Question

Prove that square of any positive integer is of the form 4m, 4m+1 for some integer m.

Answer 1

Let positive integer is of form

$\alpha =2p+q,(0\le q\le 1)$

$a=2p$ or $a=2p+1$

So, $a^2=4p^2=4m$ (let $p^2=m$)

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

$\Rightarrow a^2=4p(p+1)+1=4m+1$ where, [$m=p(p+1)$]

Hence, $a^2$ eill be of form $4m$ or $4m+1$