# Show that the square of any positive

Question:

Show that the square of any positive integer cannot be of the form 6m+ 2 or 6m + 5 for any integer m.

Solution:

Let a be an arbitrary positive integer, then by Euclidâ€™s division algorithm, corresponding to the positive integers a and 6, there exist non-negative

integers q and r such that a = 6q + r, where 0< r< 6

$a=6 q+r$, where $0 \leq r<6$

$a^{2}=(6 a+r)^{2}=36 q^{2}+r^{2}+12 q r \quad\left[\because(a+b)^{2}=a^{2}+2 a b+b^{2}\right]$

$\Rightarrow \quad a^{2}=6\left(6 q^{2}+2 q r\right)+r^{2} \quad \ldots$ (i)

where, $0 \leq r<6$

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

$a^{2}=6\left(6 q^{2}\right)=6 m$

where, $m=6 q^{2}$ is an integer.

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

$a^{2}=6\left(6 q^{2}+2 q\right)+1=6 m+1$

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

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

$a^{2}=6\left(6 q^{2}+4 q\right)+4=6 m+4$

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

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

$a^{2}=6\left(6 q^{2}+6 q\right)+9$

$=6\left(6 q^{2}+6 a\right)+6+3$

$\Rightarrow \quad a^{2}=6\left(6 q^{2}+6 q+1\right)+3=6 m+3$

where, $m=(6 q+6 q+1)$ is an integer.

Case $V$ When $r=4$, then putting $r=4$ in Eq. (i), we get

$a^{2}=6\left(6 q^{2}+8 q\right)+16$

$=6\left(6 q^{2}+8 q\right)+12+4$

$\Rightarrow a^{2}=6\left(6 q^{2}+8 q+2\right)+4=6 m+4$

where, $m=\left(6 q^{2}+8 q+2\right)$ is an integer.

Case VI When $r=5$, then putting $r=5$ in Eq. (i), we get

$a^{2}=6\left(6 q^{2}+10 q\right)+25$

$=6\left(6 q^{2}+10 q\right)+24+1$

$\Rightarrow a^{2}=6\left(6 q^{2}+10 q+4\right)+1=6 m+1$

where, $m=\left(6 q^{2}+10 q+1\right)$ is an integer.

Hence, the square of any positive integer cannot be of the form 6m + 2 or 6m + 5 for any integer m.