(4q^2\right)=4Q$ where $Q=4q^2$

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

$\Rightarrow a^2=(4q+1{)}^2=16q^2+8q+1=4q(4q+2)+1=4Q+1$

where $4q+2=Q$

iii) when $r=2,a=4q+2$

$\Rightarrow a^2=16q^2+16q+4=4(4q^2+4q+1)$

Where $a=4q^2+4q+1$

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

$\Rightarrow a^2=(4q+3{)}^2=16q^2+24q+9=4(4q^2+6q+2)+1$

$=4Q+1$ when $Q=4q^2+q+1$

Thus we can see that the square of any $+ve$ integer is of the form $4Q$ or $4Q+1$ for some integer $Q$