Question

Question 4
Show that the square of any positive integer cannot be of the form 6m + 2 or 6m + 5 for any integer m.

Answer 1

Let a be an arbitrary positive integer. Then by Euclid's division algorithm, corresponding to the positive integers a and 6, there exist non - negative integers q and r such that:
$a=6q+r,where0\le r<6$
$\Rightarrow a^2=(6q+r{)}^2=36q^2+r^2+12qr$ $[\because (a+b{)}^2=a^2+2ab+b^2]$
$\Rightarrow a^2=6(6q^2+2qr)+r^2$ ...(i)

Case I
When r = 0, putting r = 0 in eq. (i), we get,
$a^2=6\left(6q^2\right)=6m$
Where, $m=6q^2$ is an integer.

Case II
When r = 1, putting r = 1 in eq. (i) we get,
$a^2=6(6q^2+2q)+1=6m+1$
Where, $m=(6q^2+2q)$ is an integer.

Case III
When r = 2, putting r = 2 in eq (i), we get,
$a^2=6(6q^2+4q)+4=6m+4$
Where, m = $6q^2+4q$ is an integer.

case IV
When r = 3, putting r = 3 in eq. (i) we get,
$a^2=6(6q^2+6q)+9$
$=6(6q^2+6q)+6+3$
$a^2=6(6q^2+6q+1)+3=6m+3$
Where, $m=(6q^2+6q+1)$ is an integer.

Case V
When r = 4, putting r = 4 in eq. (i), we get,
$a^2=6(6q^2+8q)+16$
$=6(6q^2+8q)+12+4=6m+4$
Where, $m=(6q^2+8q+2)$ is an integer.

Case VI
When r = 5, putting r = 5 in eq. (i), we get,
$a^2=6(6q^2+10q)+25$
$=6(6q^2+10q)+24+1$
$\Rightarrow a^2=6(6q^2+10q+4)+1=6m+1$
Where, $m=(6q^2+10q+4)$ is an integer.
Hence, the square of any positive integer cannot be of the form 6m + 2 or 6m + 5 for any integer m.