Show that the square of any integer is of the form 3m or 3m1

Show that the square of any integer is of the form 3m or 3m+1 but not of the form 3m+2.

Solution

Using Euclid\'s Division lemma,

Let a be any positive integer of the form 3q, 3q+1 or 3q+2

Case 1:

a = 3q

According to question,

a^2 = (3q)^2 = 9q^2

= 3m

Taking a = 3q+1

Acc. to question,

a^2 = (3q+1)^2

a^2 = 3(3q^2+2q)+1

a^2 = 3m+1

where m = 3q^2 + 2q

Now considering case 3,

Let a = 3q+2

a^2 = (3q+2)^2

a^2 = 3(3q^2+4q +1) +1

a^2 = 3m+1

where m = 3q^2 + 4q+1

So, it is possible for 3m and 3m+1 and not for 3m+2

Hence Proved