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

Solution:

Let the positive integer = a

According to Euclid’s division algorithm,

a = 6q + r, where 0 ≤ r < 6 

   

,

   

   

…(i), where,0 ≤ r < 6 

When r = 0, substituting r = 0 in Eq.(i), we get 

   

, where, m = 6q² is an integer.

When r = 1, substituting r = 1 in Eq.(i), we get

   

, where, m = (6q² + 2q) is an integer.

When r = 2, substituting r = 2 in Eq(i), we get

   

, where, m =

   

is an integer.
When r = 3, substituting r = 3 in Eq.(i), we get

   

   

, where

   

is integer.

When r = 4, substituting r = 4 in Eq.(i) we get 

   

   

, where, m

   

is an integer.

When r = 5, substituting r = 5 in Eq.(i), we get

   

   

, where, m

   

is an integer.

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

Hence Proved.