Use Euclid’s division lemma to show that the square of any positive integer is either of the form 3m or 3m+1 for some integer m

Answer – 

Let’s think a positive integer a, 

Divide the positive integer a by 3, and let r be the reminder and b be the quotient such that

A= 3b + r ——(1)

Where r = 0,1,2,3 

Case 1: Consider r = 0

Equation (1) becomes

a=3b

Squaring both the side

a2 = (3b)2

a2 = 9b2

a2 = 3 × 3b2

a2 = 3m

where m = 3b2

Case 2: Let r = 1

Equation (1) 

a = 3b + 1

Squaring on both the side 

a2 = (3b + 1)2

a2 = (3b)2 + 1 + 2 × (3b) × 1

a2 = 9b2 + 6b + 1

a2 = 3(3b2 + 2b) + 1

a2 = 3m + 1

Where m = 3b2 + 2b

Case 3: Let r = 2

Equation (1) becomes

a = 3b + 2

Squaring on both the sides 

a2 = (3b + 2)2

a2 = 9b2 + 4 + (2 × 3b × 2)

a2 = 9b2 + 12b + 3 + 1

a2 = 3(3b2 + 4b + 1) + 1

a2 = 3m + 1

 m = 3b2 + 4b + 1

∴ square of any positive integer is of form 3m or 3m+1.

So it is proved.