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

Solution:

Suppose a be any arbitrary positive integer, then by Euclid's division algorithm, corresponding to the positive integers a and 6, there exists non-negative integers a and r such that

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

$\Rightarrow a^{2}=(6 q+r)^{2}=36 q^{2}+r^{2}+12 q r$

$\Rightarrow a^{2}=6\left(6 q^{2}+2 q r\right)+r^{2} \quad \ldots \ldots(1) \quad$ where, $0 \leq r<6$

Case: 1

When r = 0.

Putting r = 0 in (1), we get

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

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

Case: 2

When r = 1.

Putting r = 1 in (1), we get

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

$\Rightarrow a^{2}=6 m+1$

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

Case: 3

When r = 2.

Putting r = 2 in (1), we get

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

$\Rightarrow a^{2}=6 m+4$

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

Case: 4

When r = 3

Putting r = 3 in (1), we get

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

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

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

 

$\Rightarrow a^{2}=6 m+3$

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

Case: 5

When r = 4

Putting r = 4 in (1), we get

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

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

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

 

$\Rightarrow a^{2}=6 m+4$

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

Case: 6

When r = 5.

Putting r = 5 in (1), we get

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

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

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

 

$\Rightarrow a^{2}=6 m+1$

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.