Show that the square of any positive integer

Solution:

Let a be an arbitrary positive integer. Then, by, Euclid's division algorithm, corresponding to the positive integers a and 4 , there exist non-negative integers $m$ and $r$, such that

$a=4 m+r$, where $0 \leq r<4$

$\Rightarrow \quad a^{2}=16 m^{2}+r^{2}+8 m r$ $\ldots$ (i)

where, $0 \leq r<4$ $\left[\because(a+b)^{2}=a^{2}+2 a b+b^{2}\right]$

Case I When $r=0$, then putting $r=0$ in Eq. (i), we get

$a^{2}=16 m^{2}=4\left(4 m^{2}\right)=4 q$

where, $q=4 m^{2}$ is an integer.

Case II When $r=1$, then putting $r=1$ in Eq. (i), we get

$a^{2}=16 m^{2}+1+8 m$ 

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

where, $q=\left(4 m^{2}+2 m\right)$ is an integer.

Case III When $r=2$, then putting $r=2$ in Eq. (i), we get

$a^{2}=16 m^{2}+4+16 m$

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

where, $q=\left(4 m^{2}+4 m+1\right)$ is an integer.

Case IV When $r=3$, then putting $r=3$ in Eq. (i), we get

$a^{2}=16 m^{2}+9+24 m=16 m^{2}+24 m+8+1$

where, $q=\left(4 m^{2}+6 m+2\right)$ is an integer.

Hence, the square of any positive integer is either of the form $4 \mathrm{q}$ or $4 \mathrm{q}+1$ for some integer $\mathrm{q}$.